Introduction To Electrodynamics Solution Manual David Griffiths
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INSTRUCTOR'S
SOLUTIONS
MANUAL
INTRODUCTION to
ELECTRODYNAMICS
Third Edition
David J. Griffiths
Errata
Instructor’s Solutions Manual
Introduction to Electrodynamics, 3rd ed
Author: David Griffiths
Date: September 1, 2004
• Page 4, Prob. 1.15 (b): last expression should read y + 2z + 3x.
• Page 4, Prob.1.16: at the beginning, insert the following figure
• Page 8, Prob. 1.26: last line should read
From Prob. 1.18: ∇ × va = −6xz x̂ + 2z ŷ + 3z 2 ẑ ⇒
∇ · (∇ × va ) =
∂
∂x (−6xz)
+
∂
∂y (2z)
+
∂
2
∂z (3z )
= −6z + 6z = 0.
• Page 8, Prob. 1.27, in the determinant for ∇×(∇f ), 3rd row, 2nd column:
change y 3 to y 2 .
• Page 8, Prob. 1.29, line 2: the number in the box should be -12 (insert
minus sign).
• Page 9, Prob. 1.31, line 2: change 2x3 to 2z 3 ; first line of part (c): insert
comma between dx and dz.
• Page 12, Probl 1.39, line 5: remove comma after cos θ.
• Page 13, Prob. 1.42(c), last line: insert ẑ after ).
• Page 14, Prob. 1.46(b): change r to a.
• Page 14, Prob. 1.48, second line of J: change the upper limit on the r
integral from ∞ to R. Fix the last line to read:
R
= 4π −e−r 0 + 4πe−R = 4π −e−R + e−0 + 4πe−R = 4π.
• Page 15, Prob. 1.49(a), line 3: in the box, change x2 to x3 .
1
• Page 15, Prob. 1.49(b), last integration “constant” should be l(x, z), not
l(x, y).
• Page 17, Prob. 1.53, first expression in (4): insert θ, so da = r sin θ dr dφ θ̂θ̂.
• Page 17, Prob. 1.55: Solution should read as follows:
Problem 1.55
(1) x = z = 0; dx = dz = 0; y : 0 → 1.
v · dl = (yz 2 ) dy = 0; v · dl = 0.
(2) x = 0; z = 2 − 2y; dz = −2 dy; y : 1 → 0.
v · dl = (yz 2 ) dy + (3y + z) dz = y(2 − 2y)2 dy − (3y + 2 − 2y)2 dy;
0
v · dl = 2
0
y4
4y 3
y2
14
−
+
− 2y =
.
(2y − 4y + y − 2) dy = 2
2
3
2
3
1
3
1
2
(3) x = y = 0; dx = dy = 0; z : 2 → 0.
0
v · dl =
2
Total:
v · dl = 0 +
14
3
−2=
v · dl = (3y + z) dz = z dz.
0
z 2
z dz =
= −2.
2 2
8
3.
Meanwhile, Stokes’ thereom says v · dl = (∇×v) · da. Here da =
dy dz x̂, so all we need is
∂
∂
(∇×v)x = ∂y
(3y + z) − ∂z
(yz 2 ) = 3 − 2yz. Therefore
1 2−2y
(3 − 2yz) dz dy
(∇×v) · da =
(3 − 2yz) dy dz = 0
0
1
1
1
= 0 3(2 − 2y) − 2y 2 (2 − 2y)2 dy = 0 (−4y 3 + 8y 2 − 10y + 6) dy
1
=
−y 4 + 83 y 3 − 5y 2 + 6y 0 = −1 + 83 − 5 + 6 = 83 .
• Page 18, Prob. 1.56: change (3) and (4) to read as follows:
(3) φ = π2 ; r sin θ = y = 1, so r =
tan−1 ( 12 ).
v · dl
=
=
1
sin θ ,
dr =
−1
sin2 θ
cos θ dθ, θ :
π
2
→ θ0 ≡
cos θ sin θ
cos2 θ
cos θ
dθ −
dθ
− 2
r cos2 θ (dr) − (r cos θ sin θ)(r dθ) =
sin θ
sin θ
sin2 θ
3
cos θ cos θ
cos θ
cos θ cos2 θ + sin2 θ
−
+
dθ = − 3 dθ.
dθ = −
3
2
sin
θ
sin
θ
sin θ
sin θ
sin θ
Therefore
θ0
v · dl = −
π/2
θ
cos θ
1 0
1
5 1
1
dθ
=
−
= − = 2.
=
2 · (1/5) 2 · (1)
2 2
sin3 θ
2 sin2 θ π/2
2
(4) θ = θ0 , φ =
π
2;
r:
√
5 → 0.
v · dl = r cos2 θ (dr) = 45 r dr.
0
0
4 r2
4 5
4
r dr =
= − · = −2.
v · dl =
√
5√
5 2 5
5 2
5
Total:
v · dl = 0 +
3π
+2−2=
2
3π
2
.
• Page 21, Probl 1.61(e), line 2: change = z ẑ to +z ẑ.
• Page 25, Prob. 2.12: last line should read
Since Qtot = 43 πR3 ρ, E =
1
4π
0
Q
R3 r
(as in Prob. 2.8).
• Page 26, Prob. 2.15: last expression in first line of (ii) should be dφ, not
d phi.
• Page 28, Prob. 2.21, at the end, insert the following figure
V(r)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0.5
1
1.5
2
2.5
3
r
In the figure, r is in units of R, and V (r) is in units of
q
4π
0R
.
• Page 30, Prob. 2.28: remove right angle sign in the figure.
• Page 42, Prob. 3.5: subscript on V in last integral should be 3, not 2.
• Page 45, Prob. 3.10: after the first box, add:
1
1
1
√
x̂
−
ŷ
+
[cos
θ
x̂
+
sin
θ
ŷ]
,
(2a)2
(2b)2
(2 a2 + b2 )2
where cos θ = a/ a2 + b2 , sin θ = b/ a2 + b2 .
F=
q2
4π0
−
q2
F=
16π0
a
1
− 2
a
(a2 + b2 )3/2
3
b
1
x̂ +
− 2
b
(a2 + b2 )3/2
ŷ .
W =
2
−q
1
q2
−q 2
q2
1 1
1 1
√
+
+ √
.
− −
=
4 4π0 (2a) (2b) (2 a2 + b2 )
32π0
a b
a2 + b2
• Page 45, Prob. 3.10: in the second box, change “and” to “an”.
• Page 46, Probl 3.13, at the end, insert the following: “[Comment: Technically, the series solution for σ is defective, since term-by-term differentiation has produced a (naively) non-convergent sum. More sophisticated
definitions of convergence permit one to work with series of this form,
but it is better to sum the series first and then differentiate (the second
method).]”
• Page 51, Prob. 3.18, midpage: the reference to Eq. 3.71 should be 3.72.
• Page 53, Prob. 3.21(b), line 5: A2 should be
1
2R .
σ
4
0R
; next line, insert r2 after
• Page 55, Prob. 3.23, third displayed equation: remove the first Φ.
• Page 58, Prob. 3.28(a), second line, first integral: R3 should read R2 .
• Page 59, Prob. 3.31(c): change first V to W .
• Page 64, Prob. 3.41(a), lines 2 and 3: remove 0 in the first factor in the
expressions for Eave ; in the second expression change “ρ” to “q”.
• Page 69, Prob. 3.47, at the end add the following:
Alternatively, start with the separable solution
V (x, y) = (C sin kx + D cos kx) Aeky + Be−ky .
Note that the configuration is symmetric in x, so C = 0, and V (x, 0) =
0 ⇒ B = −A, so (combining the constants)
V (x, y) = A cos kx sinh ky.
But V (b, y) = 0, so cos kb = 0, which means that kb = ±π/2, ±3π/2, · · · ,
or k = (2n − 1)π/2b ≡ αn , with n = 1, 2, 3, . . . (negative k does not yield
a different solution—the sign can be absorbed into A). The general linear
combination is
∞
V (x, y) =
An cos αn x sinh αn y,
n=1
and it remains to fit the final boundary condition:
V (x, a) = V0 =
∞
n=1
4
An cos αn x sinh αn a.
Use Fourier’s trick, multiplying by cos αn x and integrating:
b
b
∞
V0
cos αn x dx =
An sinh αn a
cos αn x cos αn x dx
−b
−b
n=1
∞
2 sin αn b
=
An sinh αn a (bδn n ) = bAn sinh αn a;
αn
n=1
2n − 1
2V0 sin αn b
So An =
. But sin αn b = sin
π = −(−1)n , so
b αn sinh αn a
2
V0
V (x, y) = −
∞
sinh αn y
2V0
cos αn x.
(−1)n
b n=1
αn sinh αn a
• Page 74, Prob. 4.4: exponent on r in boxed equation should be 5, not 3.
• Page 75, Prob. 4.7: replace the (defective) solution with the following:
If the potential is zero at infinity, the energy of a point charge Q is
(Eq. 2.39) W = QV (r). For a physical dipole, with −q at r and +q
at r+d,
r+d
U = qV (r + d) − qV (r) = q [V (r + d) − V (r)] = q −
E · dl .
r
For an ideal dipole the integral reduces to E · d, and
U = −qE · d = −p · E, since p = qd.
If you do not (or cannot) use infinity as the reference point, the result still
holds, as long as you bring the two charges in from the same point, r0 (or
two points at the same potential). In that case W = Q [V (r) − V (r0 )],
and
U = q [V (r + d) − V (r0 )] − q [V (r) − V (r0 )] = q [V (r + d) − V (r)] ,
as before.
• Page 75, Prob. 4.10(a):
1
r3
should be
1
r2 .
• Page 79, Prob. 4.19: in the upper right box of the Table (σf for air) there
is a missing factor of 0 .
• Page 91, Problem 5.10(b): in the first line µ0 I 2 /2π should read µ0 I 2 a/2πs;
in the final boxed equation the first “1” should be as .
• Page 92, Prob. 5.15: the signs are all wrong. The end of line 1 should
read “right (ẑ),” the middle of the next line should read “left (−ẑ).” In
the first box it should be “(n2 − n1 )”, and in the second box the minus
sign does not belong.
5
x
• Page 114, Prob. 6.4: last term in second expression for F should be +ẑ ∂B
∂z
(plus, not minus).
• Page 119, Prob. 6.21(a): replace with the following:
The magnetic force on the dipole is given by Eq. 6.3; to move the dipole
in from infinity we must exert an opposite force, so the work done is
r
r
F · dl = −
∇(m · B) · dl = −m · B(r) + m · B(∞)
U =−
∞
∞
(I used the gradient theorem, Eq. 1.55). As long as the magnetic field goes
to zero at infinity, then, U = −m · B. If the magnetic field does not go
to zero at infinity, one must stipulate that the dipole starts out oriented
perpendicular to the field.
• Page 125, Prob. 7.2(b): in the box, c should be C.
• Page 129, Prob. 7.18: change first two lines to read:
µ0 Ia s+a ds
s+a
µ0 Ia
µ0 I
φ̂; Φ =
φ̂
ln
;
=
Φ = B · da; B =
2πs
2π s
s
2π
s
E = Iloop R =
dQ = −
dQ
dΦ
µ0 a
dI
R=−
=−
ln(1 + a/s) .
dt
dt
2π
dt
µ0 aI
µ0 a
ln(1 + a/s) dI ⇒ Q =
ln(1 + a/s).
2πR
2πR
• Page 131, Prob. 7.27: in the second integral, r should be s.
• Page 132, Prob. 7.32(c), last line: in the final two equations, insert an I
immediately after µ0 .
• Page 140, Prob. 7.47: in the box, the top equation should have a minus
sign in front, and in the bottom equation the plus sign should be minus.
• Page 141, Prob. 7.50, final answer: R2 should read R2 .
• Page 143, Prob. 7.55, penultimate displayed equation: tp should be ·.
• Page 147, Prob. 8.2, top line, penultimate expression: change a2 to a4 ; in
(c), in the first box, change 16 to 8.
• Page 149, Prob. 8.5(c): there should be a minus sign in front of σ 2 in the
box.
• Page 149, Prob. 8.7: almost all the r’s here should be s’s. In line 1 change
“a < r < R” to “s < R”; in the same line change dr to ds; in the next
line change dr to ds (twice), and change r̂ to ŝ; in the last line change r
to s, dr to ds, and r̂ to ŝ (but leave r as is).
6
• Page 153, Prob. 8.11, last line of equations: in the numerator of the expression for R change 2.01 to 2.10.
• Page 175, Prob. 9.34, penultimate line: α = n3 /n2 (not n3 /n3 ).
• Page 177, Prob. 9.38: half-way down, remove minus sign in kx2 + ky2 + kz2 =
−(ω/c)2 .
• Page 181, Prob. 10.8: first line: remove ¿.
• Page 184, Prob. 10.14: in the first line, change (9.98) to (10.42).
• Page 203, Prob. 11.14: at beginning of second paragraph, remove ¿.
• Page 222, Prob. 12.15, end of first sentence: change comma to period.
• Page 225, Prob. 12.23. The figure contains two errors: the slopes are for
v/c = 1/2 (not 3/2), and the intervals are incorrect. The correct solution
is as follows:
• Page 227, Prob. 12.33: first expression in third line, change c2 to c.
7
TABLE OF CONTENTS
Chapter 1
Vector Analysis
1
Chapter 2
Electrostatics
22
Chapter 3
Special Techniques
42
Chapter 4
Electrostatic Fields in Matter
73
Chapter 5
Magnetostatics
89
Chapter 6
Magnetostatic
Chapter 7
Electrod ynamics
125
Chapter 8
Conservation
146
Chapter 9
Electromagnetic
Chapter 10
Potentials and Fields
179
Chapter 11
Radiation
195
Chapter 12
Electrodynamics
Fields in Matter
Laws
Waves
and Relativity
113
157
219
Chapter
1
Vector
Analysis
Problem
1.1
(a) From the diagram, IB + CI COSO3= IBI COSO1+ ICI COSO2'Multiply by IAI.
IAIIB + CI COSO3= IAIIBI COSO1 + IAIICI COSO2.
So: A.(B + C) = A.B + A.C. (Dot product is distributive.)
ICI sin 82
Similarly: IB + CI sin 03 = IBI sin 01 + ICI sin O2, Mulitply by IAI n.
IAIIB + CI sin 03 n = IAIIBI sin 01 n + IAIICI sin O2n.
If n is the unit vector pointing out of the page, it follows that
Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.)
IBlsin81
A
(b) For the general case, see G. E. Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
Section 8 (cross product).
Problem 1.2
The triple cross-product is not in general associative. For example,
suppose A = ~ and C is perpendicular to A, as in the diagram.
Then (B XC) points out-of-the-page, and A X(B XC) points down,
and has magnitude ABC. But (AxB) = 0, so (Ax B) xC = 0 :f.
Ax(BxC).
= + 1x + 1Y - H; A = /3;
B
= 1x + 1Y+
A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso
10 = COS-1(t) ~ 70.5288°
BxC iAx(Bxe)
z
Problem 1.3
A
k-hB
Hi B = /3.
=>cosO= ~.
y
I
x
Problem
1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
we might pick the base (A) and the left side (B):
A = -1 x + 2 y + 0 z; B = -1 x + 0 Y + 3 z.
1
2
CHAPTER
x
y
1. VECTOR ANALYSIS
Z
= I -1
2 0 1= 6x + 3y + 2z.
-1 0 3
This has the' right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its
length:
IAxBI=v36+9+4=7.
.
AXBBI = 16'7X+ '7y
3 + '7z
2
ft - IAX
AxB
I
A
Problem 1.5
x
=
Ax(BxC)
y
Z
Ax
Ay
Az
(ByCz - BzCy) (BzCx - BxCz) (BxCy - ByCx)
= x[Ay(BxCy - ByCx) - Az(BzCx - BxCz)] + yO + zO
(I'll just check the x-component; the others go the same way.)
= x(AyBxCy - AyByCx - AzBzCx + AzBxCz) + yO + zOo
B(A.C) - C(A.B) = [Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)] x + 0 y + 0 z
= x(AyBxCy + AzBxCz - AyByCx - AzBzCx) + yO + zOo They agree.
Problem
1.6
= B(A.C)-C(A.B)+C(A.B)-A(C.B)+A(B.C)-B(C.A)
Ax(BXC)+Bx(CxA)+Cx(A-xB)
So: Ax(BxC)
= -Bx(CxA)
- (AxB)xC
= A(B.C)
= o.
- C(A.B).
If this is zero, then either A is parallel to C (including the case in which they point in oppositedirections, or
one is zero), or else B.C = B.A = 0, in which case B is perpendicular to A and C (including the case B = 0).
Conclusion:Ax(BxC) = (Ax B) xC <=:=}either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
~= (4x+6y+8z)
~
zl
+ 4 + 1 = @J
= yl4
.= =
~
= !2x-2y+
- (2x+8y+7z)
;Z;
~
3x
12A
:rroblem
-
3Y
+ 3z
2A
lAI
1.8
= (cos cpAy + sin cpAz)(cos cpBy + sin cpBz) +
(a) A.yBy + A.zBz
=
COS2 cpAyBy
(- sin cpAy+ cos cpAz)(- sin cpBy+ cos cpBz)
+ sincpcoscp(AyBz + AzBy) + sin2 cpAzBz + sin2 cpAyBy - sin cpcoscp(AyBz + AzBy) +
COS2cpAzBz
= (COS2cp+ sin2 cp)AyBy + (sin2 cp+ COS2cp)AzBz
-
2
-
2
-
2
-
3
--
(b) (Ax) + (Ay) + (Az) - ~i=lAiAi
.
.d d
Th
I A2 A2 A2
IS equa s
x +
y+
z provz e
I
-
3
- ~i=l
R R
LJi=l ij
~3
= AyBy
3
+ AzBz. ./
3
(~j=lRijAj ) (~k=lRikAk ) = ~j,k
ik
.
(~iRijRik)AjAk.
=k
-
I
if
j
- {
0
if
j =I-k }
Moreover, if R is to preserve lengths for all vectors A., then this condition is not only sufficient but also
necessary. For suppose A = (1,0,0). Then ~j,k (~i RijRik) AjAk = ~i RilRil, and this must equal 1 (since we
-2 -2 -2
..
3
3
.
want Ax+Ay+Az = 1). LIkewIse, ~i=lRi2Ri2 = ~i=lRi3Ri3 = 1. To check the caseJ =I-k, choose A = (1,1,0).
Then we want 2 = ~j,k (~i RijRik) AjAk = ~i RilRil + ~i Ri2Ri2 + ~i RilRi2 + ~i Ri2Ril. But we already
know that the first two sums are both 1; the third and fourth are equal, so ~i RilRi2 = ~i Ri2Ril = 0, and so
on for other unequal combinations of j, k. ./ In matrix notation: RR = 1, where R is the transpose of R.
3
Problem
1.9
y
Looking down the axis:
x
x
z
A 1200 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az,
Ay = Ax, Az = Ay.
001
R=
1 0 0
0 1 0
(
Problem
)
1.10
(a) No change.! (Ax = Ax, Ay = Ay, Az = Az)
I
(b) A -t
-A,
I
(c) (AxB) -t
I
in the sense (Ax
(-A) X(-B)
= -Ax, Ay = -Ay, Az = -Az)
= (AxB).
That is, if C = AxB, Ie -t
C I. No minus sign, in contrast to
behavior of an "ordinary" vector, as given by (b). If A and Bare pseudo vectors, then (AX B) -t (A) X (B) =
(AxB). So the cross~product of two pseudovectors is again a pseudovector. In the cross~product of a vector
and a pseudovector, one changes sign, the other doesn't, and therefore the cross-product is itself a vector.
Angular momentum (L = rxp) and torque (N = rxF) are pseudovectors.
(d) A.(BxC) -t (-A).«-B)x(-C))
= -A.(BxC).
changes sign under inversion of coordinates.
Problem 1.11
(a)Vf
= 2xx
(b)V f
= 2xy3z4
(c)V f
= eXsinylnzx
Problem
So, if a = A.(BxC), then a -t
I
-a; a pseudoscalar
I
+ 3y2y + 4z3z
X + 3x2y2z4
y + 4x2y3z3
Z
+ eXcosylnzy + ex siny(l/z) z
1.12
(a) Vh = 1O[(2y- 6x - 18)x + (2x - 8y + 28)y]. Vh = 0 at summit, so
2y - 6x - 18 = 0
2x - 8y + 28 = 0 ==> 6x 24y + 84 = 0 } 2y - 18 - 24y + 84 = O.
22y = 66 ==> y = 3 ==> 2x - 24 + 28 = 0 ==>x = -2.
-
Top is 13 miles north, 2 miles west, of South Hadley.
(b) Putting in x = -2, y = 3:
h = 10(-12 - 12 - 36 + 36 + 84 + 12) = 1720 ft.
(c) Putting
in x
I
= 1, y = 1: Vh = 10[(2- 6 - 18)x + (2 - 8 + 28) y] = 10(-22 x + 22y) = 220(- x + y).
IVhi = 220v'2 :::J1311 ft/mile I; direction:
I
northwest.!
CHAPTER 1. VECTOR ANALYSIS
4
Problem
1.13
= (x - x') x + (y -
~
y') y + (z - Zl) Zj
~
= Vex - X')2
+ (y - y')2 + (z - ZI)2.
(a) V(~2) = tx[(X_X')2+(y_y')2+(Z_Z')2Jx+tyOy+ tzOz = 2(X_X')X+2(y_y')y+2(z_Z')Z = 2~.
(b) V(k) = tx [(x - X')2 + (y --J!.y')2 + (z - ZI)2]-! x+ tyo-! y + tz O-!
1 -J!.
,
1
,
1 -J!.
,
=-2 () 2 2( x-x ) X-2 () 2 2(y-y ) Y-2 () 2 2( z-Z ) Z
= -O-~[(x - x') x + (y - y') y + (z - Zl)Ii]= _(1/1-3)~ = -(I/1-2)i.
~
~
~
z
(c) /x(~n) = n1-n-lfi = n1-n-l(H21-x)= n1-n-lix, so V(1-n) = n1-n-l i.1
I
Problem
1.14
= +y
y
sin
if>
by sinif>: ysinif> = +y
by cos if>:z cos if>= -y
+ z sinif>; multiply
cos if>;multiply
cosif>
= -y
Z
+Z
sin if>cos if>+ Z sin2
sin cos + Z
if>
if>
if>.
COS2 if>.
= z(sin2if>+cos2if»=z. Likewise,ycosif>-zsinif>=y.
So ~ = cosif>; ~ = - sinif>; ~v = sinif>; ~~ = cosif>. Therefore
(VJ)y = U = ~~ + M~v = +cosif>(VJ)y +sinif>(VJ)z
So V I transforms
,!,
Add: ysinif>+zcosif>
(V I) z
Problem
. ,!,(V/)
=~
=~
~ !li..0:' = - sm'l'
oz
oy oz + oz oz
y + coS'l'(V/) z }
= tx (X2) + ty (3XZ2)+ tz (-2xz) = 2x + 0':'- 2x = O.
(b)V.Vb
= tx (xy) + ty (2yz) + tz (3xz) = y +
(c)V.vc
= tx (y2) + ty(2xy + Z2)+ tz (2yz) = 0 + (2x) + (2y) = 2(x + y).
V.v
qed
1.15
(a)V.va
Problem
as a vector.
2x + 3x.
1.16
= tx(-?-)+ty(~)+tz(?-)
3
5
= 0-2
= 3r-3
+ x( -3/2)()-22x
- 3r-5(x2
= tx3 [x(x2
+y2 +z2)-~]+ty
5
3
[y(X2 +y2 +Z2)-~]+tz
5
[z(x2 +y2 +z2)-~]
+ 0-2 + y( -3/2)0-22y
+ 0-2 + z( -3/2)O-22z
= 3r-3 - 3r-3 = O.
+ y2 + Z2)
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the
0 everywhere except at the origin, but at the
origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, V.v is infinite at
that one point, and zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
=
origin. How, then, can V.v = O? The answer is that V.v
Vy
= cosif>Vy+sinif>vz;
Vz = -sinif>vy +cosif>vz.
,!,
~!bl. + ~oz
,!,+ !lJL.."!'I . P b 114 .
+ !lJL..!bl.
+ !lJL..8Z . ,!, U
8y - 8y cos'l'
o'g sm'l' oy 8y
8z 8y cos'l'
8y 8y
8z oy sm'l'. se resu t m ra. . .
= ~8y COg + ~8z sin if» COgif>+ !lJL..
+ !lJL..sin
if» sin if>.
8y cos
OZ
~
- ~
(
~8z
= -~OZ
(
if>
sin
= - (-?v
~8y + ~8z = ~8y
)
(
if>
+ ~8z cosif>
= - (~~
8y
sin if>+ i:
COS2'!'
8z
(
)
if>
+ ~8:.
OZ 8z
cosif» sin if>+ (-~
) sin
if>
sin if>+
+
~
+ !lJL..8:.
8y 8z
oz 8z ) cosif>
(!lJL..~
cosif» cosif>.So
if>+ !lJL..sin
'!'cos'!'
if>+ ~8y sin2,!,'I' - ~8z sin '!'cos'!'
'I' + ~8z sin '!'cos
'I'
8y
'I'
'I' + !lJL..sin2
8z
'I'
'I'
5
-~ oy sinif>cosif> + ~8z COS2if>
(cOS2 if> + sin2 if» + ~8z (sin2 If'A. + COS2 A.
If' )
= ~8y
Problem
+ ~.
8z
.(
1.18
x0
(a) Vxva =
(b) VXVb
8x
=
(c) Vxvc
z0
3xz2
oy
- 2xz
x
y
Z
0
8y
0
oz
xy
2yz
3xz
= I !Ix
= x(O - 6xz) + y(O+ 2z) + z(3z2 - 0)
oz
8
ox
y2
Problem
y0
x2
x
v =
or v
= ~8y
y
0
oy
(2xy + z2)
= x(O -
2y) + y(O - 3z) + z(O- x)
= 1-6xz x + 2z y + 3Z2z.1
= 1-2yx - 3zy - xz.!
Z
0
oz
2yz
= x(2z - 2z) + y(O - 0) + z(2y - 2y) = [QJ
1.19
yx + xy; or v = yzx + xzy + xy Zjor v = (3x2z -
= (sin x) (cosh
Z3)
X+ 3y + (x3 - 3xz2)z;
y) x - (cos x) (sinh y) yj etc.
Problem
1.20
(i) V(fg)
= 8~:) x + 8~:) y + o~;) Z = (t~ + gU) x + (J~ + gU) y + (t~ + g¥Z)z
= J (~x+ ~y + ~z) +g(Ux+ Uy+ ¥Zz)= J(Vg)+ g(VI). qed
= !Ix (AyBz -
(iv)V.(AxB)
AzBy)
+
!ly
-
(AzBx - AxBz) + !lz (AxBy
AyBx)
B y ML
+
A
8Bz
+
B
ML
Ax ~oy
8x
z 8y
x 8y
-- Ay ~OX + Bz ~8x - Az 8Bv
8x = Bx
+A x §oz + B y ML
8z - A y 8B,.
8z - B x ~8z
ML
ML
- ~
+ Bz ~oz
+ B y ( 8z
8y
OZ
8z )
(
( -
)
ML
OA,. - A
8y
)
- B z oA,.
oy
(~8y
x
- §
8z
-Ay (8tz,.- ~) - Az (8~v- o~,.) = B. (VxA) - A. (VxB).
(v) Vx (fA)
= (O(~:%) - O(~~v»)x+ (8(~~,.)- 8(~:%») y + (8(~:v)- 8(~:,.»)Z
A ~
= (JM...+A
~- J Mox
~ x+ (J8A,.+A
~- J ~-A
8y
z 8y
8z
Y8z )
OZ
z oz
z 8x )y
z
oAz
-A
~
+ (J~+A
~J 8y
8z
y 8x
x 8y )
ML
ML
- ~
Z
= J M...
+ ~ - 8Az
x+ ( fu
~
fu
~ )Y ( ~
~ ) ]
A
)
[(
A
- [( Ay¥Z - AzU) x + (AzU - Az¥Z)y + (AxU - AyU) z]
= J(VxA)
Problem
- Ax (VI). qed
1.21
(a ) (A. \1) B
= (A x~oB,. + A y~8B,.+ A Zfu8B,.) X +
+ A x ~+A
8x
(
(b) f
= !.r
y ~+A
oy
(A z~8Bv + Ay~oBv
z.
z ~8z )
= xx+yy+zz. Let's just do the x component.
ylx2+y2+z2
[(f. \1)f]x = )-
(x !Ix+ Y!ly+ z tz) ylz2;y2+z2
+ AZfu8Bv
)Y
A
)
qed
6
CHAPTER
= ~ {x
1. VECTOR ANALYSIS
[J- +X(-~)(}-)32X] +yx [-~(}-)32Y] +ZX [-~(}-)32Z]}
= ~{; -
~ (X3+ xy2 + XZ2)} = ~ {; - ~ (X2+ y2 + Z2)}
Same goes for the other components. Hence: (f. V') f
I
=0
= ~(; - ;) = O.
I.
(c) (Va.V')Vb= (X2tx + 3XZ2ty - 2xz tz) (xyx + 2yzy + 3xzz)
= X2(yx + Oy + 3zz) + 3XZ2(xx+ 2zy + Oz) - 2xz (Ox + 2yy + 3xz)
= (X2y + 3x2z2) X+ (6XZ3- 4xyz) y + (3X2z - 6X2Z) Z
=
Problem
I
(y + 3Z2) x + 2xz (3z2 - 2y) Y - 3X2ZZ
X2
I
1.22
= :x (AxBx + AyBy + AzBz) =
(ii) [V(A.B)]x
= Ay(VxB)z
[Ax(VxB)]x
-
[Bx (VxA )]x =B y (~-~
ax
oy
a:;", By
+ Ay
+ ~Bz
°:X"
+ Az~
= Ay(°:X" - °.:vz)- Az(°fzz -~)
Az(VxB)y
~-~
z ( oz
) -B
+ A~ °:Xz +
o~z Bx
ax
)
[(A.V)B]x = (Ax tx + Ay ty + Az tJBx = Ax °:Xz+ Ay °.:vz+ Az °fzz
+ B y oAz
+ B z oAz
[(B.V )A]x = B x oAz
ax
oy
oz
So [Ax(VxB)
+ Bx(VxA)
+ (A.V)B + (B.V)A]x
+ B Yox
oA" +B y~oAz
oBz +B ( oA" - ~ +~
) +A
--
A Yox
oB" - AoBz
Y oy - A z oBz
oz + A z !ll!
ax
+A x&oBz +A y~!ilk +A ZhoBz +B x&~
=
B oAz +A
x ax
x ax
+B (-~+~+~
z
=
Tz
y ax
ax
Tv
ax
- Tv
~ +~Tv )
)
OB"
y ( ax
Tv
) +A z (-~+!ll!
Tz
Tz
B y oAz
oyZ - B oAz
oz + B z ~aX
+B ZhM...
+~Tz
[V(A.B)]x (same for y and z)
(vi) [Vx(AxB)]x
=
--
ty(AXB)z
oAz
oy
- tz(AxB)y
= ty(AxBy - AyBx) - tz(AzBx - AxBz)
By + Ax ~oy - ~oy Bx - AY oBz
oy - ML
OZ B x - Az !ll!
OZ +
+ A(V.B) - B(V.A)]x
+B
oAz +B oAz -A oBz -A oBz -A oBz +A ( OBz +~+!ll!
x ox
Y oy
z oz
x ax
y oy
z oz
x ax
oy
-~ - ML)
= B oAz +A (-~
+~
+oB" + !ll! ) +B (~ -~
y oy
x
Tx"7fX" oy
oz
x TJX
TJX
oy
oz
+ A y (-~ oy ) + A z (-~ oz ) + B z (M..
oz )
[(B.V)A - (A.V)B
--B
oAz
= [Vx(AxB)]x
oz.
oAz
oz
Bz + A x !ll!....
oz
) -B x ( 8Az
ax +~+~
oy
(same for y and z)
Problem 1.23
VU /g) = txU/g)x+ tyU/g)y+ tzU/g)z
= gU-f~ + g*-f~
+ gU-f~
A
A
=
V.(A/g)
=
A
r (!!.Lx+ !!.Ly + !!.Lz) - f (£i.x + £i.y +
2
1
~ l!g
X
ox
Y
g2
oy
A
g2
oz
Z
ax
oy
A
£i.z
oz
)] =
gVf-fVg.
g2
Qed
tx(Ax/g) + ty(Ay/g) + tz(Az/g)
g~-A,,~
g~-A.~
= g~-Az~
g2
= ~1
[g
(
OAz
ax
+
g2
+ oA"
oy +
ML
oz
+
-
g2
) (
A x ax
£i. + A y £i.
oy + A z £i.
oz )]
= gV.A-A.Vg
g2.
Q ed
oz
)
7
[Vx(A/g)].,
= ty(Az/g)-tz(Ay/g)
g~-A'lf11
u
~
9 ~ - A v~
=
2
g2
~ [gg(¥:- ¥:)
=
-
- (Az?v- Ay~)]
g(VXA)",+(AXVg)",
(same for y and z).
g2
Problem
1.24
x
y
Z
(a) AxB
=I x
2y
-2x
3z
0
3y
-
I
= x(6xz)
Qed
+ y(9zy) + z(_2X2 - 6y2)
= t.,(6xz) + ty(9zy) + tz(-2x2 - 6y2) = 6z + 9z + 0 = 15z
V.(AxB)
VxA = x (ty(3Z)- tz(2Y))+y (tz(x) - t.,(3z))+z (t.,(2Y)- ty(x)) = 0; B.(VxA) = 0
VxB = x (ty(O) - tz(-2X)) + y (tz(3y) - t., (0)) + z (t.,(-2X) - ty(3Y)) = -5 z; A.(VxB) = -15z
?
V.(AxB)
==B.(VxA)
(b)A.B = 3xy - 4xy
= -xy
x y
- A.(VxB) = 0 - (-15z) = 15z. ./
= V(-xy) = xt.,( -xy) + yty(-xy) = -yx - xy
; V(A.B)
z
Ax(VxB)= x 2y 3z 1= x(-lOy) + y(5x); Bx(VxA) = 0
I
I
(A.V)B
0
0
-5
I
= (xt., + 2yty + 3ztz) (3yx - 2xy)
= x(6y) + y(-2x)
= x(3y) + y( -4x)
(B.V)A = (3y to: - 2xty) (xx + 2yy + 3zz)
Ax(VxB) + Bx(VxA) + (A.V)B + (B.V)A
= -10yx
+ 5xy + 6yx - 2xy + 3yx - 4xy
= -yx - xy = V.(A.B).
./
(c) V X (AxB) = x (ty (-2X2 - 6y2)- tz (9ZY))+ y (tz (6xz) - t., (-2x2 - 6y2))+ z (t., (9zy) - ty (6xz))
= x(-12y - 9y) + y(6x + 4x) + z(O) = -21yx + lOxy
V.A = t.,(x) + ty(2y) + tz(3z) = 1+2 +3= 6; V.B = t.,(3y) + ty(-2x) = 0
. (B.V)A - (A.V)B + A(V.B) - B(V.A)
= Vx(AxB).
= 3yx -
4xy - 6yx + 2xy -18yx + 12xy
./
Problem 1.25
(a) 8;;;a = 2;
(b)
8;;/ = 8;~a = 0
=> I \72Ta
= 2.1
W
8271 = W
8271 = ~
8271 = -n
=>\72 n = -3Tb= -3smxsmysmz.
.
. .
I
I
(c) 8;7 = 25Tc ; 8;;;c= -16Tc ; 8;;;c= -9Tc => \72Tc
I
(d)
8;;2'"
82v
= 2 ; a;;~",
82v
= 0.1
= 8;;; = 0 => \72V.,= 2
82v
W = W = 0 ; F = 6x
2
=> \7 Vy = 6x
}Iv'v
a;;{ = 8;;~.= 8;;1 = 0 => \72vz = 0
~ 2X+6xy.1
= -21yx
+ lOxy
8
CHAPTER
Problem
1.26
(~ay
V. (VXv ) =.JL
ax
-
1. VECTOR ANALYSIS
(
a2vz
ax 8y
~
-
- ~
az
a2vz
) + (ayoz -
8y 8x
From Prob. 1.18: VXVb
82vz
oz oy
(a2v" --~8y )
. t.
al t f
d
0 b
ax 8z ) - , y equ 1 y 0 cross- enva Ives.
- ~ ) + JL ~
ax
az ax
~
ay ( az
) + JL
82v"
) + (oz ax -
'
= -2yx - 3zy - xz =? V.(VXVb)= !z(-2y) + /y(-3z) + !/z(-x) = O. ./
Problem 1.27
=
V X (Vt)
x
ya
z8
8x
fJy
fJz
I fJta
fJt
fJy
fJx
-
fJ2t
= x ( fJy fJz -
fJt
az
fJ2t
fJz fJy )
-
fJ2t
+ Y( fJz fJx -
fJ2t
fJx fJz
-
a2t
) + z ( ax fJy -
a2t
fJyax
)
= 0, by equality of cross-derivatives.
In Prob. 1.l1(b), V f = 2xy3z4 X + 3x2y2z4 y + 4x2y3z3z, so
X
a
=
VX(Vf)
fJx
2xy3z4
y
a
fJy
3x2y3z4
z
a
az
4x2y3z3
= x(3. 4x2y2z3 - 4. 3x2y2z3) + y(4. 2xy3z3 - 2. 4xy3z3) + z(2. 3xy2z4 - 3. 2xy2Z4) = O. ./
Problem
1.28
(a) (0,0, 0) ~
(1,0, 0) ~
(1,1, 0)
~
(1,0,0). x : 0 -t 1, y = z = OJdl = dx Xjv . dl
Total: J v . dl
(b) (0,0, 0) ~
(0,0, 1) ~
(0,1, 1) ~
= x2 dxj J v
. dl
= J; X2 dx = (x3/3)IA = 1/3.
(1,1,0). x = 1,Y : 0 -t 1,z = 0;dl = dy y; v . dl = 2yz dy = OJJ v . dl = O.
(1,1,1). x = Y = 1,z : 0 -t Ij dl = dz Zjv. dl = y2dz = dzj J v. dl = Jo1dz
= zlA = 1.
= (1/3) + 0 + 1 = 14/3.1
(0,0,1). x = y = 0, z : 0 -t 1;dl = dz z; v . dl = y2 dz = 0; J v . dl = O.
(0,1,1). x = O,y: 0 -t l,z = Ijdl = dyy;v.dl = 2yzdy = 2ydy;Jv.dl = J; 2ydy = y21A= 1.
(1,1,1). x : 0 -t 1,y = z = Ij dl = dxx; v. dl = x2 dx; J v. dl = J; x2 dx = (x3/3)IA = 1/3.
Total: J v . dl = 0 + 1 + (1/3) = 14/3.1
(c) x = y = z : 0 -t Ij dx = dy = dzj v. dl = X2 dx + 2yz dy + y2 dz = x2 dx + 2X2 dx + x2 dx = 4X2 dx;
J v . dl = J; 4x2 dx = (4X3/3)IA = 14/3.1
(d) f v. dl = (4/3) - (4/3)
Problem 1.29
= @]
x,y : 0 -t l,z = Ojda = dxdyzjv' da = y(z2 - 3)dxdy = -3ydxdy;Jv. da = -3J:dxJ:ydy=
-3(xl~)(fl~) = -3(2)(2) = []I] In Ex. 1.7 we got 20, for the same boundaryline (the square in the xy-
plane), so the answer is Ino: Ithe surface integral does not depend only on the boundary line. The total flux
for the cube is 20 + 12 =
Problem 1.30
[ill
J T dr =J Z2dx dy dz.
You can do the integrals in any order-here
/
Z2
it is simplest to save z for last:
[/ (/ dX)dY]dz.
The sloping surface is x+y+z = 1, so the x integral is J~I-y-Z) dx = 1-y-z. For a giv~~z, y ranges from 0 to
1- z, so the y integral is J~I-Z)(1- y - z) dy = [(1- z)y - (y2/2)JI~I-Z)= (1- z)2 - [(1- z)2/2] = (1- z)2/2 =
9
- z + (z2/2).
i-t+fo=~
(1/2)
Problem
Finally, the z integral is J; Z2(~ -
+
Z
= Jo1(; - Z3 + ~4) dz = (~ -
Z;)dz
~4+ f~)lb =
1.31
T(b)=1+4+2=7j
T(a) =0. =>IT(b)-T(a)
=7.1
VT = (2x + 4y)x + (4x + 2Z3)y + (6yz2)z; VT.dl
= (2x + 4y)dx + (4x + 2x3)dy + (6yz2)dz
(a) Segment 1: x: 0 -t 1, y = z = dy = dz = O.JVT.dl = J;(2x)dx = x21~= 1.
Segment 2: y:O-tl,
x=l, z=O, dx=dz=0.JVT.dl=Jo1(4)dy=4YI~=4.
J:VT.dl=7../
Segment 3: z: 0 -t 1, x = y = 1, dx = dy = O.JVT.dl = J;(6Z2)dz = 2z31~= 2. }
(b) Segment!:
z:O-tl,
x=y=dx=dy=O.JVT.dl=J;(O)dz=O.
Segment2: y: 0 -t 1, x = 0, z = 1, dx = dz = O.JVT.dl
~ Jo1(2)dy = 2yl~ = 2.
J.bVT.dl
Segment 3: x: 0 -t 1, y = z = 1, dy = dz = O.JVT.dl = Jo (2x + 4) dx
=
(c)x:O-tl,
y=x,
(X2
+ 4x)l~
= 1 + 4 = 5.
a
}
z=X2, dy=dxdz=2xdx.
= (lOx
VT.dl = (2x + 4x)dx + (4x + 2X6)dx + (6XX4)2xdx
1: VT.dl = J;(lOx
+ 14x6)dx
= (5X2
+
2X7)1~
+ 14x6)dx.
= 5 + 2 = 7. ./
Problem 1.32
V.v
=y +
J(V.v)dr
2z + 3x
=
=
J(y + 2z + 3x) dxdydz
3x) dx} dydz
'--t [(y + 2z)x + ~X2]~= 2(y + 2z) + 6
J {J; (2y
+ 4z + 6)dY} dz
' t
=
= JJ{J;(y + 2z +
[y2 + (4z + 6)Y]~ = 4 + 2(4z + 6)
J;(8z + 16)dz
=
(4z2 + 16z)l~
= 8z +
16
= 16 + 32 = 148.1
Numbering the surfaces as in Fig. 1.29:
= dydzx,x = 2. v.da = 2ydydz.Jv.da = JJ2ydydz = 2y21~= 8.
(ii)da = -dydzx,x = O.v.da = O.Jv.da = O.
(iii)da = dxdzy,y = 2. v.da = 4zdxdz.Jv.da = JJ4zdxdz = 16.
(i) da
(iv)da = -dxdzy,y = O.v.da = O.Jv.da = O.
(v) da
.
= dxdyz,z = 2. v.da = 6xdxdy.Jv.da = 24.
(vi) da = -dxdyz,z = O.v.da = O.Jv.da = O.
=>J v.da = 8 + 16+ 24 = 48 ./
Problem 1.33
Vxv = x(O- 2y) + y(O- 3z) + z(O- x) = -2yx - 3zy - xz.
da = dydz X,if weagreethat the path integralshallrun counterclockwise.
So
(Vxv).da
= -2ydydz.
= 7../
CHAPTER 1. VECTOR ANALYSIS
10
J(Vxv).da
= J{J:-Z(-2Y)dY}dz
y
z
y21~-Z= -(2 - Z)2
= -J:(4-4z+Z2)dz== -(S-s+!)=I-!1
(4Z-2z2+~)I:
Meanwhile, v.dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments.
y
z
~)
(3)1
~
y
= z = 0; dx = dz = O. y : 0 -t 2. Jv.dl = O.
= 0; z = 2 - y; dx = 0, dz = ~dy, y: 2 -t O.v.dl = 2yzdy.
(1) x
(2) x
Jv.dl = J202y(2- y)dy = - J:(4y - 2y2)dy= - (2y2- ~y3)I~ = - (S - ~ .S)
(3) x = y = 0; dx = dy = 0; z: 2 -t O.v.dl = O.Jv.dl = O. So §v~dl = -i. ./
= -i.
Problem1.34
By Corollary 1, J(Vxv).da
(i) da
= dydz x,
should equal ~. VXv
= (4z2 -
2x)x + 2zz.
x = 1; y,z: 0 -t 1. (Vxv).da = (4z2- 2)dydz; J(Vxv).da
= J;(4z2 - 2)dz
= (~z3 - 2z)l~= ~ - 2 = -~.
(ii) da = -dxdyz, z = OJx,y: 0 -t 1.(Vxv).da = 0; J(Vxv).da = O.
(iii) da = dxdzy, y = 1; x,z: 0 -t 1. (Vxv).da = 0; f(Vxv).da = O.
(iv) da = -dxdzy, y = 0; x, z : 0 -t 1. (Vxv).da = 0; J(Vxv).da = O.
(v) da = dxdyz, z = 1; x,y: 0 -t 1. (Vxv).da = 2dxdy; J(Vxv).da = 2.
=>f(Vxv).da = -~ + 2 = ~. ./
Problem 1.35
(a) Use the product rule V X(fA)
Lf(VXA).da=
= f(V
XA)
L Vx(fA).da+
-
A x (V f) :
LrAX(Vf)].da=
ifA.dl+
LrAX(Vf)]'da.
qed.
(I used Stokes' theorem in the last step.)
(b) Use the product rule V.(A x B)
IvB,(VXA)dr=
= B.
Iv V.(AxB)dr+
(VxA) - A. (VxB) :
IvA,(VXB)dr=
(I used the divergence theorem in the last step.)
t(AXB).da-+
IvA,(VXB)dr.
qed.
11
Problem
1.361 r = ,jx2 + y2 + Z2;
Problem
1.37
0
= cos
-I
(
'
y'x'+y'+"
)
;
q, = tan-I
m.
There are many ways to do this one-probably the most illuminating way is to work it out by trigonometry
from Fig. 1.36. The most systematic approach is to study the expression:
r
= x x + y Y + z z = r sin 0 cos q, x + r sin 0 sin q, y + r cos 0 z.
If I only vary r slightly, then dr = fj-,.(r)dris a short vector pointing in the direction of increase in r. To make
it a unit vector, I must divide by its length. Thus:
8r
!k
8r
f= I~I; {h 1;1; J>=1;1'
!¥i= sinOcosq,x + sinOsinq,y +cosOz; 1!¥i12= sin20cos2q,+sin20sin2q,+cos20
~ = rcosOcosq,x+rcosOsinq,y
-rsinOz;
~ = -rsinOsinq,x
+ rsinOcosq,y; 1~12
= sinO cosq,x
+ sinOsin q,y + cosO Z.
f
1~12
= r2 sin2 Osin2
q,
+ r2
sin2 OCOS2 q,
=}IIi = cosOcos q, x + cosOsinq,y - sinOz.
J> = - sinq,x + cosq,y.
Check: f.f = sin2 0(COS2q, + sin2q,)+ COS2
0 = sin20 + COS2
0 = 1, .{
1i.J>=-cosOsinq,cosq,+cosOsinq,cosq,=O,.{
etc.
sinOf = sin2 0 cosq,x + sin2 Osin q,y + sin OcosOz.
cosOIi = COS2Boos q,x + COS20 sin q,y - sinOcosOz.
Add these:
(1)
sinOf+cosoli
= +cosq,x + sinq,y;
=
(2)
J>
-sinq,x+cosq,y.
Multiply (1) by cosq" (2) by sinq" and subtract:
I
x = sinOcosq,f + cosO cos q,Ii -
sinq,J>.1
Multiply (1) by sinq" (2) by cosq" and add:
Iy
cosOf
sinO Ii
= sinO cosOcos q,x
= sinOsinq,f
= 1.
= r2cos20cos2q,+r2cos20sin2q,+r2sin20
+ cosOsinq,1i + cosq,J>.
+ sin OcosO sin q,y + COS2
0 z.
= sinO cosO cosq,x + sinO cosO sinq,y - sin2 0 z.
Subtract these:
Iz=cosOf-sinoli.1
I
= r2 sin2 0..
= r2
12
CHAPTER
Problem
1.38
= ~ tr(r2r2) = ~4r3 = 4r
(a) V'VI
f(V
1. VECTOR ANALYSIS
= J(4r
'VI )dr
J vl.da
(b) V'V2
=
=
sin (Jd(J d4>f)
= ~ tr (r2~)= 0
J v2.da = J (~f)
=
)(r2 sin (Jdr d(J d4»
J(r2f).(r2
=>
(4) JoRr3dr Jo" sin (Jd() n" d4>= (4)
=0
J(V'v2)dr
I
= J sin
(r2 sin (Jd(J d4>f)
( ~4)(2)(211")
=1411" R41
r4 Jo" sin(J d(JJ:" d4>= 411"R4./ (Note: at surface of sphere r
=
R.)
I
(J d(J d4>
= 1411".1
They don't agree! The point is that this divergence is zero except at the origin, where it blows up, so our
calculation of J(V 'V2) is incorrect. The right answer is 411".
Problem
1.39
=
V.v
..
~ tr(r2 rcos(J) + rs~nOt(J(sin(Jrsin(J) + rs~nO:4>(r sin(Jcos 4»
~
A.)
cos (J+ -J:rSIn (J r 2 sin (Jcos (J+ -r s In (J r sin (J( - sin 'I'
=
=
r:13r2
3cos(J + 2cos(J.- sin 4>= 5cos(J
= J(5 cos (J -
J(V.v)dr
= (~3)
sin
- sin 4>
r2 sin (Jdr d(J d4>
4»
= JoRr2 dr Jo£ [J:"
-
(5 cos (J sin 4» d4>] d(J sin (J
~211"(5 cos (J)
(1011") Jo;; sin (Jcos (J , d(J
~
--15" 3 R 3 .
;;
s;n2°
2
1
0
-
1
2
I
Two surfaces-one
the hemisphere:
Jv.da = J(rcos(J)R2
sin(Jd(Jd4>
da
= R2
sin (Jd(J d4>f; r
= R3 Jo;;sin(Jcos(Jd()
= R;
4>: 0 -t 211",(J: 0 -t
J:" d4>= R3
other the flat bottom: da = (dr)(rsin(Jd4»(+8) = rdrd4>8 (here (J
v.da = f(r sin (J)(r dr d4» = JoRr2 dr J:" d4>= 211"~3
.
Total: v.da = 11"R3 + ~11"R3 = ~11"
R3. ./
J
U) (211") = 1I"R3.
= ~). r:
0 -t
R,
4>:0 -t 211".
J
= (cos(J+ sin(Jcos4»f + (-
Problem 1.40 IVt
\72t
~.
=
sin(J+ cos(Jcos4»8 + Si~(J(- si~(Jsin4»cb
v.(Vt)
= ~ tr (r2(cos
(J+ sin(Jcos4») + rs~n (Jto (sin(J( - sin (J+ cos(Jcos 4») + rs~nO:4>(= ~ 2r(cos (J + sin (Jcos 4» + rs~n (J(-2 sin(J cos~ + COS2(Jcos4>- sin2 (Jcos 4» - rs~n
=
r s~n0 [2 sin (JJos (J+ 2 sin2 (Jcos 4>- 2 sin (JJos (J + COS2(JCOS4>- sin2 (Jcos 4>- cos
=
~
=>I \72t
=0
-
[(sin2 (J+ COS2(J)COS4> cos 4>]
4>]
I
Gradient Theorem: J~ Vt.dl = t(b) - t(a)
Segment 1: (J= ~, 4> = 0, r: 0 -t 2. dl = drf; Vt.dl
JVt.dl
Vt.dl
(J cos 4>
= o.
Check: r cos (J= z, r sin (Jcos 4>= x => in Cartesian coordinates t
Segment 2: (J =
sin 4»
= x + z.
Obviously,Laplacian is zero.
= (cos(J+ sin(Jcos4»dr = (0+ l)dr = dr.
= J: dr = 2.
~, r = 2, 4>:0 -t~.
= (-sin4»(2d4» =
dl
= rsin(Jd4>cb = 2d4>cb.
-2 sin 4>d4>.
JVt.dl = - Jo;;2sin4>d4>
= 2cos4>l! = -2.
13
= 2, 4J = ~; 0: ~ -t O.
dl = r dO 6 = 2 dO6; Vt.dl = (- sin 0 + COg0 COg4J)(2 dO) = -2 sin 0 dO.
= - J~2 2 sin 0 dO = 2 COg Ol~2 =. 2.
J Vt.dl
J: Vt.dl = 2 - 2 + 2 = m. Meanwhile,t(b) - tea) = [2(1 + 0)] - [O()] = 2. ./
Segment 3: r
Total:
1.41 From Fig. 1.42, S = COg 4Jx+ sin 4Jy; ~ = ~ sin 4Jx
Problem
I
+ COg 4Jyj
Multiply first by COg4J, second by sin 4J, and subtract:
S COg4J sin 4J = COS2 4Jx + COg4Jsin 4Jy + sin2 4Jx - sin 4Jcos4Jy
~
x = cos 4JS - sin 4J ~.I
Multiply first by sin 4J, second by COg4J,and add:
s sin 4J + ~cas 4J = sin 4JCOg4Jx + sin2 4Jy - sin
So
z=z
= x(sin2
4J +
I
COS2 4J)
= x.
I
So y = sin 4Js + cas 4J~.I
Problem 1.42
I
=
(a) V.v
1
4JCOg4J
x + cos24Jy = y(sin2
4J + COS24J)
= y.
z = z.1
=
=
~ts(ss(2+sin24J))+~:4>(ssin4Jcos4J)+tz(3z)
~ 2s(2 + sin2 4J) + ~ s(cOS24J- sin2 4J) + 3
4 + 2sln2 4J+ cos2 4J- sin2 4J+ 3
=
4 + sin2 4J+ COS24J+ 3
= [[]
2
5
= J(8)sdsd4Jdz = 8Jo sdsJl " d4JJodz
= 8(2) (~) (5) = r-:in=-l
~
(b) J(V.v)dr
Meanwhile, the surface integral has five parts:
2
"
top: z=5, da=sdsdifJzj v.da=3zsdsdifJ=15sdsdifJ. Jv.da=15Jo sdsJ02d4J=1511".
bottom: z = 0, da = -sdsd4Jzj v.da = -3zsds.difJ= O. Jv.da = O.
back:4J= ~, da = dsdz~j v.da = ssin4Jcos4Jdsdz= O. Jv.da = O.
left: 4J= 0, da = -dsdz~j v.da = -ssin4Jcos4Jdsdz= O. Jv.da = O.
front: s = 2, da = s d4Jdz s; v.da = s(2 + sin2 4J)sd4Jdz = 4(2 + sin2 4J)d4Jdz.
"
5
Jv.da = 4 J02 (2 + sin24J)d4J
Jo dz = (4)(11"+ ~)(5) = 2511".
So !v.da
(c) Vxv
Problem
= 1511" + 2511" = 401T../
=
(~:4>(3z)- :z (ssin ifJcosifJ))s + (tz (s(2 + sin2ifJ))- ts(3z)) ~
=
+~ (ts(S2 sin ifJcosifJ)- :4>(s(2 + sin2 ifJ))) Z
~(2ssinifJcosifJ s2sinifJcos4J) =@]
(a) 3(32) - 2(3)
(b)
COS1T
-
1.43
-
1 = 27 - 6 - 1 = 120.1
=QJ
(C)Izero.I
(d) In(-2:1- 3) = In 1 = ~
Problem 1.44
= leO + 3) = []
x) = 8(x - 1), so 1 +
(a) J~2(2x + 3H8(x) dx
(b) By Eq. 1.94, 8(1
-
3 + 2 = [I]
14
CHAPTER
= 9 (-!)2! = []
(c) J~19x2!15(X + !)dx
(d) 11 (if
a> b), 0 (if a < b).
Problem
1. VECTOR ANALYSIS
I
1.45
(a) J~oof(x) [xd~15(x)]dx = xf(x)15(x)l~oo - J~oo d~ (xf(x))15(x)dx.
The first term is zero, since 15(x)= 0 at :1:00;d~ (xf(x)) = x1x + ~~f = x1x + f.
- J~oo(x1x + f) 15(x)dx = 0 - feD) = - feD) = - J~oof(x)15(x)dx.
So, xd~15(x)= -15(x). Qed
So the integral
is
(b) J~oolex) ~:dx = f(x)O(x)l~oo - J~oofxO(x)dx = f(oo) - Jooo1xdx = f(oo) - (f(oo) - feD))
~ = 15(x).
= feD) = J~oo f(x)15(x) dx. So
Problem
(a) per)
= q153(r -
r').1 Check: Jp(r)dT
(b) per)
= q153(r-
r') - q153(r).1
I
I
Qed
1.46
= A15(r-
(c) Evidently per)
Q = JpdT
=JA15(r-
= q J153(r -
r') dT = q.
./
R). To determine the constant A, we require
R)411"r2dr
= A 411"R2.
So A = 41f~2'
I
per)
= ~15(r
- R)./
Problem 1.47
(a) a2 + a.a + a2 = 13a2.1
(b) J(r - b)2s\153(r)dT = 1~5b2 = 1~5(42+32)
= []
= 25 + 9 + 4 = 38 > 36 = 62, so c is outside V, so the integral is Izero. I
(e - (2x+ 2y+ 2Z))2 = (lx+Oy+
(-I)z)2 = 1 + 1 = 2 < (1.5)2 = 2.25, so e is inside V,
(c) C2
(d)
and hence the integral is e.(d
-
e)
= (3,2,1).(-2,0,2) = -6 + 0 + 2 = GD
Problem 1.48
J
First method: use Eq. 1.99 to write J = cr (41r(~3(r))dT = 411"e-o= 1471'.1
Second method: integrating by parts (use Eq. 1.59).
J
Problem
=
I
I
=
411"(-e-r)
= -
v
f
r2'
V(e-r)
r~ e-r411"r2 dr
1.49 (a) V.F1
I:+
I ax
0
+
fs
+
I
e-r r2
8
( )
. da. But V (e-r) = 8r e-r f = -e-rf.
00
e-r :2 . r2 sin 0 dO dfjJf
411"e-R
= 411"(-e-oo
=
411"
I°
e-r dr + e-R
za
~
ay az =-Y8x
0 x2
8
I
sinO dOdfjJ
+ e-O) = 471'../ (Here R = 00, so e-R = 0.)
= tx(O)+ ty(O)+ tz (X2)= @]; V.F2 = ~~+
xa ya
VxF1=
dT
r
2
~
(x ) =~j
V F
X 2=
~+
= 1 + 1 + 1 =m
xaa y az
ax
X
g~
ay
Y
az
z
rnl
I
=t.Qj
15
F2 is a gradient; FI is a curl
I
~
(
For A I, we want
I
Al
= ~x2 y
{}z
£b
{}y
I
U2 = ~ (X2 + y2 + Z2)
= (£b{}z - £b{}x ) = O.' ~{}x
I
)
= VxAI).
(FI
I
-
I
would do (F2
- ~{}y
(But these are not unique.)
x
(b)V.Fa=/x(yz)+/y(xz)+/z(xy)=Oj
= X2 .
y
= VU2).
A =
y
3 '
x3
Ax = Az = 0 would do it.
z
VXFa=1
/x /y /z l=x(x-x)+y(y-y)+z(z-z)=O
yz xz xy
So F3 can be written as the gradient of a scalar (Fa = VU3) and as the curl of a vector (Fa = VxAa).
fact, U3 = xyz does the job. For the vector potential, we have
I
I
£b
{} - ~{}z = yz,
~{}
{ ~-~
{}x
- £b
{}x
{}y
= xz '
=x y ,
Putting this all together: Aa
Problem 1.50
(d) ~ (a): VxF = Vx(-VU)
(a) ~ (c): §F. dl = J(VxF).
I
(c) ~ (b):
J: IF.
Az = ty2z + f(x,z);
suggesting
so
Ax = tz2x+h(x,y)j Az = -tzx2 +j(y,z)
Ay = tx2y + k(y, z); Ax = -txy2 + l(x, y) }
= t {x (z2-
y2) X + Y (x2 - Z2) Y+ Z (y2
Ay = -tYz2 + g(x,y)
- x2)z}
I
(again, not unique).
=0
(Eq. 1.44 - curl of gradient is always zero).
da = 0 (Eq. 1.57-Stokes' theorem).
J: IfF.
dl-
which suggests
dl
= J: IF.
dl +
J: IfF.
{b F. dl
ia I
=
dl
= § F . d1 = 0, so
(b F. dl.
ia If
(b) ~ (c): same as (c) ~ (b), only in reverse; (c) => (a): same as (a)=> (c).
Problem 1.51
(d) ~ (a): V.F = V.(VxW)
= 0 (Eq lA6-divergence
of curl is always zero).
= J(V.F)
=0
(a) ~ (c):
§ F . da
(c)~(b):
JIF.da-JIfF.da=§F.da=O,so
dr
(Eq. 1.56~divergence
theorem).
{ F . da = { F. da.
iI
iIf
§ F . da, da is outward, whereas for surface II it is inward.)
(b) ~ (c): same as (c) ~ (b), in reverse; (c)~ (a): same as (a)~ (c) .
Problem 1.52
In Prob. 1.15 we found that V,va = 0; in Prob. 1.18 we found that Vxvc = O. So
(Note: sign change because for
v c can be written as the gradient of a scalar; Va can be written as the curl of a vector.
(a) To find t:
(1) g;
(2)
g~
= y2
= (2xy
=>
(3) g; = 2yz
t
= y2x
+ Z2)
+ f(y, z)
In
CHAPTER 1. VECTOR ANALYSIS
16
= y2x + yz2 + g(y), so :~ = 2xy + z2 + ~ =
2xy + Z2 (from (2» => ~ = 0. We may as well pick 9 = OJthenIt = xy2 + YZ2.1
8Wy - 8 w", = -2xz.
To find W: 8W.
8y
8z - 8W.
8x = 3Z2X' '
8y - 8Wy=
8z
x,2. 8W",
8x
From (1) & (3) we get
(b)
M
= 2yz
=>
J = yz2 + g(y)
=>
t
Pick Wx = OJthen
8Wz
8x
=
-3xz2 =>Wz = _~x2Z2 + J(y,z)
8Wy
8x
=
-2xz =>Wy = _X2Z + g(y,z).
- !!.i.= O. Mayas well pick
8W.
- 8Wy
= !!.l.+
X2 - !!.i.
8 z = x2 => !!.l.
8y
8z
8y
8z
8y
J = 9 = O.
IW=-X2ZY_~X2Z2Z.1
X
y
8
8x
0
VxW=
Check:
8
8y
_X2 Z
Z
8
8z
- ;!X2Z2
2
= X (x2) + y (3xz2) + z (-2xz)..f
You can add any gradient (Vt) to W without changing its curl, so this answer is far from unique. Some
other solutions:
W
W
W
Probelm
= XZ3X -x2zyj
= (2xyz + XZ3) X+ x2y Zj
= xyzx - ~x2zy + ~X2(y - 3z2) Z.
1.53
V.y
=
=
=
18
1 8
18
22
. 2
---;--(r r cos0) + rsm
A..(-r
0 80 (sm 0 r cos t/J) ---;--r"2 -8 r
rsm 0 8 'f'
1
1
1
cos 0 r2 cos t/J+ ---;--- (-r2 cos 0 cos t/J)
"24r3
cos 0 + ---;--r
rsm 0
' rsm 0
=
2
.
cos 0 sm t/J)
.
rcosO
=--sm 0 [4smO + cost/J - cost/J] = 4rcosO.
R
J (V .y) dr
=
J (4r cos O)r2 sin 0 dr dlJ dt/J = 4 J
=
(R4) (~)(~)=~
0
Surface consists of four parts:
(1) Curved: da
R2 sinO dOdt/Jrj r
=
= R.
7r/2
J y.
da
= R4
y. da
w/2
r3 dr
J cos (}gin 0 dlJ J
0
= (R2 cosO) (R2 sinOdOdt/J) .
7r/2
J0 cos 0 sin 0 dlJ J
0 dt/J= R4 (~)
w/2
(~) = 1r~.
0
dt/J
17
=
(2) Left: da = -r dr d();Pj 4> O. v. da = (r2cos() sin 4» (r dr d()) = O. f v . da = O.
(3) Back: da = r dr d();Pj 4>= 7r/2. v. da = (-r2 cas () sin 4» (r dr d()) = -r3 cos() dr d().
R
/
v. da
(4) Bottom: da=rsindrd4>8j
1r/2
/ /
=0
r3 dr
()=1r/2.
= - (~R4)
cos()d()
0
v.da=
Total: f v . da = 7rR4 /4 + 0 Problem1.54
x
y
1r/2
/ /
=0
v. da
= _~R4.
(r2cos4» (rdrd4».
R
/
(+1)
r3 dr
tR4 + tR4 = 1r:4 .
0
cas 4>d4>
= ~ R4.
.(
z
Vxv=l:z
:y :z I=z(b-a).
So f(Vxv).da=(b-a)7rR2.
ay bx 0
v. dl= (ayx + bxy)' (dxx + dyy + dzz) = aydx + bxdYi X2 + y2 = R2 =>2xdx + 2ydy = 0,
so dy = -(x/y)
dx. So
v. dl = aydx + bx( -x/y)
For the "upper" semicircle,y = VR2 -
/
v. dl
=
-RaR2 - (a+b)x2
/
R
=
=
VR2 - X2
~R2(a
- b) sin-I
2
1
"27rR2(b- a).
dx
X2, so V
=
=
dx
t (ay2 -
bx2) dx.
a (R2-z2)-bz2
. dl =
R -z
..;w:::x'i
dx.
X'
X
{
aR2 sin-I
(-R ) -
(a + b) _-VR2
[
(x/R) -R = ~R2(a - b) (sin-I(-I)
+R 2
2
R2
- X2 + -
- sin-I(+I))
I
2
X
sin-I
= ~R2(a 2
(-R )] }I +R
b)
(-~2 - ~2 )
And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so
fv. dl = 7rR2(b - a). .(
Problem1.55
(1) x = Z
= OJ dx = dz = 0; y:
0 --t 1. v. dl =(y + 3x) dy = y dy.
I
/
0
(2)x=Oi
z=2-2Yi
dz=-2dYi
y:l--tO.
I
v . dl
/
=0
y dy
= ~.
v.dl=(y+3x)dy+6dz=ydy-12dy=(y-12)dy.
0
/
(3) x
= y = OJ dx = dy = OJ z:
v . dl
/
=I
-
12) dy
2 --t O. v. dl
= 6 dz;
(y
= - (~ - 12) = - ~ + 12.
0
/
v. dl = /
2 6dz= -12.
-R
CHAPTER 1. VECTOR ANALYSIS
18
=
-
Total: f v . dl ~ - ~ + 12 12 = [QJ
Meanwhile, Stokes' thereom says f v . dl
(Vxv)x = /y(6)
-
-/z(y + 3x)
= O.
=J (V Xv) . da. Here da = dy dz X, so all we need is
Therefore J(Vxv). da = O. .f
Problem 1.56
Start at the origin.
(1) O=~,
4>=Oj r;O-+1.
v.dl=(rcos20)(dr)=0.
Jv.dl=O.
'If/2
(2) r=l,
O=~j
=
=
1
1
~[3rcosO]i+
rsmu
3cotOi
A
- 69.
(1) Backface:da=-rdrdO4>j
(2) Bottom:
1
A
-[-6r]O+
-[-2rcosOsinO
+ 2rcosOsinO]q,
r
r
(Vxv).da=O.
da = -r sin 0 dr di)
=R f
f(V -v) dr
=
f (~) (r2sin0 dr dOd4» = (£ dr) (J sin0dOd4»= 47rR. } theoremchecks.
R
sin 0 dO d4>= 47rR.
.
So dIvergence
Evidently there is no delta function at the origin.
Vx (rnr)
-(n
. = -laar (r2rn) = -laar (rn+2)= r2
1 + 2)rnH = (n+ 2)rn-I
r2
r2
I
I
(except for n = -2, for which we already know (Eq. 1.99) that the divergence is 47ro3(r».
(2) Geometrically,
it should be zero.
Likewise, the curl in the spherical
coordinates
obviously
gives
Izero. I
To be certain there is no lurking delta function here, we integrate
over a sphere of radius R, using
?
Prob. 1.60(b): If Vx(rni)
= 0, then f(Vxv)dr
= 0 == - §v x da. But v = rni and da =
R2 sin 0 dOd4>i are both in the i directions, so v x da = O. .;
Chapter 2
Electrostatics
Problem
(a)
2.1
I Zero.!
(b) IF
=_
411"1=0
r
1 q~, where r is the distance from center to each numeral. F points toward the missing q.
Explanation: by superposition, this is equivalent to (a), with an extra -q at 6 o'clock-since the force of all
twelve is zero, the net force is that of -q only.
(c) Zero.
I
I
I
(d) 14 1r€0r
1 q~, pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as
a cancellation in pairs of opposite charges (1 o'clock against 7 o'clock; 2 against 8, etc.), with one unpaired q
doing the job, then you'll need a different explanation for (d).
Problem 2.2
E
I
(a) "Horizontal" components cancel. Net vertical field is: Ez = 4;<02~ cos O.
1
Here 1-2
= Z2
+ (~)2 ; cosO = i, so E
I
2qz
= 41r€0(Z2 +
A
(~)2)3/2 z.
x
When z» d you're so far away it just looks like a single charge 2q; the field
should reduce to E = 4 ?ru=+l
.
Let u - cosO, du - -smOdO, { O = 7T => u =- 1 }
11" (z-RcosO)sinO
(2R + z 2 - 2Rzcos 0)3/2 dO.
-11 (2R + zz2-- R~
2 zu )3/2du.
27TR2a
- 47T€O
(
0
J = 27T.
1
.2.-
zu - R
Integral can be done by partial fractions-or
look it up.
'
- -2-
27TR2a (z - R) - (-z - R)
{ Iz - RI
Iz+ RI
) [ z2 ~R2 + Z2- 2Rzu ] -1 - 47T€OZ2
= 4_
:
}.
1 q
2
4
1I"fO z
1 :\, so E = 7T€O Z z.1
For z > R (outside the sphere), Ez = 4 f ~
.. 0
z
_
I
=
For z < R (inside),Ez = 0, so I E 0.1
Problem 2.8
According to Frob. 2.7, all shells interior to the point (Le. at smaller r) contribute as though their charge
were concentrated at the center, while all exterior shells contribute nothing. Therefore:
1 QintE( r ) ---r,
47T€Or2
where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, so
E=-2-Q47T€O r2 r.
Inside the sphere, only that fraction of the total which is interior to the point counts:
Qint
Problem
(a) p
=
=
:R
!7Tr3
3"7T
3Q
r3
= R 3Q,
_
1
2.9
1"0
V. E
r3
1
so E = 4
3Q2f
7T€OR
r
= €O~ tr (r2. kr3) = €o~k(5r4) = 15€okr2.1
=
.
25
(b) By
Gauss'slaw: Qenc= €ofE. da = €o(kR3)(41rR2)=
By direct integration:
Qenc
I
41r€okR5.
= J pdT = JoR(5€okr2)(41rr2dr)
I
= 201r€ok
JoR r4dr
= 41r€okR5.,(
Problem 2.10
Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface
of this larger cube gets the same flux as every other one, so:
/
one
face
The latter is 1..q, by Gauss's law. Therefore
<0
Problem
I
/
2~
/
whole
large
cube
E. da.
E. da = -.!L.
24€o
one
face
2.11
~
GaUSsian surface: Inside: f E. da
Gr
Problem
E. da =
= E(41rr2) = -!;Qenc = 0 => IE = O.I
Gaussian surface: Outside: E(41rr2)
.
= 1..(a41rR2) =>
E
R2
=~
<0
2
r.
}
(As in Frob. 2.7.)
€or,-
2.12
Gaussian surface
j.L'E.
da
= E. 41rr2= 1..Qenc=
1..!
<0
<0 3 1rr3p. So
JE= ~prr.1
Since Qtot = ~1rR2p, E = 4';<0j&r (as in Frob. 2.8).
Problem 2.13
Gaussia
j.L'E.
~/
"
'
= E.
21rs.l = 1..Qenc
= 1..>"l.
So
<0
<0
n ,unace
IE
l
Problem
da
21r€os
= ~Sl
(same as Ex. 2.1).
2.14
.
Gaussian surface j.L'E da = E .41rr2
= 1..Qenc
= 1..
<0
<0 Jp dT
= 1.. } (kf)(f2
= 1..
k 41r Jorr r3dr = 411'k
r4 = 1I'kr4.
<0
<0 4
<0
. 'E = ~1rkr2r.
41r€o
<0
sin B dr dB d4»
CHAPTER 2. ELECTROSTATICS
26
Problem
2.15
(i) Qenc = 0, so IE = 0.1
(H) f E. da = E(47IT2) = 1-Qenc=
1<0
<0 f pdr = 1<0 f rAf2 sinOdfdOd phi
r- a
<0 Ja
= M
"df
(iii) E(411"r2)
~
<0
"'(r
- oj :.1E ~ (
f:
= 4:rok(b
- a), so
=~
df
fO
r2
~
)'.1
lEI
IE=H~)r.1
Problem
r
2.16
(i)
@
(H)
b
a
a
~
)-
Gaussiansurface
. 21I"S'[-
fE.da-E -
-- -8
ps
EE]
- to enc= f-p7rs2[.
0'
A
2fO .
f
-
1Q
= 1-p7ra2[;
f E. da = E. 211"s,[= 1-Qenc
<0
<0
2
G"""'an
,."face
E = pa s.
2fOS
Gaussiansurface
= 0;
f E. da = E. 211"s,[= 1-Qenc
<0
(Hi)
IE=O.I
\
~'-'
lEI
a
Problem
6
S
2.17
On the x z plane E
= 0 by symmetry.
Set up a Gaussian "pillbox" with one face in this plane and the
other at y.
Gaussian pillbox
1-Ayp;
f E. da = E. A = 1-Qenc=
<0
<0
IE = fO
P YY (for Iyl < d).
I
27
<0
Qenc= 1...Adp=}
IE = 100
P dy
(for y
I
> d).
E
I!!!
'0 t""."'
-d
a
y
Problem 2.18
From Prob. 2.12, the field inside the positive sphere is E+ = ~r+, where r+ is the vector from the positive
r -. So the total field is
center to the point in question. Likewise, the field of the negative sphere is - 3...L
<0
E
But (seediagram) r+
Problem 2.19
VxE
- L = d.
I
J ~PdT
= 4:100 Vx
=0
So E
(since Vx
P
= -(r+
3100
r-
- r_)
~-
+
= ~d'i
= 4:100
J [vx
(~) = 0,from
(~~)] pdT
(since p depends on r', not r)
Prob. 1.62).
Problem 2.20
x
y
Z
= k Itx ty
tz 1= k [x(O - 2y) +:9(0 - 3z)+ z(O - x)] f:. 0,
xy 2yz 3zx
so El is an impossible electrostatic field.
(1) VxE1
x
(2)VxE2
y
= k Itx
y2
ty
2xy + z2
z
tz 1=k [x(2z -
2z) + y(O- 0) + z(2y - 2y)]= 0,
2yz
so E2 is a possible electrostatic field.
z
Let's go by the indicated path:
E. dl =(y2 dx + (2xy + z2)dy + 2yz dz)k
(xo, Yo, ZO)
= z = 0; dy = dz = O. E . dl = ky2 dx = O.
Step I: y
III
= Xo, Y : 0 -+ Yo, z =
O. dx = dz = O.
E dl = k(2xy + z2)dy = 2kxoY dy.
Step II: x
fII
Step
III: x
.
E. dl
= 2kxo frioy dy = kxoY5.
= Xo,Y = Yo,z : 0 -+ Zo;dx = dy = O.
I
x
II
y
28
CHAPTER 2. ELECTROSTATICS
E. dl = 2kyzdz = 2kYozdz.
JIll
E. dl = 2Yok Jozoz dz = kyoZ5.
.
V(XO, Yo,ZO)= (xo.Yo.zo)
- J0 E dl = -k(xoY5
Check:
-vv=k[lz(xy2+yz2)
2.21
Problem
V(r) =
-
+ Yoz5), or
I
x+/v(xy2+yz2)y+/;(xy2+yz2)
(
V(r)=-Je:
4;f
0
~ ) df=
< R: V(r) = - J:: (~~)
and for r
_1
41I"fO :;\r.
r
.dl.
{ Inside the sphere (r < R) :
Soforr>R:
I
Z]=k[y2 x+(2xY+Z2) y+2yz i]=E. ,(
Outside the sphere (r > R ) : E =
Je: E
+ yz2).
V(x, y, z) = -k(xy2
q
1
4;f q(to)
0
dfr2
(
=1411"100
2R 3 - R2
)
l
r00
E = 4;fO-J/:srr.
r
=1 411"100
q !,I
J~ (4;fO -J/:sf)df = -dto
[:k-:h
(r22R2)]
.
.
8 1
- -L- 1
1
Wh en r >, R VV -- -LE -- - VV -- -L,(
411"fO
8r ( ;::) r - - 411"fO
f=2"r,so
411"fO
f=2"r.
A
A
A
Whenr < R, VV = -dto2~tr (3 - ~) r = -dto2k (-~)
Problem 2.22
r = --dtobrj
soE = -VV = 4;fO-J/:srr.,(
=
4;fO 2;8 (Prob. 2.13). In this case we cannot set the reference point at 00, since the charge itself
E
extends to 00. Let's set it at s = a. Then
V(s)
= .
r (~2: ) dB =
a
1I"fO
( ).
-- 1 2,Xln ~
411"100
a
(In this form it is clear why a = 00 would be no good-likewise the other "natural" point, a = 0.)
In l.
VV = --L2'xE..
§ = --L2,Xi§
= -E.,(
411"fO 88 ( ( a ))
411"fO 8
2.23
Problem
V(O) = -J::OE.dl
= -J:O(-!o(b;;a»)dr-Jba(-!o(r;;a»)dr-J~(O)dr = -!o(b~a)- -!o(ln(%)+a(~-i))\
!!
( b ) -1+!!
= li. { 1-!!-ln
fO
b
b
I
} = ~ln
100
(~) .
a
2.24
Using Eq. 2.22 and the fields from Prob. 2.16:
Problem
-
V(O) =
=
-
V(b)
Problem
(a)Iv
=
(~)
-
fb
Jo
E. dl
8;I: +
2q
411"100
Jz2
+ (~)2
fa
Jo
E. dl-
J.ab E.
dl = _L2fO Jo
fa S ds - e£..
b ids
2fO J.a s
~ Insl: = 1- ~:: (1 + 2ln (~) ) .1
2.25
~
= -
.
29
(b) V-I-
L
A I (
= 41UO
n x + Yz 2 + x2 ) -L
L
f -L
~ Adx
~
I
£ + YZ2 + £2
,\
=I-In
41r€0
[ -£
= ~ln
+ YZ2 + £2 ]
(L+~
2"fO
I
).
z
x
+ z2)1: =
(c) V = 4:fOJoR~;~:~~ = 4:f021r<7(vr2
~
I
(JR2 + Z2- z)
.1
A
:. E = _8V
8z Z .
In each case, by symmetry ~~ = ~~ = O.
-
(a) E = -~2q
"fO
(b) E
-- - 41rfo
A
1
{ (L+~)
-
- (c) E
1
2qz 2 3/2 Z I(agrees with Prob. 2.2a).
Z = 41r€0(z2+(~))
(-!) (z2+(~)2z 2t2
A
z
1
1
1
2 v'z2+£2
~"fO v' Z2+L 2 {
2
1
(7
1
- (-L+~)
Z
a
I } z = 12 €o [
2
z Z
1
2~
}Z--12£'\_4
-L+~-L-~
( + L2 ) -L
= --2fO { _
21 v'Ri
+z 22z -
}
A
z
1r€0 ZY Z21
A
1-
z
VR2
+ Z2 ]
+ £2
Z
A
I(agrees
.
WIt h
Ex... 2 1)
z I (agrees with Prob. 2.6).
Ifthe right-hand charge in (a) is -q, then V = 0 I, which,naively,suggestsE = - VV = 0, in contradiction
with the answer to Prob. 2.2b. The point is that we only know V on the z axis, and from this we cannot
hope to compute Ex = - ~~ or Ey = - ~~. That was OK in part (a), because we knew from symmetry that
I
Ex
= Ey = O. But
now E points in the x direction,
so knowing V on the z axis is insufficient
Problem 2.26
V(a)
(where
v'2h
1
V(b)
1
= uk In
4100
[
h
=
ch
a21rr
ch,
where ii =
1-
= 41r€0 V2 Jo
= ah.
2€0
Vh2 + 1-2-
V2h1-.
1-
Jh2 + 1-2- V2h1-ch
v'2h
Vh2 + 1-2-
V2h1-+ - h In(2V h2 + 1-2- V2h1-+ 21-- V2h) ]
V2
0
+ v2
':n In(2h+2V2h-V2h) - h - ':n In(2h-V2h)
v2
2+
21ra ~(V2h)
41r€0 V2
E.
= 1-/V2)
rv'2h
2V2€0 [
~
r
-=.--
0
=~
= 2v2€0
1-
l ( )
= _41r€0
21ra
a
a21rr
( )
=~
rv'2h
41r€0 Jo
to determine
V2 = ah
( 2 - V2
:.1V(a) - V(b)
)
4€0
In
(2 + V2)2
(
= ;€~ [1-ln(1
2
)
+ V2)] .
= ah
2€0
In(1 + V2).
.
]
= 2v2€0
~ v2':n[In(2h+V2h)
-In(2h-V2h)
]
30
CHAPTER
Problem
2.27
Cut the cylinder into slabs, as shown in the figure, and
use result of Prob. 2.25c, with z -t x and C1-t P dx:
z+L/2
J (VR2 + X2 - x) dx
z-L/2
= ...£
1. XVR2
+ x2 + R2ln (x + VR2 + X2) 2<0 2 [
1. --8do: 'r
V =!to
= I 4fo{ (z+t
(Note: -
(z
L
X2
x
Z+L/2
Iz-L/2
]
[::;: :::~:::] -2ZL}.
h/ R2+(z+t)2 -(z- t h/R2+(z-t)2 +R21n
2
+ 2") +
(z
L
- 2")
2
2
= -z
L2
~2
2
L2
- zL - "'4 + z - zL + "'4
= -2zL.)
H_H_
E = -vv
= -z-
8V
z
= --E...
R2 +
4€0{/
8z
-
L
z + - 2+
(z
( 2)
+ 2"
L)
2
+L
z+
t + VR2 + (z + t) 2
~,;,O
Problem
{2VR2 + (Z+~)'
[L-JR'+(Z+~)'
+
2VRZ
--
(Z--~)'
Z- t +
~
+ (z - t) 2
'
+ (Z -
t) 2
-2£}
+JR'+(Z-~)']Z
2.28
= 41T<0
~
J1-B..dT.
{
z
Here p is constant, dT = r2 sin e dr de d z. }
}
31
= --1!..211'. 2
411"<0
... V
But P
= Fw'
Problem
\72V
fR lr2dr
r
=~~
(R2
0
so V (z)
z
-
= <0 {lz
.P...
}
z3
3
+
R2-z2
2
}= L (R2 - )
z2
2<0
~) = -d!oR (3 - ~) ;
V (r)
I
3'
=
~
(3-
~)..I
,f
2.29
= 4;<0'V2f(*)dr
= -L- f
411"<0
Problem
z lr2dr
+
z
{f
= 4;<0f p(r') ('V2k)dr (since p is a function air',
(r ) . .,f
p(r') (-4mP (r - r') ] dr = -1...p
<0
not r)
2.30.
(a) Ex. 2.4: Eabove= -2u<0 ft; Ebelow=
--2u<0 ft
(ft always pointing up); Eabove- Ebelow = .£.ft.
,f
<0
Ex. 2.5: At each surface, E = 0 one side and E = .£.
<0 other side, so 6.E = .£..
<0 ,f
Prob. 2.11:
0
(b)
Eout
:
:
'IF"
'. .~ ../
--'
= ~f
R) E = 4;€O ~r
Outside (r
- 100
r3
qp
2
qp R2= qR2 ~=
JE2dT.
(b) W = !f
r2
.
j
Inside (r
2
{R
r4 (r 411"dr)+ Jo
00
1
)1 R
+ R6
"5
2
(R3 )
R
r5
(
r
< R) E = 4;€OWrr.
2
(411"rdr) }
1
1
q2
)I 0 } = 411"100"2
1
(
+ 5R
R
)=
13q2
411"100:5R"/
(c) W = !f { is VE. da + Jv E2dT} , where V is large enough to enclose all the charge, but otherwise
arbitrary. Let's use a sphere of radius a > R. Here V = 4;€O~'
W
= 100
2
- ~ -1 ~
(
)( )
{/
=a
100
= 2"
=
{
1
411"100
r
(~f.(~
r2 sinOdOd4>+ {R E2dT + {a
Jo
q2
14
q2
411"
1
4 2
(411"100)2
~ 11"+ (411"100)2
5R + (411"100)21I"q
1
1
1
1 q2 1
411"100"2~
{ + 5R - ~ + R }
As a --t 00, the contribution
Problem
411"100
r
JR
(-
1 ~
411"100
r
)
1
( -;: ) R }
l
.
1 3 q2
=
411"100:5R ../
from the surface integral (4;€O~)
1
= dqV = dq -411"100 ~,
r
( )
4
goes to zero, while the volume integral
dq
0~~
W
=
q R3
(q = total charge on sphere).
2
= 411"r
1
dW
(q = charge on sphere of radius r).
r3
q = 311"r3 p =
411"r2
dr p = 43qdr
311"R
qr3
411"100R3
1
3q 2
R3 r dr
( ) (
1 3q2 {R
Jo r4dr
= 411"100
R6
}
a
- 1)) picks up the slack.
2.33
dW
2 (411"r2dr)
;:
3q
= R3 r
)=
1 3q2 R5
= 411"100R65
2
dr.
1 3q2 4
411"100
R6 r dr
1
3q2
( ) ../
= 411"100:5R
33
Problem 2.34
=~f
(a) W
E2 dr.
2
W- ~
-
2
E
b
~
(41TfO
) J.a;:2(
1
) 24
=
W2 = 8;fOT,
E1.E2 = (4;fO)2 ~,
(a) aR
I
+ 100fEI
f::O
81rfO
a
El = 4;fO!-rr (r > a), E2 = 4;fO::;ir (r > b). So
(r > b), and hencef E1. E2dr = - (4;fO)2q2go
;!r471T2dr
= - 4::0b'
2
. E2
dr
= 4;R2;
aa = ~;-q
q
(b)V(O)= -
J. b 1
81TfO a;:2
-
2
= 8;fO7'
Wtot= WI + W2
Problem 2.35
r -
zero elsewhere.
-IL (~- ~b ) .
- L
2d
1rr
2
(b) WI
< r < b),
4;fO!-r (a
= ~q2 (~+! -~) = ~ (~-!).,(
ab= 4;b2'
q
I
E. dl = - f::O(4;fO!-r)dr - fba(O)dr - faR (4;fO !-r)dr - f~(O)dr
(c) ab ~ 0 (the charge "drains off"); V(O) = - f::O(O)dr- faR(~!-r
Problem 2.36
I
I
(a ) aa
I
= -~; qa
I
(b) Eout
II ab
= -4 1
aR
I
~ (t ~ -~)
~ (~ - ~)
)dr - f~(O)dr =
+
I
= qa41rR2
+ qb .
qa + qb
r, where r = vector from center of large sphere.
1rfO~ r
1
A
I
Eb = _
1 q~rb, Iwhere ra (rb) is the vector from center of cavity a (b).
41rfO
rb
qa A
-ra,
(c) lEa = 41rfOr~
(d) IZera.
= -4;b2;
qb
I
=
I
(e) aR changes (but not aa or ab); Eoutside changes (but not Ea or Eb); force on qa and qb still zero.
Problem 2.37
Betweenthe plates, E = 0; outside the plates E = a / 100 = Q / foA. So
p
= 100E2 = 100~
2
-1Q2l
-~
2 f5A2
Problem 2.38
Inside, E = 0; outside, E = -41TfOr
1 ~r; so
Eave = ~ 4;fO
Fz =
-
J fzda
~ r;
fz
cME
= a(Eave)z; a = 41T~2'
= f( 41T~2) ~ (4;fO ~ ) cas
() R2 sin () d() dq;
_
- 1 9... 2 1 . 2
1T/2 1
2
1 -9- 2 f1T/2.
( ) ( 2"sm ()) 0 - 211:fO
(9...
) ( R ) 21rJo sm ()cas () d () - 1TfO4R
2<6 41T
4R
1
I
Q2
321rR2fO .
.1
.1
34
CHAPTER 2. ELECTROSTATICS
Problem 2.39
Say the charge on the inner cylinder is Q, for a length L. The field is given by Gauss's law:
_
E. da = E . 21TS. L = .!..Qenc
= .!..Q
=> E = -92 "'EO L 81 s. Potential difference between the Cy linders is
EO
EO
I
-
V(b)
V(a)
l
=-
b
a
Problem
~ ==~...t~)' so capacitance
1
b
b
Q
()
-dB= --In
21T€oL a S
- .
21T€oL
- V(b) = ~
As set up here, a is at the higher potential, so V = V(a)
C=
l
Q
E.d1= --
a
In (~).
per unit length is In21T(~O
Ii) .1
I
n "
2.40
(a) W
= (force)x(distance) = (pressure)
(b) W
= (energy per unit
= ~ E2 A€.I
x (area) x (distance)
I
volume)x (decrease in volume) = (€O~2) (A€). Same as (a), confirming that the
energy lost is equal to the work done.
Problem 2.41
From Prob. 2.4, the field at height z above the center of a square loop (side a) is
1
4Aaz
E--- 41T€o(z2 + ~2)
J
Z2
Z.
+
-~- da
2
a22
- a--..
Here A -+ ad2a (see figure), and we integrate over a from 0 to ii:
E
l
= ~2az
41T€O
a
.
1
41T€O
du
(u
J
a2
a -+ 00 (infinite Plane ) : E
»
.
1
{ tan-l
E
z
. Let u = "4' so a da = 2 duo
-2
2
= 1T:
(
az
=-
+ z2)v2u + z2
1T€O
2
-tan-l
[Z
v2u
(
2a
tan -1
1T€O[
= k"'EO [tan-l
)
g
- tan-l(l)
Z
)] 0
1+
-
2
};
1T
2z2
- - Z.
A
4]
1!:. = ...fL. ,f
= k"'EO (1!:.2
4)
2EO
(oo) - 1!:.
4]
= tan-l vI +x - 7' and
expand as a Taylor series:
1
f(x)
0.2/4
+ Z2
+ z2
2Z
=-
a (point charge): Let f(x)
-a+aa-
+ a;) /z2 + a22
a /4
= -4az
a2
alia
0 (z2
-~- da
2
= f(O) + x!,(O) + "2x21"(0)+ ...
35
= tan
Heref( 0)
-1
1<
(1) - 4"
1<
= 4"-
- O. f ' ( ) x -
1<
1'(0) = 1.4'
1
1 v'I+X
1- -- 2(2+x).,!1+x'
1
SO
1+(1+x)2
4"-,
SO
1
f(x) = -x
4 + ( )x2 + ( )x3 + . . .
.
a2
Thus (Since
W = x«
1 a2
20"
I)E,
(
~ 1rlo 4W
1
1
q
/
zr = 41ds
d8
s d
s ds
4> only,
d8
1 ~
s ds
= C1, ¥ d4>2
( )
1 d2<1>
( )+ ¥
Sincethe first term involves s only, and the second
S ds
(r :::; R),
+ ... ,
d4>2
=
O.
each is a constant:
= C2, with C1+ C2 = O.
(r
2: R).
}
56
CHAPTER 3. SPECIAL TECHNIQUES
Now C2 must be negative (else we get exponentials for ,which do not return to their original value-as
geometrically they must- when
= -k.
Then
d=> = A cos k«jJ
+ 27r) = «jJ),
k must be an integer: k = 0,1,2,3,...
but k = 0 must be included, since = A (a constant) is OK).
s~
ds
( )=
s dS
ds
B sm k n = :f:.k.
Evidently the general solution is S(s) = ask + Ds-k, unless k = 0, in which case we have only one solution
to a second-order equation-namely,
S = constant. So we must treat k = 0 separately. One solution is a
constant-but
what's the other? Go back to the differential equation for S, and put in k = 0:
s-
d
ds
dB
ds
( )
s-
dS
= 0 => s-dS
=
constant = C => ds
ds
C
ds
= -s =>dS = C- s =>S = Clns + D (another constant).
So the second solution in this case is In s. [How about ?That too reduces to a single solution, = A, in the
= 0 into the equation:
case k = O. What's the second solution here? Well, putting k
d2
d
= 0 => d = B Cl = -alR2;
(ii)--+ al = -Eo. Therefore
(
V(s,R),
r
A
I
3€or2P.r=
I
...!...P.r
3€0
I
R3 P cos B
3€or2 '
I
( > R)
r
,
So V(r,B) =
- p{ ~
r, (r < R). }
411"€0(4/3)wR'p
R3
~
I" 'owoo,
1
(r < R).
4.13
Think of it as two cylinders of opposite uniform charge density :!:p. Inside, the field at a distance s from
the axis of a uniformly charge cylinder is given by Gauss's law: E211"se= -:OP1l"S2e
:::} E = (p/2€0)s. For
two such cylinders, one plus and one minus, the net field (inside) is E = E+ + E- = (p/2fO) (s+ - s_). But
s+ - s-
= -d,
so E
= l-pd/(2€0),where d is the vector from the negative axis to positive axis. In this case
I
the total dipole moment of a chunk of length e is P (1I"a2e)= (p7ra2e) d. So pd = P, and IE = -P /(2€0), I for
s < a.
77
Outside, Gauss's law gives E27r8£ = .1...p7ra2£
:::}E = 1!!£.
2a2!,
<0
<0 s for one cylinder. For the combination, E =
a2 !:t. - iL
, where
E+ + E- = 1!!£.
2<0
s:!:
(s+
=
-S:i: =
8?t
)
s-
d
S T -j
2
( d)(
(
ST-
2
1
=
8+ - L
(
s+
8-
)
82
= ~
82
a-
S :f: S--;2
[(
-1
12
2
)
8 +-Ts.d
4
(s . d)
d
~-
1
d
ST82
2
( )(
.
s+s~
a2
E(s)
-1
)
~-
1
d
ST82
2
( )(
1:f:-
s.d
82
)
(keepmg only 1st order terms in d).
1
= -2fO 82
=~
+ ~
2
82
s.d
82
.
)
- ~ - s-s~
) (
T "2
IT-
2 )]
82
[2(P. 8) § -
for 8
P] '
(2S(S.d)
82
-d
82
).
> a.
Problem 4.14
Total charge on the dielectric is Qtot = is O"b
da + Iv Pbdr = is P . da - Iv V.p dr. But the divergence
theorem says is p. da = Iv V.p dr, so Qenc = O. qed
Problem 4.15
(a)Pb=-v,p=-~~
r2 or
( r ) =-~; r2
r2~
O"b=P.ii=
+P.~=k/b
(atr=b),
(at r = a). }
{ -P . r = -k/a
Gauss's law:::} E = 4:<0Q;~cr. For r < a, Qenc = 0, so IE = 0.1 For r > b, Qenc = 0 (Prob. 4.14), so E
I
Fora < r < b, Qenc = (~k) (47ra2)+ I: (~) 47rf2dr
(b) fD.da
= Qfenc
= -47rka -
47rk(r- a)
= -47rkr;
= O:::}D = 0 everywhere. D = foE + P = O:::}E = (-l/fo)P,
IE = 0 (for r < a and r > b);j IE
Problem 4.16
= -(k/fOr)
r (for a < r < b).
so E
I
= -(k/for)
= 0.1
r.1
so
I
(a) Same as Eo minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14)
is-P/3fO. SoIE = Eo + ~p.1
D = foE= foEo+ ~P = Do- P + ~P, so D = Do - ~p.1
I
(b) Same as Eo minus the field of:f: charges at the two ends of the "needle"-but
away,so
!
E
= Eo.J
D
= foE = foEo = Do -
P,
SO
I
D
= Do -
these are small, and far
P .1
(c) Same as Eo minus the field of a parallel-plate capacitor with upper plate at 0" = P. The latter is
-(l/fo)P, so IE = Eo + !op.1 D = foE = foEo + P, so ID = Do.!
CHAPTER 4. ELECTROSTATIC
78
Problem
4.17
P
E
(uniform)
Problem
D
(same as E outside, but lines
continuous, since V.D = 0)
(field of two circular plates)
4.18
J
=
(a) Apply D . da
Q/enc to the gaussian surface shown. DA
metal plate.) This is true in both slabs; D points down.
~
(b) D
I
E2
FIELDS IN MATTER
= fE
::}
E
= a/fl
= aA
::}
I
D = a.1 (Note:D = 0 insidethe
2+u
in slab 1, E = a/f2 in slab 2. But f = tofT, so fl = 2fo; f2 = ~fO' lEI = a/2fo,
= 2a/3fo.1
(c) P = foXeE, so P = foXed/(fOfr) = (Xe/fr)a; Xe = fr -I::}
P = (1- f;l)a.
IPl=a/2,IIP2=a/3.1
(d) V = E1a + E2a = (aa/6fo)(3 + 4) = 17aa/6fo,!
ab = +P1 at bottom of slab (1)
(e)Pb=O;
ab = + P2 at bottom of slab (2) = a /3,
ab = -P2 at top of slab (2) = -a /3.
= a/2,
ab= -PI at top of slab (1) = -a/2;
total surface charge above: a - (a/2) = a/2,
(f) In slab 1: { total surface charge below: (a/2) - (a/3) + (a/3) - a = -a/2,
total surface charge above: a
-
+ (a/2) - (a/3) = 2a/3,
(a/2)
In slab 2: { total surface charge below:(a/3)
-
a
= -2a/3,
E -
~
..(
-
2a
..(
} ==> 1 - 2fO'
E
} ==>
]+u
-u/2
(!)
+u/2
-u/3
@
+u/3
]-u
Problem
4.19
With no dielectric, Co = Afo/d (Eq. 2.54).
In configuration (a), with +a on upper plate, -a on lower, D = a between the plates.
E = a/ fO(in air) and E = a / f (in dielectric).
2
- ~ - ~
Ca - V - d
1+1/(r
==> Ca
Co - 1 2fr
+ fr .
(
)
I
So V
= {;; ~ + 7~ = 2~~ (1 + ~)
.
I
In configuration (b), with potential difference V: E
= V /d, so a = foE = fO V / d (in air).
2 - 3fO'
I
79
P = EOXeE= EOXeV/d (in dielectric), so O'b= -EOXeV/d (at top surface of dielectric).
= EoV/d= O'f +
= O'f-
EoXeV/d,so O'f= EoV(l + Xe)/d = EOErV/d(on top plate above dielectric).
V
V
A
AEO
1 + Er
=? Cb = V = V
+ O'f2"
= 2V Eod + EOdEr = d ~
. Co
Cb = ~.
1 + Er
~l+ 0. So Cb > Ca.]
[Which is greater?. ~Co - ~Co = l+ D41fr2
= p!1fr3 => D = lpr => E = (pr/3E) r, for r < R; D41fr2 = p!1fR3 => D =
pR3/3r2=>E = (pR3/3Eor2)r, for r > R.
fO
V
pR3 1
= - } E. dl =
00
R
P fo
3EO -:;. 00 - 3E
1
pR2
}R
=
rdr
3EO
+
3E
=
""2
1
pR2
P R2
I
3EO
( + ).
1
2Er
Problem 4.21
Let Q be the charge on a length £ of the inner conductor.
f D . da
V
=
D21fs£ = Q
a
-
-
f
-
Q
V£
D
b
E.dl-
C
C
£
=>
=
I
-
In(b/a)
_
Q
= 21fS~
0;
I(
a
-Q
21fEO£
Q
= 2-1fEOS~
E
0
C
(a
Q
< s < b), E = -21fES~0 (b < r < c).
) l( )
dS
-+
S
b
+ 21fEO
(l/Er) In(cjb)"
-Q
dS
21fd
S
---
b
Q
-
21fEO£
[
In -
(a )
EO +-In
(
C
E
b')]
I
Problem 4.22
Same method as Ex. 4.7: solve Laplace's equation for V;n(s, tj)) (s < a) and Vout(s, tj)) (s > a), subject to
the boundary conditions
x
(i)
V;n
=
Vout
(ii)
E8~n
=
EO8~~ut
{ (Hi) Vout
-+
- Eos cos tj)
at s = a,
at s = a,
for S
Eot
» a.
y
FromProb. 3.23 (invoking boundary condition (Hi)):
00
V;n(s, tj)) =
2::>k(ak cosktj)+ bksinktj)), Vout(s, tj))= -Eoscostj) +
k=l
00
L
k=l
s-k(Ck
cosk + dk sink
+ dk sink b);
(a < r < b);
Vrned(r,O)= L(Alrl+r~l)1't(cosO),
= 0,
(r < a).
{ ~n(r,O)
Boundary Conditions:
{
(i) Vout
(ii) f.~ or
(Hi) Vrned
= Vrned,
=
=
f.0~ or '
0,
(r = b)j
(r = b)j
(r = a).
81
" BI
(i)
~
.-Eob cos0 + ~
(ii)
~
€r ~
(Hi)
~
A al + hi
"
= 0
al+l
Alb
hi
)
+ bl+l
~(cosO)j
]
lAlb -
I
I
(
hi
"BI
- (l + 1)bl+2 ~(cosO) = -Eo cosO- ~(l
I 1
[
"
bl+l ~(cosO) = ~
h =
~
A
-a2l+1
I
+ 1) bl+2 PI (cos 0);
I.
Fori =f.1 :
(i)
(ii)
BI
=
bl+1
(
A bl - a21+1AI
I
bl+l
1-1
[
a
2l+1
AI
bl+2
+ (l + 1)
€r lAlb
)
~
B = A
I
I
(b2l+1 -
BI
- -(l + 1)bl+2
a21+1
-
]
.
),
Bl -
~
l
-€rAI
[(
l
21+1
)
+1 b
21+1
]
+a
-~
Al
- Bl - O.
Fori = 1 :
B1
(i)
..
(ll
)
3
So -3Eob
€r
[2 (b
Vrned(r,O) =
E(r,O)
=
a3 At
(
+ 2~
Al
3
= Al
a3Al
-Eob+b2=A1b-~
3
- a
)
+ €r
(
3
)
BI-Eob
3
=A12
B1
(b
3
-a
);
3
3
3
= -Eo - 2b3 ~ -2B1 - Eob = €rAI (b + 2a ) .
3
+ 2a )] ;
b
3
~
-3Eo
2[1- (afb)3] + €r[1 + 2(afb)3]
-3Eo
Al = 2[1 - (afb)3] + €r[1 + 2(afb)3].
a3
( -;:2 )
r
cosO,
-VVrned = 12[1- (afb)3] ~E:r[1 + 2(afb)3]
{ (1+ 2r~3)cosOi'-
(1- ;:) sinoo}.
Problem 4.25
There are four charges involved: (i) q, (ii) polarization charge surrounding q, (Hi) surface charge (CTb)on
the top surface of the lower dielectric, (iv) surface charge (CT~)on the lower surface of the upper dielectric.
In view of Eq. 4.39, the bound charge (ii) is qp = -q(X~f(1 + X~), so the total (point) charge at (0,0, d) is
qt = q.+ qp = qf(1 + X~) = qf€~. As in Ex. 4.8,
(a) CTb
(b)
I
CTb
=
=
-1
qdf€~
3
€oXe _
[ 47I"€o(r2 + cF):2
I
_
€oXe 4
[
1
CTb
-_
2 -
€o
CT~
~
2€o]
(here CTb
= P.n = +Pz = €oXeEz)j
CTb - _
CT~
- _
2
2
+ cF)2 €o €o]
I
qdf€~
( here CTb= -Pz = -€OXeEz )
3
7I"€o(r2
Solvefor CTb,CT~:first divide by Xe and X~ (respectively) and subtract:
CT~
CTb
1
qdf€~
X~ - Xe = 271"
(r2 + cF)~
I
~
CTb=
I
CTb
1
qdf€~
Xe [ Xe + 271"
(r2 + cF)~ ] .
.
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
82
Plug this into (a) and solve for O"b,using €~ = 1 + X~:
' O"b (
'
qd/€~
)
(
)
";!Xel+Xe--Xe+Xe,soO"b=41T (r2 + d2)2
2
O"b -
-1
-
, O"b-
,
-I
qd
{ 41T (r2
Xe
1
-1
Xe
3
, ;
(
[
41T(r2 + d2)2 1 + Xe + Xe)/]2
1
qd
,
qd/€~
1
qd
€rX~/€~
+ d2) ~ [1 + (Xe + X~)/2] + 21T(r2 + d2) ~ } ' so O"b= 41T(r2 + d2) ~ [1 + (Xe + X~)/2]'
X x. ,
~- + )/ 2 (which vanishes, as it should, when
The total bound surface charge is O"t O"b + O"~ = 417r
qd ~ E'~ l + ( ",
(r2+d2)
x.
X~ = Xe)' The total bound charge is (compare Eq. 4.51):
=
qt
V(r)
=
(X~ - Xe )q
€~
_
2€~ [1 + (Xe + X~)/2] -
I
(
-
€r
€~ + €r
)
q
€~' I and hence
I
=~
q/€~
+
qt
(for z
41T€0{ ";X2 + y2 + (z - d)2
";x2 + y2 + (z + d)2 }
\
> 0).
I
I
I
q
since €r
~ + qt
Meanwhile,
Problem
= 7"r
€~
-
€r
[ 1 + €'r + €r ]
-
~
€~
+ €r '
V(r)
=
41T€0
I
I
1
";X2
[2q/(€~ + €r)]
(for z < 0).
+ y2 + (z d)2
\
4.26
From Ex. 4.5:
0,
D
w = -1
2
I
=
(r
Q
O'Q
{ 41Tr2r,
(r < a)
(r > a) } ,
I
1 Q2
D.EdT=--41T
-
2 (41T)2
{€
l
b
1 1
E=
--r2dr+-
a r2 r2
-r,
41Ttt
(a < r < b)
~
00
1
€o b
1
=-Q2
-dr
r2
}
1
1
Q2
1 Xe
= 81T€0{ (1 + Xe) ( ~ - b) + b } = 81T€0(1
+ Xe) ( ~ + b ) .
Q2
1
1
.
(r > b) }
{ ~r,
41T€or
1
< a)
~
I
-1
- -
81T { €
( )
r
I
b
a
1
+--
€o
00
-1
( )
r
1
b
}
83
Problem 4.27
Using Eq. 4.55: W
= !f J E2 dr.
From Ex. 4.2 and Eq. 3.103,
-I
E =
{
Wr R) }
( ) =
.
.
(
)
6"
4cos
B+
sm
B
r
smBdrdBdR = 2
=
(r < R)
3102
P z,
R
~(2cosBf+sinBO),
3for
2
211'P2R3.
27 fa
1
2
2
2
r
l
1r
00
1
(R3 p)2
B) sin BdB R 4"
1 8fa 211' a (1 + 3 COS2
r dr
11'(R3P)2
9100
1
(~ ) =
3R3
11'(R3 p)2
=-
(-COSB-COS3B)I~(-3~3)1:
9100
411'R3p2.
27100
211'R3 p2
Wtot =
9100
This is the correct electrostatic energy of the configuration, but it is not the "total work necessary to assemble
the system," because it leaves out the mechanical energy involved i~ polarizing the molecules.
Using Eq. 4.58: W
= ~ J D.E dr.
For r <: R, D = foE, so this contribution is the same as before.
Forr < R, D = foE + P = -!p + P = jp = -2fOE, so ~D.E = -2!fE2, and this contributionis
=-
now(-2) (~~ p:~3)
~~ R::2, exactly cancelling the exterior term. Conclusion: Wtot = 0.1 This is not
surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem
there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as
the "work necessary to assemble the configuration" -the latter would depend entirely on how you assemble it.
Problem 4.28
I
First find the capacitance, as a function of h:
Air part: E
= -bL
==> V
41rfOS
= ..1L
In(bJa)
41rfO'
>..
==>
2>" ==>V = 41rf
2>"In(bJa) ' }
OilPart: D = ~41rs ==>E = 41rfS
Q = >..'h + >"(f - h)
= fr>"h -
>"h+ >"f
= >..[(fr -
Q
C
Th
t
e ne upwar
d
l'
..
LOrCeIS gIven
y
The gravitational force down is F
q.
2>"ln(bJa)
4 64 ' F .
.
= -j>'"10
= >"(Xeh+ f),
>"(Xeh+ f)
=V =
b E
l)h + f)
-fa
10
>..'
where f is the total height.
(Xeh + f)
= 211'100In(bJa)
.
411'100
- 1 V 2 21rfOXe
dh - 2" In(bfa) .
1 V 2 dC
- 2"
= mg = p11'(b2 -
= fa->..= lOr>".
a2)gh.
h =
}
I
V2
fOXe
p(b2 - a2)g In(bJa)
.
84
CHAPTER 4. ELECTROSTATIC
Problem
4.29
(a) Eq. 4.5
:::}
.
.
FIELDS IN MATTER
F2
8
= (P2 . V)
PI
Eq. 3.1O3:::} EI = 4m:or
~
F2
= P2~uy
EI
PI
A
(Ed;
~y
?jr
A
() = -- 41rfoY3 z. Therefore
= _ -PIP2
d
4
d
1
1rfO [ y
z
41rfoY4 Z, or
( Y )] z = _
3PIP2
A
3PIP2
A
A
F2 = ~z
41rfor
3"
I
I(upward).
.
I
To calculate F I, put P2 at the origin, pointing in the z direction; then PI
is at -r z, and it points in the -y direction.
y
=
So FI
(PI'
V) E2
=
I
-PI 8:2y x-y-O,
- - z--r
- ; we need E2 as a function of x, y, and z.
1
From
E q. 3.104:
E2
ll
-3"
41rfOr
=
P2 . r = -P2Y'
~
E2 =
8y
8E2
8y
1
(0,0)
= ~
~~
[
?
r-
h
- P] , were
r = xx + yy + ZZ, P2 = -P2Y, and hence
A
-3Y(XX + yy + zz) + (x2 +y2 + Z2)y
41rfO [
8E2
3(P2' r)r
(x2
+ y2 + z2)5/2
A
A
-3XYX + (x2 - 2y2 + z2)y - 3YZZ
=~
41rfO [
]
A
(x2 + y2 + z2)5/2
]
-~~2Y[-3XYX + (x2- 2y2+ Z2)y - 3yzz] + ~(-3xx - 4yy - 3ZZ) ;
~
{ 2~
}
= ~ -3z
41rfO r5
z;
FI
Z =
= -PI ~41rfO 3r
r5
(
)
- 3PIP2 Z.
41rfor4
These results are consistent with Newton's third law: FI = -F2.
(b) From page 165, N2 = (P2 x EI) + (r x F2). The first term was calculated in Frob. 4.5; the second we
get from (a), using r = r y:
P2 X EI
= 41rfor3
PIP2
(
A
-X)j
r x F2
= (ry-) x
-
(_ ) = _
3PIP2
41rfor 4
Z
3PIP2
N2
41rfor 3 x; so
A
I
_
2PIP2
= 41rfor 3 X.
A
This is equal and opposite to the torque on PI due to P2, with respect to the center of PI (see Frob. 4.5).
Problem 4.30
Net force is to the right (see diagram). Note that the field lines must bulge to the right, as shown, because
E is perpendicular to the surface of each conductor.
I
I
E
85
Problem 4.31
P
= kr = k(xx
+ yy + zz) =? Pb= -V.p
Total volume bound charge:
= P.il.
I
= -k(l + 1 + 1) = 1-3k.1
= -3ka3.1
Qvol
At top surface, il = z, z = a/2j so O"b
= ka/2. Clearly,
(Jb
T~tal surface bound charge:
I
I
O"b
= ka/21
on all six surfaces.
Qsurf= 6(ka/2)a2 = 3ka3.1 Total bound charge is zero.
if
Problem 4.32
f D.da = Qfonc::}D = 4- q
~
2 rj
7rr
Pb= -V.p
=
47r(~~Xe) (V.
E
1
q
f
4 7rfO(1 + Xe) 2";
r
P
= -D
=
10
)
qXe
= fOXeE =
=
~ = -q 1 ~eXe 83(r)
=
(Eq. 1.99)j O"b P.f
f
4 7r(1 + Xe) 2"'
r
47r(1~X~e)R2;
Qsurf= (Jb(47rR2) = q 1 +XeXe .1 The compensating negative charge is at the center:
I
j
= - l qXe
PbdT
j
83(r)dT
+ Xe
= -q-1 Xe .
+ Xe
Problem 4.33
Ell is continuous (Eq. 4.29); Dl. is continuous (Eq. 4.26, with O"f = 0). So EXl =-EX2' DYl = DY2 ::}
E1EYI
= f2EY2'and hence
tan02 = EX2/EY2= EYl = E2. Qed
EXl / EYl
tan 01
EY2
El
If 1 is air and 2 is dielectric, tan O2/ tan 01 = E2/ EO> 1, and the field lines bend away from the normal. This is
the opposite of light rays, so a convex "lens" would defocus the field lines.
Problem 4.34
In view of Eq. 4.39, the net dipole moment
at the center is pi
=P -
=
1~~e p
I';Xo
= tp.
p
We want the
potentialproduced by pi (at the center) and O"b(at R). Use separation of variables:
B
00
Outside: V(r,O)=
L rl:l
1=0
_
.
lnstde:
V(r,O)= 4
1
Pz(cosO)
1 pcosO
~
(Eq.3.72)
7rEO Err
R~L = AIRI,
av
8V
8r R+
I
ar Ri
= -
or
B1 1
P
R2 - 47rEoErR2 + AIR,
""'
L)l
1
= --p.
BI
EO
or
_
1
BI
~
= R2/HAI
2pcosO
Er R3
av
= --1 (EoXeE.r) = Xe _
a
~
EO
)
(l ¥' 1)
.
p3
B1 - 47rfOf~+ AIR
+ 1)R l+2Pz(cosO)+ 47rEO
r
(Eqs.3.66,3.102)
1=0
V continuous at R ::}
{
.
00
+ LAlrIPI(cosO)
r
I
R-
-
""'
LlAIR
}
I 1
-
1
PI(cosO)
I 2pcosO
= Xe { --47rEO
lOrR3
= --O"b
EO
""'
+ ~lAIR
I 1
-
PI(cosO) }
.
86
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
-(I
+ 1) R~~2 -
Forl=l:
lAIRI-I
BI
1
-2-+---AI=Xe
R3
411"fO
= XelAIRI-I (l ¥- 1); or - (2l + l)AIRI-I
1
---+A1
2p
(
frR3
2P
-~
- AIR3 + ~
=> Al =
~
2XeP = ~ 2(fr -l)p;
411"foR3fr(3 + Xe) 411"foRafr(fr + 2)
411"fOfr
- AIR3 = _~XeP
411"fOfr
2
(
cosO +
411"fo frr2
p
P
-BI+---=---+Xe-;
AIR3
2
BI =
COSO
411"for2
~
2
411"fOfr [
)(~ + 2)
fr
(r
=>Al = 0 (£ ¥- 1).
AIR3
411"fOfr
+ XeAIR3 =>
411"fOfr
V(r,O)=
Meanwhile, for r::; R, V(r, 0) = ~P
)
411"fOfrR3
= XelAIRI-I
2
1
XeP
AIR3
411"fo fr
2
(3 + Xe) = ~XeP.
411"fOfr
1 + 2(fr - 1)
(fr + 2) ]
= ~~.
411"fOfrfr + 2
~ R).
1 prcos(} 2(fr -1)
R3
freEr + 2)
411"fO
fr - 1
P cos ()
= I 41I"for 2 fr [ 1 + 2 -fr + 2
(
r3
) ] (r::;
R).
R3
Problem 4.35
Given two solutions, VI (and EI = -VVI, DI = fEd and V2 (E2 = -VV2, D2 = fE2), define V3 ==V2- VI
(E3 = E2 - EI' Da = D2 - DI).
Iv V.(VaD3) dr = Is V3Da' da = 0, (Va= ° on S), so I(VV3) . D3 dr + I V3(V.D3) dr = 0.
= V.D2 - V.DI = PI- PI = 0, and VV3 = VV2 - VVI = -E2 + EI = -E3, so IE3' D3 dr = O.
But D3 = D2 - DI = fE2 - EEl = fE3, so f(E3)2 dr = 0. But f > 0, so E3 = 0, so V2 - VI = constant. But
at surface, V2 = VI, so V2 = VI everywhere. qed
Problem 4.36
But V.D3
I
(a) Proposed potential:
I
VCr) = Vo~.1 If so, then IE = - VV = VO~ f, in which case P
I
= fOXe Vo ~ f,
;
in the region z < 0. (P = o for z > 0, of course.) Then CTb= fOXeVo (f.ft) = 1- fO~ Vo .1 (Note: ft points out
of dielectric => ft = -f.) This CTbis on the surface at r = R. The flat surface z = 0 carries no bound charge,
since ft = z 1..f. Nor is there any volume bound charge (Eq. 4.39). If V is to have the required spherical
symmetry, the net charge must be uniform:
CTtot411"R2
=
Qtot = 411"fORVo(since Vo = Qtot/411"foR), so CTtot= foVo/R.
CT
I
- {
=
Therefore
(fOVo/ R), on northern hemisphere
(fOVol R)(l + Xe), on southern hemisphere' }
=
=
(b) By construction,
CTtot CTb+CTI
foVo/R is uniform (on the northern hemisphere CTb= 0, CTI foVo/R;
on the southern hemisphere CTb= -foXeVo/R, so CTI= fVo/R). The potential of a uniformly charged sphere is
Vo =
Qtot
411"for
= CTtot(411"R2)
= fOVo
R2 = VoR.
411"fOr
R for
r
(c) Since everything is consistent, and the boundary conditions (V
Prob. 4.35 guarantees that this is the solution.
./
= Vo at
r = R, V -+ 0 at 00) are met,
87
(d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to ft,
so we'd get bound charge on this surface, spoiling the symmetry.
Problem 4.37
~
Eext = 27r€08 8. Since the sphere is tiny, this is essentially constant,
F
J (1 + Xe!3
-
-Xe
=
1 + Xe/3
) (~
2
~ ~
-
) ( ) (
(~ ) 2-~7rR38 - - (~ )
€oXe
27r€08
47r2€0
d8
833
Xe / 3 Eext (Ex. 4.7).
and hence P = 1 +€oXe
8dr
27r€08
-
€oXe
1 + Xe/3
~
)( ) (! ) ( ) J
-1
2no
8
8
dr
82
)..2R3 8
7r€083 .
3 + Xe
Problem 4.38
The density of atoms is N = (4/3)7rR3'The macroscopic field E is Eself + Eelse, where Eself is the average
fieldover the sphere due to the atom itself.
p
= o:Eelse
=} P = No:Eelse.
[Actually,it is the field at the center, not the average over the sphere, that belongs here, but the two are in
fact equal, as we found in Prob. 3.41d.] Now
1 p
Eself = - 47r€0R3
(Eq. 3.105), so
1
E=-
0:
47r€0 R3 Eelse
+ Eelse
=
0:
(
1-
47r€oR3
)
Eelse
=
(
1-
NO:
3€0
)
€r
-
Eelse.
So
P=
No:
(1 - N o:/3€0) E = €oXeE,
and hence
No:/€o
Xe = (1 - No:/3€0)'
Solving for a:
Xe
No:
No:
- -3 Xe= €o
€o
No:
=} -
€o
Xe
(1+ _3 ) = Xe,
or
€o
Xe
3€0
a = N (1 + Xe!3) = Ii
Xe
(3 + Xe' But Xe = €r - 1, so 0:=
3€0
1
Ii (z+2 ) .
qed
Problem 4.39
Foran ideal gas, N = Avagadro's number/22.4liters = (6.02 x 1023)/(22.4x 10-3)
= 2.7 X 1025.
(2.7x 1O25)(47r€0x 1O-3O),8/€0= 3.4 X 10-4,8, where,8 is the number listed in Table 4.1.
(3 = 0.667, No:/€o = (3.4 x 10-4)(0.67) = 2.3 x 10-4, Xe = 2.5 X 10-4
He: (3= 0.205, No:/€o = (3.4 x 10-4)(0.21) = 7.1 x 10-5, Xe = 6.5 X 10-5
No:/€o
=
H:
Ne: (3 = 0.396, No:/€o= (3.4x 10-4)(0.40)= 1.4x 10-4, Xe= 1.3X10-4
Ar: (3 = 1.64, N 0:/ €o = (3.4 X 10-4)(1.64) = 5.6 x 10-4, Xe = 5.2 X 10-4
..
agreementISqUItegood.
}
CHAPTER 4. ELECTROSTATIC
88
Problem
4.40
(a)
(u)
-
=
J~:E ue-u/kT du - (kT)2e-u/kT
J-pE
PE e-u/kT
du
-
= N(p);
p
- l]I~~E
pE
-pE
l
[e-pE/kT - ePE/kT] + [(pE/kT)e-pE/kT
kT
e-pE/kT
kT - pE
[ epE/kT
A
= (pcosO)E = (p.
-
+ (pE/kT)ePE/kT]
epE/kT
}
PE
+ e-PE/kT
ePE/kT
=
[-(u/kT)
-kTe-U/kT
{
P
FIELDS IN MATTER
- e-pE/kT
E)(E/E)
]
= kT
- pE coth
-
( kT ) .
-(u)
= -(u)(E/E);
P
= Np
=
pE
I
Np { coth
PE
kT
(
kT
) - pE } .
Asx --+0, y = (~+ f - ~; +... )-~ = f-~; +... ~
0, so the graph starts at the origin, with an initial slope of 1/3. As x --+00, y --+coth(oo) = 1, so the graph
Lety ==P/Np,
x ==pE/kT. Theny
goes asymptotically to y
= cothx-1/x.
= 1 (see Figure).
.E...
np'
11
"""""""""""""'"
.
pe/kT
(b) For small x, y :::::::
kx, so;; p :::::::
-f!-r, or P :::::::
~E
= €oXeE => P is proportional to E, and Xe
= ~~Np2
.
For water at 20° = 293 K , p = 6.1 X 10-30 em' 'volume
N = molecules= molecules
X molesX
!\rams.
mole
gram
volume
.
(O.33Xl029)(6.1Xl0-30)2
- j"1;)l
- 0 33 1029.
12 T bl 4 2
X
, Xe - (3)(8.85xl0-12)(1.38XlO-23)(293)
- ~
a e . gives an
6
1
N -- (6 .0 X 1023) X ( 18
) x-.(10 )
experimental value of 79, so it's pretty far off.
For water vapor at 100° = 373 K, treated as an ideal gas, v~~r::e= (22.4 X 10-3) X (~~~) = 2.85 X 10-2 m3.
(2.11 x 1025)(6.1x 10-30)2
-
N = 6.0 X 1023
,
2.85 X 10-2 = 2.11 X 1025. Xe = (3)(8.85 x 10-12)(1.38x 10-23)(373)= 15.7x 10 3.1
Table 4.2 gives 5.9 x 10-3, so this time the agreement is quite good.
Chapter 5
Magnetostatics
Problem 5.1
Sincev x B points upward, and that is also the direction of the force, q must be positive. To find R, in
I
I
terms of a and d, use the pythagorean theorem:
(R - d)2 + a2 = R2
=?
R2
-
2Rd + d2 + a2 = R2
=?R =
a2 + d2
.
2d
The cyclotron formula then giyes
r"'",,~
RV
{
p = qBR = IqB (a2 2d
+~)
Problem 5.2
The general solution is (Eq. 5.6):
y(t)
= CI cos(u;t) + C2 sin(u;t) + ~t + C3; z(t)
= C2 cos(u;t) -
CI sin(u;t) + C4.
(a) y(O) = z(O) = OJ y(O) = E/ Bj i(O) = O. Use these to determine CI, C2, C3, and C4.
y(O)= 0 =? CI + C3 = OJ y(O) = u;C2 + E/B = E/B =? C2 = OJ z(O) = 0 =? C2 + C4 = 0 =? C4 = 0;
i(O)= 0 =? CI = 0, and hence also C3 = O.So y(t) = Et/ B; z(t) = 0.1 Does this make sense? The magnetic
forceis q(v x B) ==-q(E/B)Bz
==-qE, which exactly cancels the electric force; since there is no net force,
I
the particle moves in a straight line at constant speed. ..(
(b) Assumingit starts from the origin, so C3 = -CI, C4 = -C2, we have i(O)
= 0 =? CI = 0 =? C3 =
0;
y(O)= 2~ =? C2u; + ~ = 2~ =? C2 = - 2~B = -C4; y(t) = - 2~B sin(u;t) + ~ t;
E
E
E.
E
z(t) = - 2u;B cos(u;t) + 2u;B' or y(t) = 2u;B [2u;t- sm(u;t)]; z(t) = 2u;B [1 - cos(u;t)]. Let (3 == E/2u;B.
Then y(t) = (3 [2u;t - sin(u;t)];
z(t) = (3 [1 - cos(u;t)]; (y - 2(3u;t) = -(3 sin(u;t), (z - (3) = -(3 cos(u;t) =?
(y- 2(3VJt)2
+ (z - (3)2= (32. This is a circle of radius (3 whose center moves to the right at constant speed:
Yo =2(3VJt; Zo = (3.
.
.
(c) z(O) = y(O)
E
=B
E
=? -ClUJ
=B
E
=? CI
= -C3 = - u;B j
89
E
E
C2u; + B = B =? C2 = C4 = O.
CHAPTER 5. MAGNETOSTATICS
90
y(t)
=-
E
E
E
wB cos(wt) + Bt + wB; z(t)
E.
= wB
E
sm(wt). y(t)
Let /3 ==EjwB; then [y - /3(1 + wi)] = -(3cos(wt),
of radius
{3 whose
center
is at Yo
= (3(1 +
= wB
z = {3sin(wt)j [y - {3(1+ wtW +
r.vt), Zo = O.
z
4~A
Problem
E
= wB sin(wt).
z2 = {32. This is a circle
[1 + wt - cos(wt)] j z(t)
y
-/3
(c)
5.3
(a) From Eq. 5.2, F = q[E + (v x B)] = 0 =>E = vB => v
q
v
(b) FromEq. 5.3, mv = qBR =>;;, = FiR =
Problem
m
~
I
= ~ .1
5.4
Suppose f flows counterclockwise (if not, change the sign of the answer). The force on the left side (toward
the left) cancels the force on the right side (toward the right); the force on the top is laB = lak(a/2) =
lka2/2, (pointing upward), and the force on the bottom is laB = -lka2j2 (also upward). So the net force is
F = Ika2 z.1
Problem 5.5
I
(a) K
I
(b) J
Problem
= 21ra
1 ,
I
because the length-perpendicular-to-flow
= ~s =>1 =
J da
=a
/
!s ds de/>= 21ra
s
/
ds
= 21raa
=>
a
= 21ra
1 ;J
=
1
21ras
1 .1
5.6
(a) v = r.vr,so K
Problem 5.7
I
:
/
is the circumference.
=~
= f7r.vr.1
(b) v
= r.vrsin8~
/ (a;:)
=>
I
J
= pr.vrsin8~,where
I
p ==Qj(4j3)1rR3.
/
Iv pr dr =
r dr = (V . J)r dr (by the continuity equation).
Now product rule #5
says V . (xJ) = x(V . J) + J . (Vx). But Vx = x, so V. (xJ) = x(V . J) + Jx. Thus Iv(V' J)xdr =
Iv V . (xJ) dr - Iv Jx dr. The first term is Is xJ . da (by the divergence theorem), and since J is entirely
insideV, it is zeroon the surfaceS. ThereforeIv(V' J)xdr = - Iv Jx dr, or, combining this with the y and
z components, Iv(V'
Problem 5.8
J)rdr
= - Iv J dr.
Or, referring back to the first line,
(a) Use Eq. 5.35, wIth
B=
. z = R,82 = -81 = 45°, and four sIdes:
.
(b) z
= R, 82= -81 = ;, and n sides:B = ~~~ sin(1r/n).
ii= /
~.
I~oll
J dr.
Qed
91
.
n/-LoI 7r
(c) For small B, sm B ~ B. So as n -t 00, B -t 2;R, ;: = I/-LOII
2R (same as Eq. 5.38, wIth
. z = 0).
Problem 5.9
()
(a) The straight segments produce no field at P. The two quarter-circles give B
=
I
¥ (~- i) I(out).
(b) The two half-lines are the same as one infinite line: :;~j the half-circle contributes ~~'.
SoB =I~~ (1 + ~) I (into the page).
Problem 5.10
.
/-LaI
(a) The forces on the two sIdes cancel. At the bottom, B = 27rS => F
top, B =
_
/-LaI
27r( S a )
.
/-LoI2a
+
=
/-LoI2a
Ia = _
_
2
( ) 2 (up).
/-LOI
7rS
7rS
At the
/-LoI2a2
=> F = 27r( S + a ) (down). The net force IS 27rS( S + a) (up).
(b) The force on the bottom is the same as before, /-LoI2/27r
(up). On the left side, B = /-LoIz;
27rY
dF = I(dl x B) = I(dx x + dyy + dz z) x (~;~ z) = ~~:( -dx y + dy x). But the x component cancels the
/-L 12
correspondingterm from the right side, and Fy = -~
/-La12
Fy
= - 2y37r
rq In (
s/J3 + a/2
s/ J33
I
)
= -rq2y37r In ( 1+ _2S )
12
2
is /-L;7r ['1 -
J3
I
force on the triangle
In
- dx. Here y = J3x, so
Y
.
..
27r 1s/V3
J3a
/-La12
(s/V3+a/2) 1
J3a
( 1 + ~ )] .
I
. The force on the right sIde IS the same, so the net
y
a
a
.""
~
S
:\ 600 :
x
s
V3
z
Problem 5.11
Use Eq. 5.38 for a ring of width dz, with I -t nI dz:
a2
B = /-LonI
3/2 dz. But z = acotB,
2
(a2 + Z2)
_ J
a
so dz = -~dB,
sm B
So
1
and
~~B
3/2 =~.
(a2 + z2)
a
~z
Z
dz
a2 sin3 B
.
(h - /-LonI
- /-LonI)~ sm
B - COSB1) .
B -- /-LonI
BdB-- /-LonI
cos BI - (COS2
( -a dB) - -2
a3 sin2 B
2
2
01
2
J
J
Foran infinite solenoid, B2= 0, BI= 7r,so (cosB2- cosBd = 1 - (-1) = 2, and B =
I /-LonI.I../
92
CHAPTER 5. MAGNETOSTATICS
Problem
5.12
,\2 2
Magnetic attraction per unit length (Eqs. 5.37 and 5.13): 1m = Po - v .
211' d
Electric field of one wire (Eq. 2.9): E
1
_
= 21I'EO
1 ~. Electric
S
,\2
~
1
Ie = -211'EOd
-. They balance when pov2
= -,
EO
repulsion per unit length on the other wire:
..
= v ~.EOPO
or v
Puttmg m the numbers,
v = \/(8.85 x 10-12)(411'x
1
10-7) = 13.00 X 108m/s.1 This is precisely the speed of light(!), so in fact you could
never get the wires going fast enough; the electric force always dominates.
G
Problem 5.13
f
(a)
(b) J
B . dl
= ks;
= B 211's= polenc=> B =
1=
l
a
Jda =
0
l
a
0
0,
for s < a;
poI:i..
_
{ 21I'S'f', £or s > a. }
211'ka3
ks(211's)ds= - 3
31
=> k =~.
Ienc =
211'a
PoI S2
---;!
211'a cjJ, for s
i
0
s
Jda =
i
s
ks(211's)ds=
0
A
211'ks 3
-
s3
I
a
Problem
B. dl
Pol
{ -211'scjJ,
A
for s
> a. }
5.14
-y
By the right-hand-rule, the field points in the
At z = 0, B = O. Use the amperian loop shown:
f
=
= 13, for s < a; Ienc= I, for s > a. So B
3
< a;
= Bl = polenc = polzJ
so IB =
{
=>
I
B
-poJa~,
for z > +a;
+I'oJay,
1m z
= -PoJzy
I
direction for z > 0, and in the +y di.rection for z < O.
(-a < z < a). If z > a,Ienc = polaJ,
}
> -a.
-1
~y
z{~ampedan
loop
~
l
Problem
5.15
The field inside a solenoid is ponI, and outside it is zero. The outer solenoid's field points to the left (-2),
whereas the inner one points to the right (+z).
Problem
So: (i) B
I
= poI(n1 -
n2)
z,
I
(ii) B
I
= -poln2
z, (iii) B
I
I
= 0.1
5.16
From Ex. 5.8, the top plate produces a field poK/2 (aiming out of the page, for points above it, and into
the page, for points below). The bottom plate produces a field poK/2 (aiming into the page, for points above
it, and out of the page, for points below). Above and below both plates the two fields cancel; between the plates
they add up to poK, pointing in.
(a) B = poO'V (in) betweem the plates, B = 0 elsewhere.
(b) The Lorentz force law says F = J(K x-B) da, so the force per unit area is f
I
I
I
I
to the right, and B (the field of the lower plate) is poO'v/2, into the page. So
I
=K
x B. Here K = (TV,
1m= PO0'2V2/2
(up).!
93
(c) The electric field of the lower plate is O'/2Eo; the electric force per unit area on the upper plate is
Ie = O'2/2Eo (down). They balance if POV2 = I/Eo,or v = 1/..,ftOiIO = c (the speed of light), as in Prob. 5.12.
Problem 5.17
We might as well orient the axes so the field point r lies on the y axis: r = (0,y, 0). Consider a source point
at (x',y',z') on loop #1:
I
I
dl'
I
I
~ = -x' x + (y - y') y - z' z; dl' = dx' X+ dy' y;
z
X
Y
0 = (-z' dy') X + (z' dx') y + [(y - y') dx' + x' dy'] z.
X ~ = dx'
dy'
-x' (y - y') -z'
Pol dl' x ~
Pol (-z' dy') X + (z' dx') y + [(y- y') dx' + x' dy'] Z
dB1=--=471"
1-3
471"
[(x')2+ (y - y')2 + (z')2]3/2
.
Now consider the symmetrically placed source element on
loop #2, at (x', y' , - z'). Since z' changes sign, while everything else is the same, the x and y components from dB1 and
dB2 cancel, leaving only a z component.
qed
With this, Ampere's law yields immediately:
B
=
z
r
ponI Z, inside the solenoid;
outside
{ 0,
y
(the same as for a circular solenoid-Ex. 5.9).
For the toroid, N/271"s= n (the number of turns per unit
length), so Eq. 5.58 yields B = pon1 inside, and zero outside,
consistent with the solenoid. [Note: N/271"s = n applies only
if the toroid is large in circumference, so that s is essentially
constant over the cross-section.]
Problem 5.18
J
It doesn't matter. According to Theorem 2, in Sect. 1.6.2, J . da is independent of surface, for any given
boundary line, provided that J is divergenceless, which it is, for steady currents (Eq. 5.31).
Problem 5.19
I
I
(a)
P
= volume
charge = charge.
atom
atoms. moles. grams
mole gram volume
e
N
M
d
=
=
=
=
p= (1.6x 10-19)(6.0 x 1023)(~.~)
(b) J
1~, = pv =>v =
= 7I"S2
is astonishingly small-literally
= (e)(N)
charge of electron
Avogadro's number
atomic mass of copper
density of copper
=
=
(~ )
M
1.6 X
(d), where
10-19
C,
6.0 X io23 mole,
'7 64gm/mole,
- 9.0gm/cm3.
= 11.4 X 104 C/cm3.1
1 = 71"(2.5x 10-3)(1.4
1
7I"S2p
x 104) = 9.1 X 10-3 cm/s, or about 33 cm/hr. This
I
I
slower than a snail's pace.
Po 1112
(c) From Eq. 5;37, 1m = 271" d
( )
=
(471"
X271"
10-7)
= 2 X 10-7 N/cm.
I
I
94
CHAPTER 5. MAGNETOSTATICS
(d) E
= 211"fOd
~~;
fe = ~
211"fO
= ~~
A1A2
= 3.00 x
d
V2 211"fO
30
C2
108m/s. Here
=
fe
(1.1 x 1025)(2 x 10-7)
Problem
5.20
c2
/-Lo
ltI2
v2
211"
d
( )
( ) ( ) ( )=
i = ;2 = (9.'1 ) =
f,
C == l/yfO/-LO
=
ltI2
d
X 1010
2
C2 fm,
V2
where
11.1 x 1025.1
X 10-3
= 12 X 1018 N/cm.1
Ampere's law says V x B = /-LoJ. Together with the continuity equation (5.29) this gives V . (V x B) =
. J = -/-Loop/ot, which is inconsistent with div(curl)=O unless p is constant (magnetostatics). The other
Maxwell equations are OK: V x E = 0 =? V . (V x E) = 0 (./), and as for the two divergenceequations, there
is no relevant vanishing second derivative (the other one is curl (grad), which doesn't involve the divergence).
/-LoV
Problem 5.21
~
At this stage I'd expect no changes in Gauss's law or Ampere's law. T4e divergence of B would take the
form V . B = o.OPm,where Pm is the density of magnetic charge, and 0.0 is some constant (analogous to (0
I
. and
I
/-Lo).The curl of E becomes V x E
I
= !3oJm,where Jm is the magnetic current density (representing the,
I
flow of magnetic charge), and!3o is another constant. Presumably magnetic charge is conserved, so Pm and Jm
satisfy a continuity equation: V. Jm = -oPm/ot.
As for the Lorentz force law, one might guess something of the form qm[B + (v x E)] (where qm is the
magnetic charge). But this is dimensionally impossible, since E has the same units as vB. Evidently we
need to divide (v x E) by something with the dimensions of velocity-squared. The natural candidate is
c'
~
IF = q, IE + (v x B)] + qm [B - ~(v x E)] .I'n this fo'm the magnetic aoalog to Coulomb',
1/<01'0'
law reads F = 0.0qml
411" r ;m2 f, so to determine 0.0we would first introduce (arbitrarily) a unit of magnetic charge,
then measure the force between unit charges at a given separation. [For further details, and an explanation of
the minus sign in the force law, see Prob. 7.35.]
Problem 5.22
f I Z dz = /-LoIZ t2
A = /-Lo
411"
Iz.
= /-LolA
-z
411"
B
=
[I
n
V x A
/-Lol
2
(Z + v2Z + 8 )]! Zl = -411" ln
-
411"
[ Z2
+
V(Z2)2
_/-LoI8
Z2 - V(Z2)2 +82
1
411" [ (Z2)2 - [(Z2)2 + 82] V(Z2)2
( ~)
- /-LoI8 -
-
411"
..
82
or, sInce smlh =
/-LOI(sin()2
411"8
-
+ 82
Zl
v(zd2 + 82
sin()l)
V(Zl)2
+ 82
V(Z2)2
- Zl-
+ 82
Zl
and sm()2=
;p I (as in Eq. 5.35).
Z
-
+ 82
v(zd2
+ 82
+1
]
1
Zl
-
8
+ v(zd2
+82
;p
1
/-LoI
V(Z2)2 + 82
,
;p
+ 82 V(Zl)2 + 82]
;p
+ 82]
Z2
- 411"8[ V(Z2)2 + 82
Z2
~Z
'--""'"
+ 82 ]
z? - [(zd2 + 82] v(zd2
V(Zl)2
.
A
z
8
- 1-
Z2
[ V(Z2)2
+
[ Zl
1
= - oA;p = _/-LoI
-
=
Z2+V(Z2)2+S2
Z2
08
-
v
dz
411" JZ1 yz2 + s2
-
Zl
v(zd2
+ 82]
(j;
'
I
95
Problem 5.23
a
1
-
k
A",=k~B=VxA=--s as (sk)z=-z;s
Problem
fLo
a
1
fLo
[
-- a
s
-
k
(
-s )] cp=
5.24
-~V . (r
=
V. A
V xr
- J=-(VxB)=1
=0
= -~ [B. (V x r) - r. (V x B)] = 0, since V x B = 0 (B is uniform) and
x B)
1
(Prob. 1.62). V x A
=0
(r. V)B
and V
.B
~
-
zcp.
fLoS
=0
1
= -2V
= -2
x (r x B)
[(B. V)r - (r. V)B
+ r(V. B) - B(V. r)]. But
(since B is uniform), and V . r = ~~ + ~~ + ~: = 1 + 1 + 1 = 3. Finally,
= Bxx+Byy+Bz z = B.
(B.V)r = (Bx :x + By :y + Bz :z) (xx+yy+zz)
So VxA
= -~(B-3B)
= B.
qed
Problem 5.25
.(a) A points in the same direction as I; and is a function only of s (the distance from the wire). In cylindrical
coordinates,then, A = A(s) Z, so B = V x A = -
aA = _fL
o1 ~ (the
aaAs ~ = fL27rS
field of an infinite wire). Therefore
oI, and A(r) = _fL2o1
7r In(s/a) z (the constant a is arbitrary; you could use 1, but then the units
aAz
= o.,( v x A = --aAz cp= - cp= B. ,(
look fishy). V. A =
as
2~
_a
aS
z
(b) Here Ampere's law gives
B=
-
aA
21!'R2 cpo as
J.LoIs
fLoI s
=-
21!' R2
- fLoI-
f B. dl = B 27rs =
~
A
=-
27rS
fLolenc
fLoI
2
41!'R2 (s
-
= fLoJ
-
7rS2
= fLo 7r~2
...
7rS2
= fL~:2
2
b ) z. Here b IS agalll arbItrary,
.
.
except
that
Slllce A
-~~ In(R/a)
= - :;~2 (R2 - b2),which means that we must pick a and b such that
- fLoI ( 2
for s ~ R;
47rR2 S - R2) Z,
2In(R/b)= 1- (b/R)2. I'll use a = b = R. Then A =
must be continuous at R,
fLoI
{ - 21!'In(s/ R) Z,
for s
;:::
R. }
Problem 5.26
K
= Kx
~ B
z
= xfLoK
2 Y (plus for z < 0, minus for z > 0).
A is parallel to K, and depends only on z, so A = A(z) x.
x
y
z
aA
a/ax a/ay a/az =8y=xfL~
B=VxA=1
JA= -¥Izi
x
A(z)
I
0
0
z
K
y.
y
~ill do the job-or this plus any constant.
x
Problem 5.27
(a) V.A
= ~;
/ v. (~)
dr'. V. (~) = ~(V .J) +J. V (~). But the first term is zero, because J(r')
isa functionof the sourcecoordinates, not the field coordinates. And since~ = r - r', V (~) = - V' (~).
So
96
V.
CHAPTER 5. MAGNETOSTATICS
(~) = -J.V' (~). But V'. (~) = ~(V"J)+J.V'
J
.
, J
d
h
()
(~), and V'.J = 0 in magnetostatics(Eq.5.31).So
Po
()
J
v,.
J
()
,
Po
dr = - 47r
f
J
d
-;;. a,
f
V. -;; = - V.
-;; , and hence, by the Ivergence t eorem, V .A = - 47r
-;;
where the integral is now over the surface surrounding all the currents. But J = 0 on this surface, so V.A = 0./
(b) V x A = ~:
J V 1x (~) dr' = ~; j
(- )
a function of f), and V
~:
(c) \72A =
J
[~(V x J) - J x V (~)] dr'. But V x J
4
po JX4
,
= -2'IJ. (Eq. 1.101),so V x A = -47r ~ IJ. dr = B. ./
IJ.
\72
(~)
dr'. But \72
= 47r
Po
Problem
5.28
=
Pol
j
J(r') [-4m53(1t)] dr'
f B . dl = -lb
(~) = J\72 (~)
= -poJ(r).
. dl = -[U(b)
VU
27rS Jy.
For an infinite straight wire, B = Pol
-VU
= Pol
V(e/» = Pol ~~e/> Jy= B.
27r
27r s lie/>
value; it works (say) for 0 ::;
e/>
(once again, J is a constant, as far as differenti-
= -47r(53(1t)(Eq. 1.102).
c
./
- U(a)] (by the gradient theorem), so U(b) f U(a).
I
qed
7r would do the job, in the sense that
U =- P2oIe/>
I
But when
< 27r,but
e/>
advances by 27r,this function does not return to its initial
at 27r it "jumps" back to zero.
Problem
5.29
Use Eq. 5.67, with R...,.+f and a ...,.+pdt:
A
-
=
B
=
R - - Powp .
B ;;.,
r dr + -rsm
'f'
r dr
3
r
r5
R2
r2
1
POWP .
Powp .
r
2
2
smB [ r2 5 + 2' (R - r) ] Eq. 5.86. Qed
98
CHAPTER 5. MAGNETOSTATICS
Problem
5.34
(a)m=Ia=lhrR2i.1
(b) B ~
I~; I :~2
(2 cos0 f + sin
0
8) . .
~
(c) On the z axis, 0 = 0, r = z, f = i (for z > 0), so B ~
i (for z < 0, 0 = 11",
f = -i, so the field
is the same, with Izl3 in place of Z3). The exact answer (Eq. 5.38) reduces (for z » R) to B ~ /LoIR2/2IzI3,
so they agree.
Problem 5.35
I
I
For a ring, m = I1I"r2. Here I --+av dr = aUJrdr, so m = foR1I"r2aUJr
dr = 11I"aUJR4
/4.1
Problem 5.36
The total charge on the shaded ring is dq a {211"R
sin O)R dO.
The time for one revolution
is dt
=
=
211"
/UJ.
SO the
\"
current
in the ring is I = ~~ = aUJR2 sin 0 dO. The area of the ring
is 1I"(Rsin 0)2, so the magnetic moment of the ring is dm
=
z
RsinO
{aUJR2sin 0 dO)1I"R2sin2 0, and the total dipole moment of the
shell is
m = aUJ1I"R4fo" sin3 Ode = (4/3)aUJ1I"R\ or 1m = ~aUJR4 i.1
The dipole term in the multi pole expansion for A is there. .
~
- /Lo 411" R 4sin 0 ;., - /LoaUJR4sin 0;.,
h h
A
lore dip - 411"
3aUJ
7
'P 3
r2 'P, w IC IS
also the exact potential (Eq. 5.67); evidently a spinning sphere
produces a perfect dipole field, with no higher multipole contributions.
Problem 5.37
The field of one side is given by Eq. 5.35, with s --+
..JZ2+ {w/2)2 and sinO2=
B
=
/LoI
-
sinOl =
w
(w/2)
..Jz2
+ w2/2
z
j
. To pick off the vertical
411"..Jz2 + {w2/4)..Jz2 + (w2/2)
component, multiply by sin 4>=
sides, multiply by 4: IB
= /LoI
(w /2)
; for all four
..Jz2 + (w/2)2
W2
211"(z2 + w2/4)..Jz2 + w2/2
i. IFor
z » w, B ~ /L;:~2 i. The field of a dipole 1m = IW2, for
W/2
points on the z axis (Eq. 5.86, with r --+ z, f --+ i, 0 = 0) is
B-/Lomi
..(
- 211"
z3 .
Problem 5.38
The mobile charges do pull in toward the axis, but the resulting concentration of (negative) charge sets up
an electric field that repels away further accumulation. Equilibrium is reached when the electric repulsion on
a mobile charge q balances the magnetic attraction: F = q[E+ (v x B)] = 0 => E = -(v x B). Say the current
I
99
is in the z direction: J
= p_vz
(where p- and v are both negative).
f B . dl =
/
2~O(p+ + p-)ss
E. da
=-
B 27rS
= l1oJ7rS2 ~
= 110P2-vs
B
= E27rsl = .!..(p+
+ p_)7rs21 ~
fO
[(vz) x (110P2-VS
J;)]
= ~Op_V2ss
E
J;;
=_
21fO (p+ + p-)ss.
~ p+ + p-
= p-(fOI10V2)
= p- (~:) .
~;,
Evidently p+ = -p- (1- ~:) =
or p- = _,2p+. In this naive model, the mobile negative charges fill a
smaller inner cylinder, leaving a shell of positive (stationary) charge at the outside. But since v « c, the effect
is extremely small.
Problem 5.39
(a) If positive charges flow to the right, they are deflected down, and the bottom plate acquires a positive
charge.
I
I
= qE ~ E = vB ::} V = Et = vEt, with the bottom at higher potential.
(c) If negative charges flow to the left, they are also deflected down, and the bottom plate acquires a negative
charge. The potential difference is still the same, but this time the top plate is at the higher potential.
(b) qvB
I
I
Problem 5.40
From Eq. 5.17, F = I J(dl x B). But B is constant, in this case, so it comes outside the integral: F
=
I (Jdl) x B, and J dl = w, the vector displacement from the point at which the wire first enters the field to
thepoint where it leaves. Since wand B are perpendicular, F = I Bw, and F is perpendicular to w.
Problem 5.41
The angular momentum acquired by the particle as it moves out from the center to the edge is
L=
/
~~ dt
=
/
N dt
=
/
(r x F) dt
=
/
r x q(v x B) dt
=q
/
r x (dl x B)
= q [/ (r.
B) dl-
/
B(r. dl)] .
But r i~ perpendicular to B, so r. B = 0, and r. dl = r. dr = !d(r. r) = !d(r2) = rdr = (lj27r)(27rrdr).
SoL = - 2: foR B 27rrdr = - 2: / B da. It followsthat L = - -!; ,where
=J
B da is the total flux.
In particular, if = 0, then L = 0, and the charge emerges with zero angular momentum, which means it is
goingalong a radial line.
qed
I
I
Problem 5.42
From Eq. 5.24, F =
J (K
x Bave) da. Here K = av, v = wR sin BJ;, da = R2 sin BdBd4J,and
Bave= !(Bin + Bout). From Eq. 5.68,
100
CHAPTER 5. MAGNETOSTATICS
Bin
=
~J.100"R(VZ=~J.100"R(V(cosBf-sinBO).FromEq.5.67,
Bout
=
V XA = V X
=
J.10R4(VO"
B
3r3 (2 cos r
(
~
J.1oR(VO"
B aye =
3
.
.
Sin2 B
SinB
f - ~~
~~
0
r or ( r )
[ r smBoB ( r2 )
J.1oR4(VO"
3
]
.
cos B r + sm B O) ( smce r
~
J.1oR(VO"
(2
3
~
.
= R) .
~
(4 cosB r - SIn.BO)
6
=
r2
~
+ sm BO) =
~
J.10R(VO"
K X Baye
~) =
J.10R4(VO"sinB
(O"(VRsin B)
(
6
~
~
)[
fjJ x (4cosBf
- sinBO)
J.10
?
] = 6(O"(VR)-(4cosBO
'
+ sinBf)sinB.
Picking out the z component of 0 (namely, - sin B) and of f (namely, cosB), we have
(K x Baye)z = -~0(O"(VR)2sin2BcosB, so
f
Fz = -~o (O"(VR)2
R2
Problem
= -~o
sin3 BcosBdBdcj;
= _~(O"(VR2)2
(O"(VR2)2
211"Ci:4 B) C2, or F
I
z.1
5.43
( a) F
= ma = qe( V
X
B)
J.10qeqm
= -411"r
~
'
J.10qeqm
( v x r;)
a
= 411"
- ~mr
(v x r ) .
1 dId
dv
dv
Buta,v=2dt(v,v)=2dt(v2)=vdt'
So dt =0.
qed
r V
r dr
J.1oqeqm d
J.1oqeqm
dQ
J.1oqeqm
- - - 0+
- --( c) - - m (v x v ) + m (r x a ) [r X ( v x r J (b) BecauseaJ..v,a.v=O.
dt
411" dt
(r )
= J.1oqeqm~[r2v - (r. v)rJ - ~ + !...~(~)
411"
r3
r
r2 dt
.
{
~
~
(d) (i) Q . fjJ = Q(z
r
= r f,
~
}
411"r3
= J.1oqeqm ~
411" [ r
J.1oqeqm
r
r2 dt
)
- (f 'v)f - ~ + f 2(r .V) = O. v"
r
~
r
~
2r
r
~
411" (f. fjJ). But z . fjJ= f.
. fjJ) = mer x v) . fjJ-
(
411"
fjJ
]
~
= 0, so
(r x v) . fjJ = O. But
and v = dl
dt = r f + rB 0 + r sin B~ ~ (where dots denote differentiation with respect to time), so
f
r xv
0
~
= I rOO
I
= (-r2sinB~)0+
(r2B)~.
r rB r sinB~
Therefore (r x v) . ~
(ii) Q . f
Q
cos
B
=-
= Q(z.
(from
(e) v2
f)
= mer
= Q(z.O) = mer x v) .0.
, so
- Q sIn B =
= r2 + r
2.
2
. 2'
B.
.
t
IS cons ant, so too IS Q . qe
= 0, so
d
J.1o:~qm
(f.O). But z.O = -sinB, f.O = 0, and (r x v).O = -r2 sinB~
2 . Bl
l
Q
-mr
sm 'f/ => 'f/ = -:2
mr
2
qed
x v) . f - J.1o:~qm(f. f). But z. f = cosB, and (r x v) J.. r ~ (r x v) . f
.
J.1oqeqm
J.1oqeqm
A d
411" ' or Q = - 411"cos B. n smce
(iii) Q.O
'
(1))
= r2B= 0, so B is constant.
Q
J.1oqeqm
.
WIth k ==-m = - 41I"mcbs. B .
k
= 2'
r
.
k
B2+ r sin Bcj;, but B = 0 and cj;= 2'
r
2
2
2
2 k2
so r = v - r sin B4r
=v
2
-
k2 sin 2 B
r2 .
101
dr
()
= 1-2 = V2 -
2
defy
(f)
r(efy)
~2
dr
/J
r
2
(vr /k)
. 2 ()
- sm
0 sm ()]'
= cos [(cjJ- AcjJ)'
Problem
=
(ksin()/r)2
(k2/r4)
Vr
r
/
=
2
[( k )
2
dcjJ* cjJ-
.
-
2
sm
- .
2
rv ( k )
2 ()
sm.
1
vr
.
sec-1 - '
; sec[(cjJ - cjJo)sm()]
()
()
sm
k sm
cjJo
()]
(
= --:--
where A ==- ~~~~n()
41fmv
I
Vr
= /
. dr
'defy
)
vr
= -k sm. () '
or
.
5.44
Put the field point on the x axis, so r = (s, 0, 0). Then
J.lo (K x 4)
B = da- da = RdcjJdz' K = K
~
~'
,
/
= (s -
K(- sincjJx + coscjJy);'"
RcoscjJ) X - RsincjJy
z
A
--
R, so Bz = J.loK(-1+1)
2
2
HereK
= nI, so B
I
= J.LonIz(inside),
= O.
and O(outside) (as we found more easily using Ampere's law, in Ex. 5.9).
I
5. MAGNETOSTATICS
CHAPTER
102
Problem 5.45
Let the source point be r'
=
-
Rcosu:1~-1j
+ z2 - 2Rzu -
fl
=
.
sinBdB VR2 + z2 - 2RzcosB -
10
=
= L1
= -~
y
11"
2
Letu::cosB,
h
- Z]
- 2z
2RzcosB + R2 COS2B) = R2
-
- 2RzcosB. So
Bz
R2
..jR2+ Z2
z
(z
- RcosB) .
]
sin2B=1-u2.
(R2/2)(1 - u2) - z + RU du
VR2 + Z2 - 2Rzu
]
+ 15] .
1
VR2 + Z2- 2Rzudu
-
= - 3Rz
1
3/2
(R2 + Z2 - 2Rzu)
1
-1
3Rz [ (R2 + Z2 - 2Rz)3/2 - (R2 + Z2 + 2Rz)3/2 ] = -~3Rz [(z - R)3 - (z + R)3]
1
2
-- 3& (z3- 3Z2R + 3zR2 - R3 - z3 - 3z2R - 3zR2 - R3) = -(3z2
+ R2).
~
1
1
1-1 ..jR2 + Z2-
1
2Rzu
1
du=--vR2+Z2_2Rzu
Rz
1
-1
1
2
Rz
z
=--[(z-R)-(z+R)]=-.
104
CHAPTER 5. MAGNETOSTATICS
h
1
=
1-1
= =
UZ
.../RZ +
-
ZZ
2Rzu
1
60 R3 z 3 [8(RZ + ZZ)Z
- 60~3Z3 {[8(RZ
-
du
1
+ 4(RZ+ zZ)2Rzu+ 3(2Rz)ZuZ)VRZ + ZZ- 2Rzu -1
1
+ ZZ)Z + 8Rz(RZ + ZZ)+ 12RZzZ)(Z - R)
[8(RZ + ZZ)Z - 8Rz(RZ
+ ZZ) + 12RZzZ) (Z + R)}
1
= - 60R3z3 {z [16Rz(RZ + ZZ)]- R [16(RZ+ zz)z + 24RZzZ)}
=
-
4
=
(
1 16R
60R3 z3
(
RZzZ
5
2
Z Z
4
-15RZz3 -"2R z - R
Bz
=
4
)=
15z3
2
Z
J.1.oRpw-(3z
[ 3z
= J.1.oRpw
2z+
(
Z
5 Z
R +"2Z . /4
= Z J-I
Z
4
(
)
Rz 2
Problem
J.1.0PW
15z3'
Rz
{I
{I
du
5
Z
(
= 2Zj
+ R ) - --2 z + --2 15z3 R + -z
2
2Rz
-3z
-
Rz
-Z
+
2R5
=
)
+ Z4 - R4 - 2Rzzz - z4 - ~RZzZ
2R4
-15z3
+
Rz
-3z
udu
= O.
) - 2z
]
)
- 2z
Q
But P
Z
= R J-I
/5
J.1.oQUJRZ A
= (4/3)l'OR3'so B =
101'Oz3Z.
5.48
J ;
B = J.1.o/ dl' Iz.. Iz. = -Rcos4Ji. + (y - Rsin4J)y + zz.
41'0
~
(For simplicity I'll drop the prime on 4J.)
~z = Rz cosz 4J+ yZ - 2Ry sin 4J+ Rz sinz 4J+ zZ = Rz + yZ+ zZ - 2Ry sin 4J. The source coordinates (x', y', z')
satisfy x' = R cos4J~ dx' = - R sin 4Jd4Jj y' = R sin 4J ~ dy' = R cos4Jd4Jjz' = 0 ~ dz' = O. So dl' =
-Rsin4Jd4Ji. + Rcos4Jd4Jy.
.
i.
Y
Z
dl' X Iz.= I -Rsin4Jd4J
Rcos4Jd4J 0 1= (Rzcos4Jd4J)i. + (Rz sin4Jd4J)
y + (-Ry sin 4Jd4J+ Rz d4J)z.
-Rcos 4J
Bx
= J.1.o/Rz1 Z".
41'0
0
(y - Rsin 4J) z
~
cos4J d4J
= J.1.o1Rz
(RZ + yZ + ZZ - 2Ry sin 4J)3/2
1
41'0 Ry VRz
+ yZ + ZZ
-
Z".= 0,
2Ry sin 4JI 0
since sin 4J = 0 at both limits. The y and z components are elliptic integrals, and cannot be expressed in terms
of elementary functions.
B x-a,
.
B y- _J.1.o/Rz
41'0
Problem
d4J
10 Z". (RZ + yZ + sin4J
zZ - 2Ry
3Z'
sin 4J) /
5.49
. B z = J.1.o/R
_
41'0
10
i
.
.
J.1.011 dli xi.
From the BIOt-Savart law, the field of loop #1 IS B = 4
~;
1'0 1
~
F
= lz1
1;
dlz x B
= J.1.0/Ilz
4
1'0
i1
11;
dlz X (~I X i).
~
Now dlz X (dli X i)
(R - ysin4J) d4J
21r
(RZ + yZ + ZZ
-
3Z'
2Ry sin 4J) /
the force on loop #2 IS
.
.
= dlI(dlz .i) - i(dli . dlz),
so
105
J.lo
F
=-
47rhI2
4
1.2(dl1 . dl2) -
{ff
f f
(dh'4)
dl1
1.2
}
The first term is what we want. It remains to show that the second term is zero:
= (X2 -
x + (Y2-
yd Y+ (Z2- Zl) z, so V2(1/1,) = 00X2 [(X2- xd2 + (Y28
2
0
2
2
2
2 -1/2
+
X2 - Xl ) + (Y2 - Y1) +
X2 - Xl ) + ( Y2 - Y1) + ( Z2 - Zl ) ]
Y+
[(
[(
0
8~
~
4
(x2-xd~
(Y2-Y1)~
(z2-zd~
-t
4
=xyz = -- = --. So
-.d}z = V2
It
Xl)
_
~
1.3
1-3
2 in Sect. 1.3.3).
Problem 5.50
1-3
1-3
f
1-2
yd2 + (Z2- Zl)2r1/2 X
2 -1/2
~
(Z2 - zd ]
z
1
f (- ) .dl2 = 0 (by Corollary
1.2
1.
qed
=
_-.!:..p. For dielectrics (with no free charge), Pb = -V. P
fO
1
(Eq. 4.12), and the resulting potential is V(r) = 47rfO P(r']1. .4 dr'. In general, P = fOV . E (Gauss's law),
Poisson's
equation
(Eq. 2.24) says \72V
_
!
!
.
1
E(r') .4
sothe analogy ISP --+-foE, and hence V(r) = - 47r
1.2 dr'. qed
[Thereare many other ways to obtain this result. For example, using Eq. 1.100:
V.
V(r) =
j
V(r')83(r -r')dr'
(~) = -V'. (~) = 47r!53(-t) = 47r!53(r -
= -~47r
!
4
()
v(r')V',
1.2
=~
47r
dr'
!
r'),
4 . [V'V(r') ] dr' -
1-2
~
47r
f v(r')
!
.
1
(Eq.1.59). But V'V(r') = -E(r'), and the surface mtegral --+0 at 00, so V(r) = - 47r
before.You can also checkthe result, by computing its gradient-but it's not easy.]
Problem 5.51
A = -HB x f).
(a) For uniform B, f;(B x ell)= B x f; dl = B x r
= J.loI
(p, so
27rS
f B x dl = (J.lOI8 27ra
)
J.loI 8 w
27rb
= -r x B fOl>'d>'= I-~(r x B).I
J.loI
J.loI
(d) B = - cjJ; B(>.r)= cjJj A =
27rS
27r>.s
.4
1.2
dr', as
I =1=
I
(b) B
E(r')
4 .da'
1.2
=
(~ - ~)
J.lolw
27r
a
b
8
=1=
O.
(c) A
~
~
J.loI
--(r
27rs
~
1
10
x cjJ)
1
J.loI
>.- d>' = --(r
>.
~
x cjJ). But r here is the
27rs
J.loI
- 27rs s(s X cjJ)+ z(z x cjJ) , and
~
vector from the origin-in
cylindrical coordinates r
(§x~)=z, (zx(p)=-s.
SOIA=~(ZS-sz.1
= S s + z z.
So A
=
[
~
]
The examples in (c) and (d) happen to be divergenceless, but this is not the case in general. For (letting
f;
>'B(>.r)d>',for short) V . A = -V. (r x L) = -[L . (V x r) - r . (V xL)] = r. (V x L), and
V x L = fOl>'[V x B(>.r)]d>'= f; >.2[VAx B(>.r)]d>'= J.lo >.2J(>.r)d>',so V . A = J.lor. fOl >.2J(>.r) d>', and
L ==
f;
it vanishes in regions where J = 0 (which is why the examples in (c) and (d) were divergenceless). To construct
an explicit counterexample, we need the field at a point where J =1=
a-say,
inside a wire with uniform current.
CHAPTER 5. MAGNETOSTATICS
106
Here Ampere's
law gives B 211"s= /-LoIenc= /-LoJ1I"S2=> B = /-L~JS 4>, so
t
A =
/-LoJ
=
V.A
/-LoJ
-r x Jo ,\
(2 )
~~(S2Z)
8s
6
I
B
=
_4 1
=
/-Lo
411"
r13[3(m
13[3(p . f) f
11"100
r
-
p]
. f) f - m]
= -Ts(r
= /-LoJ
+ ~(-S2)
8z
[s
Conclusion: (ii) does not automatically yield V . A
Problem 5.52
(a) Exploit the analogy with the electrical case:
E
/-LoJ
~
'\sljJd'\
6
]
/-LoJs
x ljJ) = ~(zs
~
(s ) =
~2SZ
(Eq. 3.104)
(Eq. 5.87)
=F
= - VV, withV = _411"100
1 pr ~ f
sz).
O.
(Eq. 3.102).
= - VU, (Eq. 5.65).
~~.I
I
.
3
~
-
= 0.1
Evidently the prescription is p/€o -t /-Lom: U(r) =
(b) Comparing Eqs. 5.67 and 5.85, the dipole moment of the shell is m
.
/-LoJz
~
= (411"/3)VJaR4
z (which we alsogot
/-LoVJaR4cos()
.
m Prob. 5.36). USIng the result of (a), then, U(r) =
3
r2
for r > R.
Inside the shell, the field is uniform (Eq. 5.38): B = ~/-LoaVJR
z, so U(r) = -~J:toaVJRz + constant. We may
I
I
= -~/-LoaVJRrcos() for r < R.
= R): Uin(R) =
as well pick the constant to be zero, so U(r)
I
I
[Notice that U(r) is not continuous at the surface (r
-~/-LoaVJR2
cos() =FUout(R) =
i/-LoaVJR2cos(). As I warned you on p. 236: if you insist on using magnetic scalar potentials, keep away from
places where there is current!]
(c)
B
3r2
= -/-LoVJQ
1- 411"
R [(
5R2
8U
atjJ
=
18U
r 8()
8U
8r
0
=>
(
-
6r2
(
1-
-
5R2
)
.
1 au
= - VU = - -au
r - - - () 8r
r 8()
~
~
sm ()()
]
.
6r2
1- 5R2
)( )
( )( )
411"R
/-LoVJQ
sm() => U(r,()) = -
3r2
J.LoVJQ
=
-
~
-
411"R
1-5R2
6r2
1- 5R2
( )( )
( )( )
411"R
411"R
rcos() + f(r).
r3
/-LoVJQ
cos()=>U(r,())=-
r-5R2
cos() + g«()).
Equating the two expressions:
J.LoVJQ
-
(
)(
411"R
6r2
/-LoVJQ
)
1- 5R2 rcos() + f(r) = -
( )(
411"R
( )
411"
R3
r3 cos()
+ f(r)
)
1- 5R2 rcos() + g«()),
or
/-LoVJQ
r2
= g«()).
1
8U
-ljJ.
r sin () 8tjJ
U(r, (), tjJ) = U(r, ()).
/-LoVJQ
-
)
~
cos () r
~
107
But there is no way to write r3 cos() as the sum of a function of () and a function of r, so we're stuck. The
reason is that you can't have a scalar magnetic potential in a region where the current is nonzero.
Problem 5.53
/Lo J
(a) V . B = 0, V x B = /LoJ,and V . A = 0, V x A = B =>A = -Ij. dr', so
411"
j
j
V. A = 0, V x A = B, and V. W = 0 (we'll choose it so), V x W = A => W = -};
~ dr'.1
(b) W will be proportional to B and to two factors of r (since differentiating twice must recover B), so I'll
try something of the form W = ar(r . B) + .8r2B, and see if I can pick the constants a and .8 in such a way
that V . W = 0 and V x W = A.
I
2
2
8x
8y
8z
V. W= a [(r. B)(V . r) + r. V(r. B)] +.8 (r (V. B) + B. V(r )] . Vr = 8x + 8y + 8z = 1 + 1 + 1 = 3;
V(r. B) = r x (V x B) + B x (V x r) + (r. V)B + (B. V)r; but B is constant,so all derivativesof B vanish,
andV x r = 0 (Prob. 1.62), so
V(r . B)
= (B . V)r = (Bx :x + By:y + Bz :z) (x x + YY+ z z) = Bx x + By Y+ Bz z = Bi
= (x :x + Y:y + z :z) (x2 + y2 + z2) = 2x x + 2y Y+ 2z z = 2r. So
V . W = a [3(r . B) + (r . B)] + .8 [0 + 2(r . B)] = 2(r . B)(2a + .8), which is zero if 2a + .8 = o.
V x W = a [(r. B)(V x r) - r x V(r. B)] +.8 (r2(V x B) - B x V(r2)] = a [0 - (r x B)] +.8 [0 - 2(B x f)]
V(r2)
= -(r
x B)(a
- 2.8) = -~(r
x B) (Prob. 5.24). So we want a - 2.8 = 1/2. Evidently a - 2(-2a)
= 5a = 1/2,
~
= 1/10;.8 = -2a = -1/5. Conclusion: W = (r(r. B) - 2r2B] .1 (But this is certainly not unique.)
(c) V x W = A => J(V x W) . da = J A. da. Or fW. dl =
A . da. Integrate around the amperian loop shown, taking
or 0:
I
J
W to point parallel to the axis, and choosingW = 0 on the
axis:
r
= 10 (~ ) lsds = ~2
/LonI
-Wi
Iw= -~zl
1:'
R
I'Or s
> ,
/LonI821.
(8< R).
- Wl = /LonIR21
4
(usmg Eq. 5.70 for A).
r )s
+1 (
/LonI
2
R
W
R2
Id -= /LonIR21 /LonIR21 1 ( /R)
8
4
+
2
n 8
= _/LonIR2
[1+ 2In(8/R)]
4
'
,
z 1(8> R).
Problem 5.54
Apply the divergence theorem to the function [U x (V x V)], noting (from the product rule) that
V. [U x (V x V)]
!
= (V
x V). (V xU) - U. [V x (V x V)]:
V . [U x (V x V)] dr
=
j {(V x V) . (V x U) - U . [V x (V x V)]} dr = f [U x (V x V)] . da.
Asalways,suppose we have two solutions, B1 (and Ad and B2 (and A2)' Define B3 = B2 - B1 (and
A3==A2 - Ad, so that V x A3 = B3 and V x B3 = V X B1 - V X B2 = /LoJ - /LoJ = O. Set U = V = A3
in the above identity:
CHAPTER
108
!
- A3'
{(V x A3)' (V x A3)
= f
[A3
X
specified
(V
x A3)]
(in which
= f
. da
case
hence B1 = B2.
[V x (V x A3)]} dr
(A3
X
B3)
. da.
But
=!
either
{(B3)'
A
is
(B3) - A3'
specified
= 0), at the surface. In either case f(A3
B3
(in
x B3)
5. MAGNETOSTATICS
[V x B3]}
which
case
=
.da
O.
dr
A3
So
!
=
!(B3)2dr
=
0), or else B
is
(B3)2 dr = 0, and
Qed
Problem 5.55
From
"
Eq. 5.86, Btot = Bo z - J.L
1f:r (2cos()r+sin()6).
4om3o
B
~
lore
.
This
= B0 Z .r
~
r
J.Lorno
~
(
)
-
2
()
411"r3 cos
= (B0 -
by Bo = ~;~~, or
r = R, given
is zero, for all,(),when
There-
J.Lomo
()
) cos.
211"r3
1/3
R =
( ~;;~ )
.
I
Evidently
no
field lines cross
this sphere.
Problem 5.56
(a)
1= (2~W)= ~:;
L
m = '2
Q MwR2
wR'
(b) Because
= 2M'
Q
I
a=1I"R2;m=
m
(
Q
= ~~
g is independent
2m 2
Problem
=~ =
4m
.,
d h
L=RMv=
.
= 2M)L,an t e gyromagnetlcratioIS
I
of R, the same
(or any other figure of revolution):
(c) m
~:1I"R2Z= ~WR2Z.
(1.60
I
g
=
MwR2;L= MwR2Z.
~
LNJ
ratio applies to all "donuts", and
hence
2~.1
x 10-19)(1.05
x 10-34)
4(9.11
x 10-31)
=14.61 X 10-24 Am2.1
5.57
(a) Bave =
3
--
411"R3
f
(3/4~1I"R3! B dr = 411"~3!(V x A) dr =
J- dr I x da =
A x da = - - 3 J.Lo
411"R3 411"
!
f {! 11-
}
{f ;
- (4:~oR3
J x
da} drl. Note that J depends on the
source point r/, not on the field point r. To do the surface
integral, choose the
(x,y,z) coordinates so that
rl liesonthez
axis (see diagram). Then 11= VR2 + (ZI)2 - 2RzICOS(), while
da = R2 sin ()dOd4>r. By symmetry, the x and y components
must integrate to zero; since the z component of r is cos (), we
have
z
to the entire sphere
109
f~
da =
z
1-
cos ()
J VR2
R2
sin() d() d4> =
=
27rR2Z
.
=
f
1
+ (Z')2
-1 VR2
27rR2Z
{
=
3(Rz')2
{ [R2 + (Z')2
""3Z
{
,
-
47r
Z -
""3r
A
47rR3
( )
+ (Z')2 + Rz'] IR (
,
'
+ (z' ) 2 -
r
(r
}
-1
1
2Rz' - [R2 + (Z')2 - Rz'] VR2 + (Z')2
+ 2Rz'}
[R2 + (Z')2 - Rz'] (R + z')}
z'l-
< R) ,
> R).
}
3/Lo 47r
= - (47r)2R33
J (J xr
,
/Lo
3/Lo 47r 3
b ThIS tIme r > R, so Bave = - (47r)2R3 3 R
from the source point to the center (~= ~r'). Thus
J (J x
Bave
1
J
) aT' = - 47rR3 (J xr') dT'. Now m = :2j(rxJ)
qed
,
2Rz'u
'
,
3(z')2 Z = ""3 (r')3 r,
/Lo 2m
(Eq. 5.91), so Bave = 47r R3'
.;
VR2
+ Rz'] VR2 + (z')2 -
,
Fornowwe want r' < R, so Bave
d().
- 2Rz' cos()
1
+ 2Rz'u]
47r R3,
A
cas () sin ()
+ (z')2
VR2
du
3(2Rz')2
- [3(~~)2z] {[R2
47r
Jo
- 2Rz'u
2 [2(R2 + (z')2)
-
27rR2Z
=
U
r
27rR2 Z
+ (z')2 - 2Rz' cos()
Let u ==cas (), so du = - sin ()d().
r'
)
(r')3
= Been.
d
'
/Lo
T = 47r
qed
,
J x Ii.
J~
dT
dT, where ~ now goes
Problem 5.58
(a)Problem5.51givesthe dipolemomentof a shell: m = 4; (1wR4Z. Let R -7 r, (1 -7 P dr, and integrate:
m = 47r
""3wpz Jo
{R r 4 dr = ""3wPT
47r
R5 z.
A
(b)
B
ave
/Lo 2m -
But p = (4/3)7rR3'
Q
so m = SQwR
1
I
2
z.
A
I
/Lo 2Qw A
- 47rR3 - 47r 5R Z.
(c) A ~ /Lomsin()
47r
J
A
r2
$=
(d) UseEq. 5.67, with R
/Lo QWR2 sin()
47r
5
r2
-7 r,(1 -7 par,
$.
and integrate:
A = /Lowpsin () $ {R r4 dr = /Low
3
r2
Jo
~
sin () R5
3 47rR3 r2
5
$=
/Lo QwR2 sin ()
I
47r
5
r2
$.
This is identical to (c); evidently the field is pure dipole, for points outside the sphere.
(e) According to Frob. 5.29, the field is B =
/L:;~[(1- :~:
)
cas ()f
- (1 - :~2 ) sin() {j]. The average
110
CHAPTER 5. MAGNETOSTATICS
obviously points in the z direction, so take the z component of r (cos B) and {J (- sin B):
Eave
1
411"R(4/3)1I"R3
J-LoUJQ
=
1r
3J-L UJQ
1
3J-LoUJQ 3
811"
R4 R
1r
10
!(
- "55R2
2
(
+ (3
cos B +
R3
COS2B
6r2
1 - 5R2
)
.
2
(
2
6 R5
- "55R2
)
3
- 3cos
1r
B) 0
1
J-LoUJQ
2
.
sm B] r smOdr d(}dif>
)
]
sin2 B sin BdB
.
3J-LoUJQ1
7. 2
75 cos B + 75 sm B sm BdB = 811"
R 75
16
J-LoUJQ
2001l"R (-7cosB
)
)
[ 1 - 5R2
r3
3 R5
[( 3"
(411"~2)2211" 0
3r2
1r
1
0
.
2
(7 + 9 cos B) sm BdB
J-LoUJQ
= 2001l"R(20) = 1O1I"R(same
as (b)). ./
Problem 5.59
The issue (and the integral) is identical to the one in Prob. 3.42. The resolution (as before) is to regard
Eq. 5.87 as correct outside an infinitesimal sphere centered at the dipole. Inside this sphere the field is a
delta-function, Ac53(r), with A selected so as to make the average field consistent with Prob. 5.57:
Bave
Problem
-
1
-
(4/3)1I"R3
!
3
Ac5 (r)dr
-
-
3
J-Lo
2m
.
- 2J-Lom
- 411"R3A - 411"R3 =} A -~.
2J-LO
3
The added term IS "3mc5
(r).
1
5.60
00
(a) I dl -+ J dr, so
!
IA = :; n=O
L rn~1 (r,)n Pn(cosB)J
dr.
!
J-LO
J-LO d
J dr =
41I"r
41I"rdPt (Prob. 5.7), where P is the total electric dipole moment. In magneto statics, P is constant, so dp/dt = 0, and hence Amon = O. Qed
(c) m = Ia = ~I §(r X ill) -+ m = ~ J(r X J) dr. ,Qed
Problem 5.61
For a dipole at the origin and a field point in the x z plane (if>= 0), we have
(b) Amon
B
=
=
-
41I"rn; (2 cos Br + sin B9) = 4J-LO
1I"rn;
J-LO
[2 cos B(sin B x + cos B z) + sin B( cos B X - sin B z)]
n; [3 sin B cos ()x + (2 COS2B - sin2 B) z].
4 1I"r
J-LO
Here we have a stack of such dipoles, running from z =
-L/2 to z = +L/2.
Put the field point at s on the x
z
axis.
The x components
cancel (because of symmetrically placed dipoles above and below z = 0), leaving B =
J-Lo
L/2 (3 COS2B - 1)
3
dz, where M is the dipole mo4 2M z
_
10
11"
=
m
h
y
r
= I1I"R2 = (CTvh)1I"R2 =
.
3
.
8
1
= 1I"CTUJR. Now smB = -,r so r =
ment per unit length: m
M
£/2
CTUJR1I"R2h
"3"
8
-8 cotB =} dz = ~
sin B dB. Therefore
sin3 B
~;8
z
=}
=
x
-£/2
111
B
=
9m
J1-
sin3
()
8
1
o (7raCI.JR3)zrr/2 (3cos2()-I)~~d()=
27r
8
sm
J1-oaCI.JR3
-
~
228
8
But sin()m=
J82
+ (L/2)2'
9m
3
z(-cos()+cos())
1
rr/2
and cos() =
m
°228
()
z
J1-oaCI.JR3
=
228
9m
J1- aCI.JR3
11r/2 (3cos2()-I)sin()d()
2
A
J1-oaCI.JR3
COS()m(1-COS()m)Z=
-(L/2)
J82 + (L/2)2'
so B =
I
-
.
228
COS()msm2()mZ.
J1- aCI.JR3 L
°
~
4[82 + (L/2)2]3/2 z.
~
Chapter 6
Magnetostatic
Fields in Matter
Problem 6.1
N
= m2
X
_
A A
B 1; B 1 = J.Lo1
41rr"3 [3(mi. r ) r - ml ] ; r
N =--J.Lomlm2
(yxz )
47rr~
A
'downward
I
A
J.Lomlm2A
A
= --4 7rr~x.
Hereml
A
A
= y; ml = mlZ;
A
m2
= m2Y'
B
J.Lo ml
1
= --41rr3z,
J.Lo(abI)2A
= 1ra2 1 , m2 = b2 I . Sa IN = -- 4
~x.r
A
.
.
.
I F ma I onentatIOn:
(-z).
Problem 6.2
elF = I dl X B; tIN
(r X ill)
= O.
= r X elF = Ir X (dI X B). Now (Prob. 1.6): r X (dI X B) + dl X (B X r) + B X
But d[r X (r X B)] = dr X (r X B) + r X (dr X B) (since B is constant), and dr = dI, so
dl X (B X r) = r X (dI X B) - d[r X (r X B)]. Hence 2r X (dI X B) = d[r X (r X B)] - B X (r X dI).
dN =!I {d[r X (r X B)] '- B X (r X dI)}. :. N = !I {§d[rX (r X B)] - B X §(r X dI)}. But the first term
is zero (§d(... ) = 0), and the second integral is 2a (Eq. 1.107). So N = -I(B X a) = m X B. qed
Problem 6.3
(a)
I~
Accordingto Eq. 6.2, F = 27rIRBcos(). But B =
A so B cas () =
l!Q.
4".[3(ml.r~r-mtl
r
' and B cas () = B. Y
,
~~ [3(ml . i)(i. y) - (mi' y)]. But ml . Y = 0 and
i .Y
~~3ml
F = 27rIR~~3ml
sin q,cos q,. Now sinq,
But IR27r = m2, so F = :!.!!!!.
2". mlm2
~, r
= sinq"
while ml
sin q, cas q,.
= ~, cosq, = yr2
or, since z = r: F
I
=-
21r r4m2 z.
3J.Loml
A I
113
J.LOm1 m2.
r, so F = 321r
(b)F = V(m2. B) = (m2. V)B = (m2:z) [~z\(~(ml. z)z - ml)]
2ml
ml cos().
- R2/r, so F = 3~mlIR2~.
while for a dipole, R«
~
.i =
= ~mlm2z
r4
d~(z\),
~
-3:\ z
:. Bcos()
=
114
CHAPTER 6. MAGNETOSTATIC
Problem
dF
6.4
= J {(dyy)
-
X B(O,y,O) + (dzz) X B(O,t,z)
= J {-(dY y) X JB(O, y, t) - B(O, y, On +(dz z)
~
(dyy)
X
:::}Jt2
y
~ t8B
8y
B
= mOO
{
8Bz
~!!..!i...
8y
~
z
Y
8y
8Bx
~
x
-
1
But m. B
= mBx
Therefore F
Problem 6.5
0
8Bz
8y
8z
8Bx
~
=m x-+y-+z[ 8x
8y
y
z
,y,
Z
1
0
~!!..!i...
8z
8z
O ~t~Bl
]
z,
o ooandJdz~B
,
{8y
8y
B).
I
. -+8By
(usmgV.B=Otownte
8y
8z
8Bz
8z
(since m = mx, here), so V(m. B) = mV(Bx) = m (8fxzx +
= V(m.
y O,O,z
.
~t~B
y 0,0,0 ]
I
= m y 8Bx - x8By- x8Bz - z8Bx
}
.
8Bx
8z
8z }
= --8Bx
8x
~y
+
).
8fzz
z).
Qed
z
= 110Joxy (Prob.
5.14).
m. B = 0, so Eq. 6.3 says F
I
(b)
(dzz) X B(O,O,z)}
JB(O, t, z) - B(O, 0, zn}
zx8 B_YX8
.
8Y
8 z } [ NotethatJdy~Blo
{
X
F
-
X B(O,y,t)
~ t8B
8z
(a) B
FIELDS IN MATTER
= 0.1
m. B = mWoJox,so IF = mol1oJox.!
y
(c) Use product rule #4: V(p. E)
= p x (V x E) + E x (V x p) + (p . V)E + (E . V)p.
But p does not depend on (x, y, z), so the second
and fourth terms vanish, and V x E = 0, so the
first term is zero. Hence V (p . E) = (p . V)E. Qed
This argument does not apply to the magnetic analog,
since V x B =I0.' In fact, V(m. B) = (m. V)B + 110(mx J).
(m. V)Ba = mofx(B) = mol1oJoY,(m. V)Bb = mo-/y(l1oJoxy) = O.
Problem 6.6
Aluminum, copper, copper chloride, and sodium all have an odd number of electrons, so we expect them to
be paramagnetic. The rest (having an even number) should be diamagnetic.
Problem 6.7
Jb
= VxM = 0;
Kb
=M
X ii
= M(p.
The field is that of a surface current
Kb
= M (P,
~fi
but that's just a solenoid, so the field
!outside is zero, and inside B
I
= 110Kb = 110M. Moreover,
it points upward (in the drawing), so B
I
= 110M.!
115
Problem 6.8
VxM
= Jb = ~!
= 3ksz,
s vS (sks2)z = ~(3ks2)Z
s
Kb = M X Ii
= ks2(J> X s) = -kR2z.
Sothe bound current flows up the cylinder, and returns down the surface. [Incidentally, the total current should
be zero ... is it? Yes, for JJbda = JoR(3ks)(27rsds) = 27rkR3, while JKbdl = (-kR2)(27rR) = -27rkR3.]
Sincethese currents have cylindrical symmetry, we can get the field by Ampere's law:
B
Outside the cylinder
. 27rS = polenc= Po18 Jbda = 27rkpoS3=} B = POkS2J> = paM.
Ienc
I
I
= 0, so B = 0.1
I
Problem 6.9
I
- --
B
B
Kb = M X Ii = M J>.I
(Essentially a long solenoid)
(Essentially a physical dipole)
(Intermediate case)
[The external fields are the same as in the electrical
case; the internal fields (inside the bar) are completely
different-in fact, opposite in direction.]
Problem 6.10
Kb = M, so the field inside a complete ring would be PoM. The field of a square loop, at the center, is
givenby Prob. 5.8: Bsq = V2poI/7rR. Here I = Mw, and R = a/2, so
B
- V2poMw - 2V2poMw.
sq - 7r(a/2) 7ra
'
net fieldin gap: IB = paM(1 - 2:: w) .
116
CHAPTER
Problem
MAGNETOSTATIC
6.
FIELDS IN MATTER
6.11
As in Sec. 4.2.3, we want the average of B = Bout + Bin' where Bout is due to molecules outside a small
sphere around point P, and Bin is due to molecules inside the sphere. The average of Bout is same as fieldat
center (Prob. 5.57b), and for this it is OK to use Eq. 6.10, since the center is "far" from all the molecules in
question:
/lo
Aout
= -47f
.
M
~ ~2
J
d
T
outside
5.89-where
m = !7f R3M. Thus the average Bin is 2/loMj3. But what is
left out of the integral Aout is the contribution of a uniformly magnetized sphere, to wit: 2/loMj3 (Eq. 6.16),
The average of Bin is '1; (iW )-Eq.
and this is precisely
Problem
what Bin puts back in. So we'll get the correct macroscopic
field using Eq. 6.10.
qed
6.12
(a) M = ksz; Jb = VxM = -kepi Kb = M X ft = kRep.
B is in the z direction (this is essentially a superposition of solenoids). So
I
B
= 0 outside.!
= Bl
=
Use the amperian loop shown (shaded)-inner side at radius s:
/loIenc= /lo [fJbda + Kbl] = /lo[-kl(R - s) + kRl] = /lokls.
fB.
dl
:.1 B
= /loksz inside.
I
(b) By symmetry, H points in the z direction. That same amperian loop gives fH
since there is no free current here. So H
I
M = ksz, so B = /loksz.
Problem
=0
I,
and hence B
I
I
In the cavity, H=
]
:oB,soH=
:0 (Bo-~/loM)
= 0; inside
= Bo -
~/lOM, with the sphere removed.
I
The field inside a long solenoid is /loK. Here K = M, so the field of the bound current on
the inside surface of the cavity is /loM, pointing down; Therefore
K'~
(~)Kb
= 0, so B
= 0,
=Ho+M-~M=>IH=Ho+~M.1
I
(c)
Outside M
/loIfenc
6.13
(a) The field of a magnetized sphere is ~/loM (Eq. 6.16), so B
(b)
= /loM.1
. dl = HI =
H
= ~(Bo
/lo
B
= Bo -
/laM)
/laM;
= ~Bo
/lo
I
M
=>
I
H
= Ho.1
This time the bound currents are small, and far away from the center, so B
I
while H
= :0 Bo = Ho + M =>
I
H
= Bo,
I
= Ho+ M.I
[Comment: In the wafer, B is the field in the medium; in the needle, H is the H in the medium; in the
sphere (intermediate case) both Band H are modified.]
117
Problem 6.14
M:
~
; B is the same as the field of a short solenoid; H
= ;0B - M.
Problem 6.15
"Potentials":
{
Win(r,O) = EA,r'p'(eosO),
(r R).
E r~l p,(eos0),
Boundary Conditions:
{
(i) Win(R,O)= Wout(R,O),
(ii) _8~;ut IR+ 8~n IR= MJ. = Mi. r = M eos0.
(Thecontinuity of W follows from the gradient theorem: W(b)
- W(a) =J: VW.
ifthetwopointsare infinitesimallyseparated,this last integral-+ 0.)
(i) => A,R' = RIf.!.l=> B, = R2'+1A"
{ (ii) => E(l + 1) :.!.2P,(eosO) + EIA,R'-2P,(eosO)
dl
= - J: H. ill;
= M cosO.
Combining these:
~
M
L.J(21+ I)R' IA,p, (eos0) = M cosO,so A, = 0 (I '11), and 3Al = M =>Al = 3.
M
M
M.
1
ThusWin(r,O)= 3reosO = 3z, and henceHin = -V~n = -3z = -3M, so
B = ~(H+M) =~ (-~M+M)
=1~JLOM.I./
CHAPTER 6. MAGNETOSTATIC
118
Problem
fH.
FIELDS IN MATTER
6.16
= Ilene = I, so H = 2~8cP.
dl
Jb
= VxM = ~~
S as
B
= JLo(l+ Xm)H =
( )
SXmI
JLo(l+Xm)
I
A
I
2.,'1>.. M
XmI
.
211"s z = [QJ
xn
Kb = M
=
A
I
Z,A
-21Ta
XmI
{ 21Tbz,
~
.
2~s
S - a'
XmH
~ I xmJ.1
at
'
at r = b.
.
Total enclosed current, for an amperian loop between the cylinders:
I + XmI211"a= (1 + Xm)I,
21I"a
Problem
f
so
B' dl
= JLolenc = JLo(1+ Xm)I::}
B = JLo(12+
11"
SXm)I cPo,(
6.17
From Eq. 6.20: fH.
dl
I(S2/a2),
{I
= H(211"s) = Ilene=
(s < a);
(s > a).
/La(1+Xm)Is
H =
Jb
Kb
h
= XmJI
2~:2' (s < a) ,
(s » a }
{ -L
21T8 '
(Eq. 6.33), and JI = ~,
so Jb
I
;;:
so B = JLH=
I
= ::;
I
(s
21Ta2,
< a)j
(S > a).
21TS'
{ I!:SJl.
(same direction as I).
= M X fi. = XmH X fi. ::} Kb =
(opposite direction to I).
= Jb(ira2) + Kb(211"a)= XmI - XmI = [Q] (as it should be, of course).
I
I
Problem 6.18
By the method of Prob. 6.15:
For large r, we want B(r,O) -+ Bo = Boz, so H = /La
...LB -+ ...LBoz,
and hence W -+
/La
LBorcosO.
/La
"Potentials"
LBoz
/La
=
:
Win(r,O) = L:Alrlpl(cosO),
{ Wout(r,O) = -~aBorcosO+L:r~l~(cosO),
(r < R)j
(r>R).
Boundary Conditions:
(i)
Win(R,O)= Wout(R,O),
-
-
I/. 8Wout
I/. 8W;n
{ (ii)
,..0 8r R + ,..
8r R - 0.
(The latter follows from Eq. 6.26.)
I
I
(ii)
::}
JLo[:0 BocosO+ 2)1 + 1):~2 ~(COSO)]+ JLLIAIRI-I
Pt(cosO) = O.
For 1 ¥- 1, (i) ::} BI = R21+IAI, so [JLo(l+1) + JLl]AIRI-1 = 0, and hence AI = O.
For 1 = 1, (i) ::} Al R = /La
L BoR + Bd R2, and (ii) ::} Bo + 2JLoBd R3 + JLAI = 0, so Al = -3Bo
3Bo
3Boz
Win(r,O) =-(2 JLo+JL
)rcosO=-(2
B-
H-
JLo+JL
).
3Bo
Hin=-VWin=
1 + Xm
3/-LBo -
- /-L - (2JLo+ /-L)-
I
A
(2/-Lo+JL
)z=
(1 + Xm/3 )B o.
/ (2/-LO
3Bo
(2JLo+/-L
).
+ /-L).
119
By the method of Frob. 4.23:
= XmHo = I'o(~';m)
Bo. This magnetization
2 Xm
2
Bo = -~Bo
3 1 + Xm
3
(where ~ ==~l+Xm' )
Step 1: Bo magnetizes the sphere: Mo
the sphere given by Eq. 6.16:
2
3
B1 = -/-LoMo= Step 2: B1 magnetizes
the sphere:
the sphere an additional
2
B2
amount
2
This sets up an additional field in
2
2~
= 3"/-LoM1 = 3"~Bl =
= ~B1.
1'0
M1
sets up a field within
( )
3"'
Bo,
etc.
The total field is:
2
B
= Bo + B1
2
[
+ B2 + .. . = Bo + (2~/3)Bo + (2~/3) Bo + . .. = 1 + (2~/3) + (2~/3) +...
1
3
3 + 3Xm
- 3(1 + Xm)
B 1 - 2~/3 - 3 - 2Xm/(1+ Xm) - 3 + 3Xm- 2Xm- 3 + Xm ' so
I
] Bo =
Bo
(1 - 2~/3)'
1 + Xm
(1 + Xm/3) B o.
Problem 6.19
~m = -~:::B;
M = ~?" = - 4~:~B,
where V is the volume per electron. M
2 2
=
XmH (Eq. 6.29)
= J.!O(~Xm)
B (Eq. 6.30). So Xm = - 4"m:v /-Lo. [Note: Xm « 1, so I won't worry about the (1 + Xm)
term; for the same reason we need not distinguish B from Belse, as we did in deriving the Clausius-Mossotti
equation in Frob. 4.38.] Let's say V = t1l"r3. Then Xm = -
~ (4~:r).
I'll use 1 A= 10-10 m for r.
Then Xm = -(10-7) (4(9~1(~':OX_13~~t:J~10))
= 1-2 X 10-5, which is not bad-Table 6.1 says Xm = -1 X 10-5.
However,I used only one electron per atom (copper has 29) and a very crude value for r. Since the orbital
radius is smaller for the inner electrons, they count for less (D.m rv r2). I have also neglected competing
paramagnetic effects. But never mind. .. this is in the right ball park.
Problem 6.20
Place the object in a region of zero magnetic field, and heat it above the Curie point-or simply drop it on
a hard surface. If it's delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it
back and forth many times; each time you reverse the direction, reduce the field slightly.
Problem 6.21
I
(a) Identical to Frob. 4.7, only starting with Eqs. 6.1 and 6.3 instead of Eqs. 4.4 and 4.5.
(b) Identical to Frob. 4.8, but starting with Eq. 5.87 instead of 3.104.
(c) U = -~~[3COSO1 COS02- COS(02- 01)]m1m2. Or, using COS(02- Od = COS01COS02- sin 01 sin O2,
/-Lomlm2
U = -~
411" r
.
.
(smOl sm02 - 2COSOl COS02) .
Stableposition occurs at minimum energy: g~
= g~ = 0
g~ = 1'°4:~;;'2(cos 01sin O2+ 2 sin 01 cas O2) = 0 => 2 sin 01 cos O2 = - cas 01 sin O2;
{ g~ = 1'°4:~;;'2(sin 01 cas O2+ 2 cos 01 sin O2)= 0 => 2 sin 01cos O2= -4 COS01 sin O2,
6. MAGNETOSTATIC FIELDS IN MATTER
CHAPTER
120
Thus
= sin
sin ()1 cos ()2
()2 cos ()1
= sin()2 = 0:
cos ()1 = COS ()2 = 0 :
Either sin()1
= O.
or
{
Which of these is the stable minimum? Certainly not @ or @-for
know m2 will line up along B1. It remains to compareG)
~CD~
t t
or ~@f-
t
or
@
I
+
@
these m2 is not parallel to B1, whereas we
(with ()1 = ()2
= 0) and@
(with ()1 =
rr/2, ()2= -rr/2):
U1 = /.L°4:~~2
(-2); U2 = /.L°4:~~2
(-1). U1 is the lower energy, hence the more stable configuration.
~
, Conclusion: They line up parallel, along the line joining them: ~
.
(d) They'd line up the same way: ~
Problem
6.22
F
~
(I dl)x Bo + II
= If dl x B = I
~
~
~
~
dl x ((r. Vo)Bo]- I (I dl) x ((ro. Vo)Bo]= II dl x ((r. Vo)Bo]
(because f dl = 0). Now
(dl X BO)i
= L Eijkdlj(Boh,
and (r. Va) =
j,k
Fi
L
=
I
=
L
I L
k,l,m
J,k,l
Eijk
I
[I rl dlj]
Lemma 2:
EijkEljmam(VO)l(Boh
{
j,k,l,m
-
=
(OilOkm
- OimOkl) am(Vo)l(Bo)k
so
1
1 rl dlj = Lm Eljmam
{ Lemma 1:
((VoMBoh]
L rl(Vo)l,
LEijkEljm
= Oillhm
(proof belOW). }
- OimOkl (proof belOW).
}
j
= IL (ak(VoMBoh- ai(Voh(Bo)k]
k
I((Vo)i(a.Bo)-ai(Vo.Bo)].
But V 0 . Bo = 0 (Eq. 5.48), and m = I a (Eq. 5.84), so F = Vo(m. Bo) (the subscript just reminds us to take
the derivatives at the point where m is located). Qed
Proof of Lemma 1:
Eq. 1.108 says f(c . r) dl = a x c = -c x a. The jth component is L:p f cprp dlj = - L:p,m EjpmCpam' Pick
cp = Opl(Le. 1 for the lth component, zero for the others). Then f rl dlj = - L:m Ejlmam = L:m Eljmam. Qed
Proof of Lemma 2:
= 0 unless ijk and ljm are both permutations of 123. In particular, i must either be I or m, and k
must be the other, so
EijkEljm
L EijkEljm ==AOilOkm + BOimOkl.
j
To determine the constant A, pick i = I = 1, k = m = 3; the only contribution comes from j
E123E123
To determine B, pick i
= 1 = AOllO33 + BO13031 = A => A = 1.
= m = 1, k = I = 3:
E123E321
= -1 = AO13031+ BOllO33= B => B = -1.
So
L
j
EijkEljm
= OilOkm -
OimOklo
Qed
= 2:
121
Problem
6.23
(a)The electricfieldinsidea uniformlypolarizedsphere, E = - 3~0P (Eq. 4.14)translates to H = - 3~0 (JLoM)
-l M. But B = JLo (H + M). So the magnetic field inside a uniformly magnetized sphere is B = JLo (-l M + M)
I
~/-LOM
I
=
=
(same as Eq. 6.16).
(b) The electric field inside a sphere of linear dielectric in an otherwise uniform electric field is E = 1+~e/3 Eo
(Eq.4.49). Now Xe translates to Xm, for then Eq. 4.30 (P = EoXeE) goes to JLoM = JLoXmH, or M = XmH
(Eq. 6.29). So Eq. 4.49 =} H = 1+x1m/3Ho. But B ~ JLo(1+ Xm)H, and Bo = JLoHo (Eqs. 6.31 and 6.32),
so the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field is
.
1 + xm
B
1
Bo
Bo (as In Prob. 6.18) .
/-La
(1 + Xm ) = (1 + Xm /3)' JLo or B = 1 + Xm /3
(
I
)
(c) The average electric field over a sphere, due to charges within, is Eave = - 4';'0 -h. Let's pretend the charges
are all due to the frozen-in polarization of some medium (whatever p might be, we can solve V.p = -p to find
the appropriate P). In this case there are no free charges, and p = IP dT, so Eave = - 4';<0b IP dT, which
translates to
1 1
1
Have =
- 47rJLo R3
!
JLoMdT
=-
47rR3m.
-~p
But B = JLo(H+ M), so Bave =
+ JLoMave,and Mave = !~3' so lEave = **,
I in agreement
with Eq. 5.89. (We must assume for this argument that all the currents are bound, but again it doesn't really
matter, since we can model any current configuration by an appropriate frozen-in magnetization. See G. H.
Goedecke, Am. J. Phys. 66, 1010 (1998).)
Problem 6.24
Eq. 2.15:
E
=
p {4';<0 Iv #z dT'}
Eq. 4.9 :
V
=
p.
Eq. 6.11:
A
=
JLo€oM x { 4';<0 Iv #z dT' }
{ 4';<0
(for uniform charge density);
Iv #z dT' }
Fora uniformly charged sphere (radius R):
(for uniform polarization);
{
(for uniform magnetization).
Ein
=
Eout
=
p -t-r)
.
~o R3
P ~ 3<0~r
of a uniformly
magnetized
sphere is:
)
(Ex. 2.2).
= -t-(p. f),
~n
Sothe scalar potential of a uniformly polarized sphere is: { V,out
and the vector potential
(.Prob. 2.12),
~
-A,
m
{ A out
10 R3 (p
3<0~
. r ),
=
x r)
'
(M X r ),
~(M
~~
--
3
R3
~
~
(confirmingthe results of Ex. 4.2 and of Exs. 6.1 and 5.11).
Problem 6.25
(a) Bl
= ~~z
(Eq. 5.86,with ()= 0). So m2.B1 = -~r;:.
F = V(m.B) (Eq. 6.3) =} F = !1z[-~r;:]
z=
3t~~2z. This is the magnetic force upward (on the upper magnet); it balances the gravitational force downward
(-mdgz):
3JLom2
3JLom2
27rZ4
-
mdg
= 0 =} z = [ 27rmdg]
I
1/4
.
l
CHAPTER 6. MAGNETOSTATIC FIELDS IN MATTER
122
(b) The middle magnet is repelled upward by lower magnet and downward by upper magnet:
3pom2
21TX4
-
3pom2
21Ty4
- md9 = O.
The top magnet is repelled upward by middle magnet, and attracted
3J.Lom221Ty4
I
1
I
S bt
t . . 3/ R) is
B
(
)
1
411" 2J.to+ J.t r3 [3(m. f)f - m]
.
Problem 6.28
The problem is that the field inside a cavity is not the same as the field in the material itself.
(a) Ampere type. The field deep inside the magnet is that of a long solenoid, Bo ~ J.toM. From Prob. 6.13:
Sphere: B = Bo -- ~J.toM = iJ.toM;
Needle: B = Bo - J.toM= 0;
B = J.toM.
{ Wafer:
(b) Gilbert type. This is analogous to the electric case. The field at the center is approximately that midway
between two distant point charges, Bo ~ O. From Prob. 4.16 (with E --t B, 1/Eo --t J.to,P --t M):
Sphere: B = Bo +'M = iJ.toM;
Needle: B = Bo = 0;
Wafer:
B = Bo + J.toM= J.toM.
{
In the cavities, then, the fields are the same for the two models, and this will be no test at all. Yes. Fund it
with $1 M from the Office of Alternative Medicine.
I
I
Chapter 7
Electrodynamics
Problem 7.1
(a) Let Q be the charge on the inner shell. Then E = 4;EO
~f in the spacebetweenthem, and (Va- Vb)=
L
fa J dr - ...sL(1 - 1 )
41rEO Q Jb r~
- 41rEO a b'
- JbfaE. dr --
I. =I
J da
1=
(b) R
= Va -I
v"
E. da = u Q
u
= u 47r€o(Va-
100
100 (I/a
-
=
v,,)
I/b)
- Vb)
(I/a - I/b)
47rU (Va
1
.
!- ! .
=~
47ru a
b
( )
I
(c)For large b (b » a), the second term is negligible,and R = I/47rua. Essentially all of the resistance is in
theregionright around the inner sphere. Successiveshells, as you go out, contribute less and less, because the
1
2
= 2 1rua
cross-sectionalarea (47rr2)gets larger and larger. For the two submer ged spheres, R = 4 1rua
(one R as
_
_
the current leaves the first, one R as it converges on the second). Therefore I
Problem 7.2
(a) V
= Q/C = IR.
= V/ R =
I
27ruaV.I
Because positive I means the charge on the capacitor is decreasing,
~~ = -I = - RICQ,so Q(t) = Qoe-t/RC. But Qo= Q(D)= avo, so Q(t) = CVoe-t/Rc.1
I
Hence let)
dQ
1
=
= --'-dt = CVi0 -e-t/RC
RC
(b)W=I!cVl.1
~ (- ~C e-2t/RC)
~
o
-e-t/RC
R'
roo
roo
[=
~CV02.
IQ(t)
=>
= avo
In(Q
roo
./
(c) Vo= Q/C + IR. This time positive I means Q is increasing: ~~
1
-liCdt
~2
The energydeliveredto the resistoris Jo Pdt= Jo 12Rdt= ~ Jo e-2t/RCdt=
- aVo) = -
(1 - e-t/RC)
.1
1
RCt + constant
let)
= ~~ = avo
=>
Q(t)
= avo
(Rlce-t/RC)
125
= I = RIC(CVo - Q)
+ ke-t/RC. But Q(O) = 0 =>
=
I
~ e-t/Rc.1
=>
k
Q
~~Vo
=
= -avo, so
CHAPTER 7. ELECTRODYNAMICS
126
roo
V,2
V,2
~ Jo e-t/RCdt= ~ (-RCe-t/RC)lo = ~RC=ICV02.1
V,2
rOO
(d) Energy from battery: Jo VoIdt=
00
Since I(t) is the same as in (a), the energy delivered to the resistor is again I ~CV02
.1 The final energyin
the capacitor is also ~ cvl,
I
so half the energy from the battery goes to the capacitor, and the other half
I
I
to the resistor.
Problem 7.3
I
(a) I = IJ . da, where the integral is taken over a. surface enclosing the positively charged conductor. But
J = O'E, and Gauss's law says IE. da = (10
Q, so I = a IE. da = tiQ. But Q = CV, and V = IR, so
6£]_
0
1= !LCIR, or R = a C '
(0
(b) Q
= CV = CIR
time constant
is T
=>
~
= RC =
Problem 7.4
qed
= -I = - JcQ => Q(t)
I
~
1= J(s) 27rsL => J(s)=I/27rsL.
V = -la E.dl= -~(a-b).
Problem 7.5
I=~;
P=I2R=
r+R
= Qoe-t/RC
1
' or, since V = Q/C, V(t) = Voe-t/RC. The
E=J/O'=I/27rsO'L=I/27rkL.
SOIR= ~.I
1
2R
£2R . dP _£2
(r + R)2' dR [ (r + R)2 - (r + R)3 ] = 0 =>r + R = 2R => R
I
= r.1
Problem 7.6
£ = f E . dl = Izero Ifor all electrostatic fields. It looks as though [; = f E . dl = (a /£o)h, as would indeed
be the case if the field were really just a/Eo inside and zero outside. But in fact there is always a "fringing
field" at the edges (Fig. 4.31), and this is evidently just right to kill off the contribution from the left end of
the loop. The current is
Problem 7.7
(a) £
= -~r
~
= -BZ~~ = -Blv; £ = IR =>II = B~v.1 (Never mind the minus sign-it just tells you the
direction of flow: (v X B) is upward, in the bar, so downward through the resistor.)
(b)F=IlB=~,
(c) F
B2Z2V
I
to the left.
I
I
I
= ma = m-dv
=>-dv
=> v = voe--;;;]'f"t.
B2Z2
dt = --vB2Z2
R
dt = -( -)v
Rm
82,2
I
I
(d) The energy goes into heat in the resistor. The power delivered to resistor is 12R, so
dW
-
dt
=I2R=
B2Z2V2
-R=
R2
B2Z2
-v2e-2at
R
0'
wherea=-.
-
00
The total energy delivered to the resistor is W
= amv5 10
B2Z2
dW
mR'
(it = amv5e-2at.
-2at
e-2atdt
= amv5 ~ 2
- ao
00
1
1
1
= amv5 _
2a = 2-mv5.
I
127
Problem 7.8
s+a
..
/-LoI
(a) The field of long wire IS B
A
= 211"8
cp,so
=
s+a
,and
dt
211"dt
8
The field points out of the page, so the force
the far side it's also to the right, but here the
(b) [;=_d=_/-Lola~ln
( )
!
/-LoI
!
B. da = 211"
s
1
8
/-Lola
-;(ad8) = ~
+a
( ).
In ---;-
I
~dS
_~d8 = /-Lola2v.
dt
211" 8 + a dt
8 dt
211"8(8
+ a)
on a charge in the nearby side of the square is to the right. In
field is weaker, so the current flows counterclockwise.
(
d8 =v,so_/-Lola
)
I
(c) This time the flux is constant, so [; = 0.1
Problem 7.9
Since V.B = 0, Theorem 2(c) (Sect. 1.6.2) guarantees that fB
boundary line.
Problem 7.10
= B . a = Ba2 cos ()
I
I
Here () =
. da is the
(view from above)
wt, so
[= -4Jt = -Ba2( - sinwt)w;
B=~~
1[;= Bwa2 sin wt.1
Problem 7.11
[; = Blv
downward:
= IR
=>
I
dv
= IJiv => upward
= m dv
dt;
mg - Rv
B2l2
-
1
= dt => --In(g
a
g-av
same for all surfaces with a given
dt
dv
magnetic force
=9 -
- av)
= IlB
= B;t v. This opposes the gravitational force
a=
av, where a =
mR'
- B2l2
= t + const. =>9 -
9 - aVt = 0 =>Vt = 9
I
B2[2'
mgR
I
av = Ae-at; at t = 0, v = 0, so A = g.
av = g(l - e-at); v = ~(1- e-at) = vt(l - e-at).1
I
At90%of terminal velocity,v/Vt = 0.9 = 1 - e-at =>e-at = 1 - 0.9 = 0.1; In(O.l) = -at; In 10 = at;
t = lln
10,or
a
I
t90%
= Vt9 In 10.
.
Now the numbers: m = 41].4.l,where 1]is the mass density of aluminum, A is the cross-sectional area, and
I is the length of a side. R = 4l / Ao-, where 0- is the conductivity of aluminum. So
P = 2.8 X lO-s n m
- 41]Alg4l - 161]g- 16g1]p
Vt - Ao-B2[2- o-B2 SoVt = (16)(9.S)(2.7Xi03)(2.SXIO-S)
= 11.2
cm/s;
B2 ,an
I t90%
d
9
= 9.8 m/s2
1]= 2.7 X 103kg/m3
.
}
{ B= IT
= 1.2~.lSO-2In(10) = 12.8ms.!
If the loop were cut, it would fall freely, with acceleration g.
CHAPTER 7. ELECTRODYNAMICS
128
Problem
7.12
a
2
q,= 7r (2)
Problem
= TBo
E
7ra2
7ra2UJ
= -di = TBoUJsin(UJt).I(t) = Ii =
J B dx dy = kt2 loa dx loa y3 dy = ~ kt2 a5 .
l
.
4R Bo sm(UJt).
E
=-
~~
= -I ~kta5 .1
7.14
Suppose the current (1) in the magnet flows counterclockwise (viewed from
above), as shown, so its field, near the ends, points upward. A ring of
pipe below the magnet experiences an increasing upward flux, as the magnet
approaches, and hence (by Lenz's law) a current (find) will be induced in it
such as to produce a downward flux. Thus Iind must flow clockwise, which is
opposite to the current in the magnet. Since opposite currents repel, the force
on the magnet is upward. Meanwhile, a ring above the magnet experiences
a decreasing (upward) flux, so its induced current is parallel to I, and it
attracts the magnet upward. And the flux through rings next to the magnet
is constant, so no current is induced in them. Conclusion: the delay is due
to forces exerted on the magnet by induced eddy currents in the pipe.
~
pIpe-+-
~
falling
magnet
J~\
-find
ring
Problem
I
cos(UJt); E
7.13
q, =
Problem
dq,
7ra2
B
7.15
..
..
/LonIZ,
h
B
n t e quasIstatIc approxImatIOn,
= { 0,
(s < a);
(s > a).
Inside: for an "amperian loop" of radius s < a,
q, = B7rS2= /LonI7rs2;
f
dq,
dI
E. dl = E27rs = -di = -/Lon7rs2dt;
I
2 dI
dt cjJ.
E = - /Lons
.~
A
I
Outside: for an "amperian loop" of radius s > .a:
dI
q, = B7ra2 = /LonI7ra2; E27rs = -/Lon7ra2 dt;
Problem
E = _/Lona2dI
~
A
dt cjJ.
7.16
(a) The magnetic field (in the quasistatic approximation) is "circumferential".
in a solenoid, and hence the field is longitudinaL
I
This is analogous to the current
--lrr-
I
(b) Use the "amperian loop" shown.
Outside,B = 0, so here E = 0 (likeB outside a solenoid).
.
So § E dl = El = - ~r =
!!.
... E = _l!:JJ..df
211"
dt In ( s ) . But
/Lo10UJ
so IE
=-
27r
- ft fB
df
dt
a
sin(UJt)In -S z.
()
.da = - ft f:
= -LoUJsinUJt'
A
f~:, l ds!
I
z
129
Problem 7.17
(a) The field inside the solenoid is B
= /LonI.
So ~
= 7ra2/LonI
~ £
= -7ra2/Lon(dIJdt).
In magnItude,
then, £ = 7ra2/Lonk. Now £ = IrR, so Iresistor= 7ra2/Lonk
R
.
.
B is to the right and increasing, so the field of the loop is to the left, so the current is counterclockwise, or
I
I
Ito the right, through the resistor.
I
(b) LH) = 27ra2/Lonl; 1
= dt
dQ = Ii
£ = -lidt
1 d~
~ D.Q= IiD.~,
ill magmtude. SO
1..
I
D.Q = 27ra2/Lonl
R
.
I
Problem 7.18
J. -
'¥
-
fB
. d a,. B --
/LO1 :i.. .To.- /Lola
0/,
'¥ 27rs
27r
.
dQ
2a ds
1a
S
-- /LolaIn 2., c..co R -- dQ R -.Lloop
27r
dt
T
= _/L~:~2 dl
IQ =
~
- d~ -- - /LoaIn 2 dl .
dt
27r dt
1/L~::2.
The field of the wire, at the square loop, is out of the page, and decreasing, so the field of the induced
current must point out of page, within the loop, and hence the induced current flows counterclockwise.!
I
Problem 7.19
..
..
~;p
In the quasistatIc approxImatIOn B =
271"8' (inside toroid)',
,
(outside toroid)
{ 0,
(Eq.5.58). The flux around the toroid is therefore
~
= /LoNI
27r
ra+w !hds
}a
= /LoNlh
S
27r
In 1 +
(
W
a
)~
/LoNhw I.
27ra
d~
dt
= /LoNhw
27ra
dI
dt
= /LoNhwk
27ra
Theelectricfield is the same as the magnetic field of a circular current (Eq. 5.38):
B
= J.LoI
a2
2 (a2 + Z2)3/2 z,
with (Eq. 7.18)
I~_~d~
=_Nhwk.
/Lodt
27ra
SoE=/Lo
2
_NhWk
(
27ra
) (a2 + Z2)3/2z=
a2
I
-/LO Nhwka z.
47r(a2 + Z2)3/2
Problem 7.20
(a) From Eq. 5.38, the field (on the axis) is B = ¥
.
I
IS IIi
(b2+~22)3/2Z,
so the flux through the little loop (area 7ra2)
/LO7r/a2b2
= 2(b2 + z2)3/2 . ,
=~
(b) The field (Eq. 5.86) is B
?-(2 cos0 r + sin0 9), where m = l7ra2. Integrating over the spherical "cap"
(bounded by the big loop and centered at the little loop):
~
=
!
B. da = ~: 1;:2 !(2COSO)(r2sinOdOdc/J)= /Lo~a2 27r 16 cosO sin 0 dO
CHAPTER 7. ELECTRODYNAMICS
130
where r
= Vb2 + Z2 and
sin B
e
. 2
Evidently cI>= /10r7ra 8m2(J 0 =
I
= b/r.
2
I
J.lO7rIa2b2
I
I
2(b2 + Z2)3/2 ' the same as in (a)!!
J.lO7ra2b2
= M12h
(c) Dividing off I (cI>1
Problem
7.21
E = --
dcI>
dI
=
-M dt
dt
= M21h):
cI>2
M12
I
= M21 = 2(b2 + z2)3/2'
= -Mk.
It's hard to calculate M using a current in the little loop, so, exploiting the equality of the mutual inductances,
I'll find the flux through the little loop when a current I flows in the big loop: cI>= MI. The field of one long
01
wire is B = l!:Q2.
01 =} cI>1= l!:Q2.
07rIa In 2, so the total flux is
27r J.a2a la
2 7rS
S ds = /12
if,.
'£'
= 2 if,. = J.loIaln2
7r
'£'1
=} M =
=} ,,=
7r
J.loaln2
I"
,
7r
J.lokaln21'
ill magnltu
.
de.
Direction: The net flux (through the big loop), due to I in the little loop, is into the page. (Why? Field
lines point in, for the inside of the little loop, and out everywhere outside the little loop. The big loop encloses
all of the former, and only part of the latter, so net flux is inward.) This flux is increasing, so the induced
current in the big loop is such that its field points out of the page: it flows I counterclockwise.!
Problem
7.22
= /-LonI =} cI>1=
B
/-LonI7rR2(flux through a single turn).
In a length 1 there are nl such turns, so the
total flux is cI>= J.lon27rR2
Il. The self-inductance is given by cI>= LI, so the self-inductance per unit length is
1.c=J.lon27rR2.1
Problem 7.23
The field of one wire is B1
= ~ ~,
so
cI>
=2. ~
d-f
. 1 ff ~s = /1~IlIn (d;f).
The € in the numerator is
negligible (compared to d), but in the denominator we cannot let € -+ 0, else the flux is infinite.
L
= J.lolln(d/€)
I.
7r
Problem
Evidently the size of the wire itself is critical in determining L.
7.24
(a) In the quasistatic approximation B
= 2J.lO;p.
7rS
= /-L
20I
7r
SO cI>1
(b ~hds
Ja s
= J.loIh
In(b/a).
27r
This is the flux through one turn; the total flux is N times cI>1:cI>= J.lO;:h In(b/a)Io COS(LVt).So
E
==
I
d
In(2)(0.5)(27r60)sin(LVt)
dcI>
t = J.loNh
27r In(b/a)IoLVsin(LVt) = (47r x 1O-7~(1O3)(1O-2)
7r
E 2.61 X 10-4 .
( )
)
/
. volts, where LV= 27r60 = 377 s. Ir = R =.
2.61 x 10- 4 sin(LVt) (ill
500.
sm LVt
I
= 15.22 x 10-7 sin(LVt)I (amperes).
(b) Eb
Therefore
= -L ~;; where (Eq. 7.27) L = /1°I:rr2h
In(b/a) = (47rXl0-7~~O6)(10-2)
In(2) = 1.39 x 10-3
Eb = -(1.39 x 10-3)(5.22 x 10-7 LV)COS(LVt)= 1-2.74 x 10-7 COS(LVt)I (volts).
(henries).
131
. amphtudes:
.
RatiOof
Problem 7.25
2.74
10-7
2.61 X
x 10-4
= I 1.05 x 10-31= /-LoN2lu.J
27rR In (b/ a).
With I positive clockwise, £ = -L~:
~
= -lcQ
Q = CV, so A
= Q/C, where Q is the charge on the capacitorj
I
= _VJ2Q, where VJ = k. The general solution is Q(t) = AcosVJt + BsinVJt.
1(t) = ~ = -AVJsinVJt + BVJsinVJt. At t = 0, 1= 0, so B = 0, and
= ~,
so
At t = 0,
= CVj
1(t)= -CVVJsinVJt= I-v/f
sin
(Jfc)
'.
If youput in a resistor, the oscillation is "damped". This time -L~: =
§ +1R, so L~
+R~ + bQ =0.
For an analysis of this case, see Purcell's Electricity and Magnetism (Ch. 8) or any book on oscillations and
waves.
Problem 7.26
(a) W
= ~L12.L = /-Lqn27rR2l(Prob. 7.22)
I
W
= ~/-Lon27rR21I21.
(b) W = ~ f(A. I)dl. A = (/-Lon1/2)R~, at the surface (Eq. 5.70 or 5.71). So WI = ~JL02nI
R1. 27rR,for one
turn. There are nl such turns in length l, so W = ~/-Lon27rR21I2.,(
(c) W = 2~0 JB2dr.
B = /-Lon1,inside, and zero outside; Jdr = 7rR2l, so W = 2~0/-L6n2127rR2l
=
!J.ton27rR2l12.,(
(d) W = 2~0 [JB2 dr - f(A X B) . da]. This time B2 dr = /-L6n2
127r(R2 - a2)l. Meanwhile,
A X B = 0 outside (at 8 = b). Inside, A = JL02nI
a ~ (at 8 = a), while B = /-Lon1 z.
Ax B= ~/-L6n212a(~X z)
J
~
points inward ("out" of the volume)
I
S
i;
B) .da = J(~/-L6n212as)'[ad4>dz(-s)] = -~/-L6n212a227rl.
W -- -L
1I2n2127r(R2 - a2) l + 112n2127ra2l ] - 111 n2 12R27rl ,(
21'0 (,..0
,..0
- 2,..0
.
f(A
X
..21
Problem
7.27
B = /-Lon1;
27rs
W
,L= ~n2hln(b/a)
Problem
= ~2/-LoJ B2dr = ~/-L6n212
2/-Lo 47r2
I
J 82
~ =
()
a
1
47r/-Lon2 12 h In(b/a).
(same as Eq. 7.27).
7.28
2
f
= /-Lon212
h27rln
87r2
~hrd4>d8
2
/-Lo18
. dl = B(27r8 ) = /-Lo1enc= /-Lo1s( / R) => B = 27rR2 .
W = 2B2 dr = ~ /-L612 (R s2(27r8)l ds = /-Lo12l(84) R = /-Lol12 = ~L12.
2/-LoJ
2/-Lo47r2R4 10
47rR4 4 0
167r
2
B
I
So L
= ~l, and.c= L/l =
I
/-Lo/87r,
I
independent of R!
Problem 7.29
(a) Initial current: 10 = £o/R.
(b) P
So -L ~~ = 1R =>~~ = - ~ I =>1= 1oe-Rt/L, or 1(t)
I
= 12R = (£0/ R)2 e-2Rt/L
w = £6 roo e-2Rt/ L dt = £6
R 10
R
R
= ~ e-2Rt/L = d:.
(- ~e-2Rt/ L)
2R
1
00
0
= £6(0 + L/2R) = ~L (£0/R)2 .
R
2
= ~e-Rt/L.I
CHAPTER
132
(c) Wo
7. ELECTRODYNAMICS
= ~LI5=! (£ofR)2. .(
Problem 7.30
= ~r\Id3(al
CP2= B1 . a2 =
4 n.
(a) B1
.i)i
- a1]' since m1 = I1a1. The flux through
loop 2 is then
4n
1M
fJ-o
3[3(a1.i)(a2 . i) - a1 . a2].1
(b) £1 = - M %,
d:; 11= -£lIt = M It %. (This is the work done per unit time against the mutual emf in
loop I-hence the minus sign.) So (since It is constant) WI = MItI2, where 12 is the final current in loop 2:
fJ-O
~ It [3(a1
.
.i)(a2 .i) - a1 . a2] = MIl.
.
=
.
= ~[3(m1
i)(m2 i) - m1 m2].1
Notice that this is opposite in sign to Eq. 6.35. In Prob. 6.21 we assumed that the magnitudes of the dipole
moments were fixed, and we did not worry about the energy necessary to sustain the currents themselves-only
the energy required to move them into position and rotate them into their final orientations. But in this
problem we are including it all, and it is a curious fact that this merely changes the sign of the answer. For
I
W
commentaryon this subtle issuesee R. H. Young,Am. J. Phys. 66,
there,
Problem
1043 (1998), and
the referencescited
7.31
The displacement current density (Sect. 7.3.2) is Jd
radius s,
B . ell= B . 2ns
yC
Problem
= fO ~~ = f = ~ z. Drawingan "amperianloop"at
I
s2
fJ-oIs2
= fJ-oIdenc = fJ-o~na . ns2 = fJ-oI2"
a =>B = _
2nsa 2 j
B
= 2na2
fJ-oIs
A
B 2ns = fJ-oIa2 => B = 2na2 s B = ~sfJ-o
.Qm + L!::>.ip,where !::>.QmIS the total magnetIc charge
passing through the surface, and!::>.ip is the change in flux through the surface. If we use the flat surface, then
tJ.Qm= qm and !::>,ip
= 0 (when the monopole is far away, ip = 0; the flux builds up to /loqm/2 just before
it passes through the loop; then it abruptly drops to -/loqm/2, and rises back up to zero as the monopole
disappears into the distance). If we use a huge balloon-shaped surface, so that qm remains inside it on the far
side, then !::>.Qm
= 0, but ip rises monotonically from 0 to /loqm' In either case,
/ = /l01m.1
- --
II:""
CHAPTER
134
Problem
7.37
V
1
V
aD
a
E = d:::}Jc = aE = pE = pd' Jd = at =
at
a
=
(fE)
7. ELECTRODYNAMICS
fVO
Va COS(271"vt)
fat
d
[
]
.
= d
[-271"vsm(271"vt)].
The ratioof the amplitudes istherefore:
Jc
Vo
=
Jd
Problem
The
d
1
pd 271"VfVo
=
271"Vfp
=
12.41.1
7.38
potential
and
field in this configuration
are
identical
1= !J.da=a
where
=
[271"(4x 108)(81)(8.85 x 1O-12)(O.23)r1
the integral
is 0"'"'
the hemisphedca1
surface
jnst
to those
in the
upper
half of Ex.
3.8. Therefore:
!E.da
outside
the sph",e.
Qm....
~
But I can with impunity close this surface:
~
(because E = 0 down there
anyway-inside a conductor).
~.......
So I = a E.da = !l...Qenc
= !l...
ae da, where ae is the electric charge density on the surface ofthe hemisphereEO
EO
to wit (Eq. 3.77) ae = 3fOEo cosO.
J
J
a
1= -3fOEo
~
!
cosOa2sinOdOdcjJ = 3aEoa2271"
~/2
sin 0 cosO dO = 3aEO7l"a2.
~
0
,
~
Sin2(}
2
But in this case Eo
1
'
~/2_!
0
-
2
= Void, so 1= 3a7l";oa2.1
I
z
Problem 7.39
Begin with a different problem: two parallel
wires carrying charges +>' and ->. as shown.
Field of one wire: E
Potential
or V(y, z)
=_
2
>.
~EOS
of combination: V
-
>.
4~EO In
y
8; potential: V
=
-- 2 >' In (s / a) .
~EO
x
= 2;EO In(L/s+),
{
(Y+br+Z2
(y-b 2+Z2
}.
Find the locus of points of fixed V (Le. equipotential surfaces):
e4~EOV/>'
==J..t=
(y+b)2+z2
=:::}J..t(y2- 2yb + b2 + Z2)
(y - b)2 + Z2
y2 (J..t- 1) + b2(J..t- 1) + z2 (J..t- 1) - 2yb(J..t+ 1)
(y - b(3)2+ z2 + b2 - b2(32= 0
=0
=:::}
=:::}
= y2 + 2yb + b2 + Z2;
y2 + Z2 + b2 - 2yb(3= 0
(y - b(32)+ Z2 = b?((32- 1).
((3 ==~
~ ~) ;
135
This is a circle with center at Yo = b(3 = b(I<+I ) and radius = b
2 - 1 =
,
1<-1
Y IF - .1
This suggests an image solution to the problem at hand. We want Yo =
determine the parameters b, J.l,and>' of the image solution:
~
b l:d:!
= ( It-I ) = J.l + 1.
~=~
a
4a2 J.l
J.l
= 4a2 -
= (J.l+
2::1:y4(1-
.
radius
2a2)2 - 4
= 2a2 -
!.
.
..
per Unit Iengt h IS ~
J
I
da
=
0'
E. da
= O'..!..Qenc= ~>.l.
EO
EO
411'0'Vo
(2a2 -
~
1 ::I:2av a2 - 1)
,
'
.
Wh lCh sIgn d 0 we want, ? Suppose
- ~ +.. .
= 2a2 -1::1: 2a2 1- ~
2a2
8a4
[
J
() = 2a2 -1::1: 2a2yl-1/a2
= 2a2(1::1: 1) -
r;;--.
4a2 + 4a4 - 1 = 2a2 - 1::1:2ay a2 - 1;
'
.
.
. Th at ' s t h e Ime ch arge m t he Image pro blem.
0'>.
the cylindersare far apart, d » a, so that a » 1.
It-I
= 0;
1::1:yl-
!
= -z = -EO = In
(It-I)'
a -
2J.l+ 1 ===} J.l2+ (2 - 4a2)J.l + 1
~
1 ::I:2ava2
- 1)
(2a2 -
V
d, radius = a, and V = Vo.These
~= a.
Call
2"fji
411'EOVo
= In
1=
Th e current
= J.l2+
1)2
2
47n:oVo I
\
= n J.l===} 1\
>.
2b.,jii
It-I
b /(1<2+21<+1)_(1<2_21<+1)= 2b.,jii.
=
(1::1: 1) =F ~::I:'"
4a
4a2 - 2 -1/2a2
+'"
(+ sign),
(- sign),
~ 4a2
{ -1/4a2
The current must surely decrease with increasing a, so evidently the + sign is correct:
i
=
411'0'Vo
In (2a2 -1
d
+2aya2
-1)'
where a =~.
Problem 7.40
(a) The resistance
of one disk (Ex. 7.1) is dR
disk. The total resistance is
R
=
!!..
{L
11'10
pL
=
1
2 dz
[(bI:a)z+a]
b-
=
!!..
= :A
( ){
dz, where r = (bI:a) Z + a is the radius of the
-1
~
11' b-a
= ~
[(bI:a)z+a]
L
} 0
I
=
pL
1I'(b-a)
[
-
1
(b-a+a)
+
~
aJ
a
( ) =~rPLl
1I'(b-a) ~
(b) In Ex. 7.1 the current was parallel to the axis; here it certainly is not. (Nor is it radial with respect to
the apex of the cone, since the ends are fiat. This is not an easy configuration to solve exactly.)
(c) This time the flow is radial, and we can add the resistances of nested spherical shells: dR
{(J
(J
A = 10 r2 sin () d() d= 211'r2( - cos()) 10 = 211'r2(1 - cos()),
~b
~
L
= ~ dr, where
CHAPTER
136
p
=
R
p(b-a)
27rab
p(b
that if b - a
(1
r2
To
P
rb- ra
.,
.
Now
27r(I-cosO) ( rarb )
-cosO)
=
a
=-b = .
ra
rb
slOB.
=
b-a
./L2
-a)2
L
and cosO
+ (b -a)2
---.
./£2
+ (b - a)2
1
+
[./£2
(b
- a)2 - L]
+
(b
.
1 (b -a)2
«
L,
PL
b'
7ra
I
sinO. .ButsmO
27rab
[Note
-1 dr=
Tb
27r(I-cosO)
7. ELECTRODYNAMICS
then
./£2
2
£2
skbk sin(k4»,
(s <
a);
s-kdk
(s
-
a)2
S:!
L [1 +
'"
]
p(b
-a)2
1
27rab (b- a)2j2L
,and R =
-
as in (a).]
Problem
7.41
00
~n(s,4»
L
=
k=l
From
Prob. 3.23,
00
Vout(s,4»
L
=
sin(k4»,
>
a).
k=l
(We don't need the cosine terms, because V is clearly an odd function of
Let's start with ~n, and use Fourier's trick to determine bk:
= ~o:
~akbkSin(k4»
1: sin(k4»sin(k'4»d4>
1:
)
' and hence ~n(s,
= ajs
for r
00
= 1m k=l
L
1
>
00
= 1m k=l
L
1
Now In (ReiD)
= InR + iO, so S = -0,
.
k
6 k (-~ )
sm(k4».
00 1
-w2 + -w3 - -W4... = - ~ -( -w)k
3
4
6k'
Re (1 + xei
=w -
2
4»
So
s
~ k=l
f ~ ( -~f sin(k4».Bothsumsareof the formS
But In(l + w)
tan 0
At s = a, ~n = Vout= Vo4>j21r.
1: Sin(k4»sin(k'4»d4>=~ 1: 4>sin(k'4»d4>.But
~akbk
=>
4>.)
x(ei-e-i
H(1 + xei
+ e-i.
137
Vo
~n(s,~)
Conclusion:I
j
Vr
Vout(S,~) --
s sin ~
-1
= -;-tan
(a+scos~ ) ,
~
-1
( + acos~) '
a sin
'Tr
Q.tan
S
(s < a);
(s > a).
(b) From Eq. 2.36, u(~) = -100{a~:ut Is=a- a~n Is=a}'
aVout
8S
Vo
1
= -;Vo
= - -;a~n
Vo
as = -;Vo
= -;a~n
as
I
(s+acost/J) ]
1
[ +
{
(
S2
2
asin.p
(-aSin~)
+ acos~)2
}
asin~
+ 2as COB
~+ a2
1
(
)
(
(a +
)
asin~
+ 2ascos~ + a2
= - aVant
as
s=a
=
I
;
Vo
[(a+SCOS~)Sin~-SSin~cos~]
2
ssin.p
1
{ [ + a+scost/J ]
s2
asin~
Va
= --;- [ (s+ acos~)2+ (asin~)2]
(s
s=a
asin~
= -;- [ (a+ SCOS~)2 + (ssin~)2
scos~)2
]
}
).
sin~
2'Tra 1 + cos~
Vo
(
),
so u(~)
= fOVO
=
sin~
'Tra (1 + cos~)
fOVO tan(~/2).
'Tra
I
Problem 7.42
(a) Faraday's law says VxE =
(b) Faraday's
~~ = 0 => B(r) is independent of t.
§ E. ell = -d~/dt. In the wire itself E = 0, so ~ through
-~~, so E = 0 =>
law in integral form (Eq. 7.18) says
the loop is constant.
(c)Ampere-Maxwell=>
V xB =
at the surface.
JLoJ+
JLofo~~' so E = 0, B = 0 =>J = 0, and hence any current must be
(d) From Eq. 5.68, a rotating shell produces a uniform magnetic field (inside): B = !JLoulAJai.So to cancel
such a field, we need ulAJa= -2Po'
3 Bo Now K = uv = ulAJasinOcp,so I K = -2j;;
3Bo sinOcp.I
Problem 7.43
~
~
(a) To make the field parallel to the plane, we need image monopoles of the same sign (compare Figs. 2.13
and2.14),so the image dipole points I down(-z).1
(b) From Prob. 6.3 (with r -+ 2z):
F
= I3JLo m2
2'Tr (2z )4 .
!
3JLo m2
2'Tr (2h)4
= Mg => h =
I
2
3JLom2
1/4
(2'TrMg)
.
CHAPTER
138
m
(c) Using
B
Eq. 5.87, and
= ~; (r~)3 ([3(mz.
referring to the figure:
r1) r1
- mz]+ [3(-mz. r2)r2 + mz]}
A
A
A
= 3/-lom3 [(z.A r1
) r1 - ( z. r2 ) r2.]
411" (r1 )
A
3J.lom
8(
A
A
A
B
)
11"r1
= - 23~O~3
3/-lomh
A
A
A
B u t z" r1
2
AA
.
()
r1 cos()
But sin () = ~,
sin() cos()r.
r
A
= -z
. r2
= cos
()
"
A
ut r1 + r2 = sm r.
= - 411"(rd3
cos r1 + r2.
= -
7. ELECTRODYNAMICS
r1
= ~,
and r1
= vr2
+ h2.
-m
A
211" (r2 + h2)5/2r.
Now B
= /-lo(K x z)
z x B = /-lo z x
=>
= /-lo [K -
(K x z)
= /-loK.
z(K . z)]
(I used the BAC-CAB
rule,
and noted that K. z = 0, because the surface current is in the xy plane.)
K
1
=
3mh
r
3mh
r
:i.
d
(
(
B)
)
/-lo z x
= - 211"(r2 + h2)5/2 z x r = - 27r (r2 + h2)5/2'1'" qe
A
A
A
Problem 7.44
Say the angle between the dipole (md and the z axis is () (see diagram).
z
~+
m1
The field of the image dipole (m2) is
B(z)
= ~; (h
:
z)3 [3(m2
.z)z -
x
m2]
_
for points on the z axis (Eq. 5.87). The torque on m1 is (Eq. 6.1)
h
~
m2
= m1
N
But m1 = m(sin()x + cos()z),
m1 x m2 = 2m2 sin ()cos ()y.
m2
/-lo
x B
= 411"~oh)3[3(m2 . Z)(m.1 x z) -
= m(sin()x - cos()z), so m2. Z =
2 .
A
2
.
+
-mcos(), m1 x Z = -msin()y,
A
N= 411"(2h)3
[3m sm()cos()y - 2m sm()cos()y)]
0
(m1 x m2)] .
/-lom2.
and
A
= 411"(2h)3sm() cos() y.
Evidently the torque is zero for () = 0, 11"/2,or 11""But 0 and 11"are clearly unstable, since the nearby
ends of the dipoles (minus, in the figure) dominate, and they repel. The stable configuration is () = 1r/2:
I
parallel to the surface (contrast Prob. 4.6).
In this orientation, B(z) = - 41Tr~;z)3x, and the force on m1 is (Eq. 6.3):
I
F
=V
2
/-lom
[ 411"(h
+ z)3 ]1 z=h
=
At equilibrium this force upward balances the weight
3
411"(h
+ Z)4 1z=h
= Mg
=>
=
M g:
~
3/-lom2
411"(2h)4
2
/-lom z
h
=
I
2
3/-lom2
(
1/4
)
411"Mg
.
3
2
/-lom
z
411"(2h)4.
139
Incidentally, this is (1/2)1/4
pla~e (Prob. 7.43b).
Problem 7.45
f
=v
X B; v
= U)a sin
= 0.84 times
B (bj f
the height it would adopt in the orientation perpendicular to the
= U)aBosin B((b X z).
t: = If. ell, and ell= a dBO.
X z) . OdB.
But O. «b X z) = z. (0 X (b) = z. f = cosB.
('/2 .
sin2B 11= 2~V(1//2 - y); F = 2. 2~V (1//2 - y)2 = mg (at terminal velocity).
Vdiamond=
mgR
4B2 (1//2 - y)
2.1 (This works for negative y as well as positive, if you replace y by Iyl.)
mgR
2 l 2.
mg R
toward the halfway mark (y
total time it takes for the square to fall is:
Vsquare
Thus. Vd,amond=
In
In
(
)
).
(B ) 4B;- (/1,;2y)' = (v2 0), the "square" falls twice as fast; then the diamond again takes over. The
2y/l
I
2
=_
v square =
tsquare
1
(assuming it always goes at the terminal velocity, which-as
the field is strong). For the diamond, t is
I
At first y -1/v2
I
the "diamond" falls faster;
m gR
B2131
we found in Prob. 7.ll-is
close to the truth, if
0
-
/
dy
=-
Vdiamond
8B2
mgR
/
- 1/v'2 - Y 2 dy
)
(
1/..!2
3
= 8B2 [ ! (1/v'2- Y) )
mgR 3
= 8B2 !~
0
= 2/2 B213 .
1/..!2 mgR 32/2
1
1
3 mgR
Sotsquare/tdiamond
= 3/2/2 = 1.06. The "square"falls faster, overall. If free to rotate, it would start out
in the "diamond" orientation, switch to "square" for the middle portion, and then switch back to diamond,
alwaystrying to present the minimum chord at the field's edge.
z
(b) F = IElj ~ = 2B I~a ya2 - X2dx (a = radius of circle).
[;= -~~ = -2Bva2 - y2~ = 2Bvva2 - y2 = IR.
= va2
I = 2~Vva2 - y2 j 1/2
Vcircle=
tcircle=
mgR
4B 2 ( 2
a
i
-a
+a
2
-y )
dy
--
V
-
y2. So F
= 4~V
(a2
-
y2)
1/2
x
= mg.
1
;
4B2
= mg
- R 1-a (a2 -
4B2
0
y2)dy
= mg
- R
(a2y -
0
1
-3 y3) -a
1
4B2
4
JY
= mg
- R (-3 a3) =
16 B2a3
3mgR'
CHAPTER 7. ELECTRODYNAMICS
140
Problem 7.47
(a) In magnetostatics
JLo
V. B = 0, V x B = JLoJ ~
= 47r
B(r)
!
J(r/) x ~
~2
dr/.
For Faraday electric fields (with p = 0), therefore,
V.E=O
(with the substitution J -t (b) From Frob. 5.50a,
,
VxE=_aB
~
at
E (r t )
.'.
=_2.~
47rat
!
B(r/t)X~dr'
~2
:0~.)
-- 2.
47r
A( r,t )
!
B(r', t) x ~ d '
- - aA .
r, so E ~2
at
[Check: V x E = - tt (V x A) = - ~~, and we recover Faraday's law.]
(c) The Coulomb field is zero inside and 4;fO f = 4;fO tT4;2R2 f
f outside. The Faraday field is - ~~'
where A is given (in the quasistatic approximation) by Eq. 5.67, with I..Va function of time. Letting w ==dJ..Jjdt,
~
=$
JLORWU
E(r, O,~,t) =
Problem
7.48
qBR
= mv
f E.dl=-dt'
dip
rsin (JA.
I{ UR:.
-r+
€or2
I'<>R:':U
3
sinO
r2
(r
< R),
(r
> R).
(Eq. 5.3). If R is to stay fixed, then qR~~ = m ~~ = ma = F = qE, or E = R~~.. But
dip
soE27rR=-dt'
1 dip
dB
so -27rRdt=Rdt'
1
orB=-'2
1
(7rR22
integral first, letting u ==ct>2 21<-<1>1
/
-<1>1
cosu
duyI- 2{3cosu -
Z2
d1:
21<
/
0
cosu
yI-
2{3cosu
du
(sincethe integral runs over a complete cycle of cosu, we may as well change the limits to 0 ~ 211").Then the
~1 integral is just 211",and
21<
M
= 411"
/-Lo ab{3211"
~0
~
21<
cosu
du
yI- 2{3cosu
= /-Lo
vab{3
~0
2
cosu
duo
yI- 2{3cosu
(a) If a is small, then {3« 1, so (using the binomial theorem)
1
yI-
21<
2{3cosu
~I+{3cosu,and
~0
21<
COS U
.
yI - 2{3cosu
du~
~0
21<
cosudu+{3
r::i7i3
and hence M
= (/-Lo1l"j2)yab{33.Moreover, (3 ~
~0
cos2udu=0+{311",
/-LO1l"a2b2
abj(b2 + z2), so M ~
I
2(b2
+ z2)3/2
I(same
.
as In Prob. 7.20).
142
CHAPTER 7. ELECTRODYNAMICS
(b) More generally,
-1 / 2
(l+f)
1
=1--f+-f
2
3 2
5 3
--f
+...=>
8
16
1.
2,8cosu
Vl-
3 2
2
5 3
3
CoS U+-,8
cos u+"',
2
2
=1+,8cosu+-,8
so
{21r
M= ~Ovab,8 { Jo
=
/-L°..jCiJJ
2
Problem
[
{21r
3
{21r
5
{21r
cosudu+,8 Jo cos2udu+2,82 Jo cos3udu+2,83 Jo cos4udu+",
0 + ,8(7r)+ ~,82(0) + ~,83(~7r) +" .
2
2
4
J
=
(1 + 15,82+
2 vab,83
/-Lo7r
I
8
}
)
+.. . .
(),84
Qed
7.53
Let q> be the flux of B through
a single loop of either coil, so that q>1= N1 q> and q>2= N2q>, Then
dq>
£2
£1 = -N dq> £2 = -N2-,
Idj'
dt
Problem
N2
so - = -,
£1 N1
Qed
7.54
~a) Suppose current II flows in coil 1, and 12 in coil 2. Then (if q>is the flux through one turn):
q>1
I
n case
I
= ItLI
-
1-
0
+ MI2
, we
h
= N1q>;
M
-
fu
dt - L 1 dt
C - ~
(b) -°1 -
.0..,
ave Nt -
+
= I2L2
q>2
'f
-
I
0
+ MIt = N2q>,
h -
h
or q>
'
= It
'
D d'
M
L1
M
L2
N1
+ 12 N1
= 12 N2
,M
-
N2' 1 2 - , we ave Nt - N2'. IVI mg, Lt TT
fu
M !ll.2.( t) ., -O2c -- ~dt -- L 2 !ll.2.
dt - vI COSW
dt + M dt
(c) Multiply the first equation by L2: LIL2%
+ L2!!ftM
= L2 VI coswt,
Vout
(d)
VI
dt
= L1
I2R
-~n = v,1 cos
wt
=
L2.
(
cQswt
VI
v,1 COgwt
L2
=- M
1
I
- I R
- - 2
= -I2R
Plug in L2!!ft
= VoutI2 = (12)2R = I(~~~2
1
So (Rn) = 2(Vl)
2
L2
(
L2
,
L2
(
(Vr)2
1
)= T (
)
- M5!Jt.
= VI coswt,
,
L2
,
~smwtcoswt+
d
qed
'
= - -N2 . The ratio of the amplitudes is N2 .
N1
N1
(e) Rn = ~nIt = (V1coswt)(L; ) ~sinwt+ Rcoswt
Pout
1
'
2
M' or L 1L 2 - M ,qe
- RW smwt => II (t) = L1 ~ smwt + R coswt
-~~ coswtR
VI
)
N2
h
M2% - MRI2 - M2!!Jt = L2V1coswt =>II2(t) = - ~~ coswt.1
Ll% + M (~~wsinwt)
dII
M
+ It
Qed
2
RCOS
wt
)
,
COS2
wt.1Average of COS2
wt is 1/2; average of sinwtcoswt is zero.
1
(L;R ) ; (Pout)= 2(Vr)
2
(L2)2
[ M2R J
1
2
(L2)2
= 2(Vr) [ L;L;,R;J
(Vr)2 L2
(Rn) = (Pout)= 2L1R .
Problem 7.55
(a) The continuity equation says !!It= - V. J. Herethe right sideis independentof t, so wecan integrate:
p(t) = (- V. J)t+ constant. The "constant" may be a function of r-it's only constant with respect to t. So,
putting in the r dependence explicitly, and noting that V.J
= -p(r,O), p(r,t) = p(r,O)t+ p(r,O), Qed
143
(b) Suppose E
= 4;'0
V.E = 10 p, and V XE
J~dr and B = ifrJJ~.tdr.
=-
We want to show that V.B
~~ ' provided that J is independent of t.
= 0,VxB = J..toJ+J..to€o
~~j
~
Weknowfrom Ch. 2 that Coulomb's law (E = 4;'0 J dr) satisfies V.E = 1op and V XE = O. Since B is
constant(in time), the V.E and V XE equations are satisfied. From Chapter 5 (specifically,Eqs. 5.45-5.48)we
knowthat the Biot-Savart law satisfies V.B = O. It remains only to check V XB. The argument in Sect. 5.3.2
carries through until the equation following Eq. 5.52, where I invoked Vi . J = O. In its place we now put
V' . J = - p:
VxB
J..to
471"
j --------
= J..toJ -
(J.
-t
V) ~2
(-JtpVI)~
dr
(Eqs.5.49-5.51)
(Eq.5.52)
Integration by parts yields two terms, one of which becomes a surface integral, and goes to zero. The other is
= ~:(-p).
~V/.J
'"
So:
J..to
= J..toJ
V xB
-
471"
j
.t .0
1
= J..toJ
~2 (-p)dr
+
J..to€o
at { 471"€0
oE
pi"
j
~3 dr
}
= J..toJ + J..tO€°Ft.
qed
Problem 7.56
(a) dE z
sin
-
471"€0
~
~
= vi Z2
+ S2
-z
e= -;
E z -
1 (-A)dz . e
2
sm
-
~j
471"fO
(z2
~
-
+
- 471"€0 [ vi Z2
S2)3/2
vt
-1
zdz
+
vt - €
S2 J vt-,
I
-Z
v't
Z
1
1
z - 471"€0
{ J(vt - €)2 + S2 J(vt)2 + S2 } .
E -~
(b)
1
E=~ r
{ J(vt
471"€010
-, ~
2€0
1
J(vt)2
a
271"SdS=~
+ S2
}
2€0
[
J(vt-€)2+s2_J(vt)2+s2
]l 0
- f)2 + a2 - J(vt)2 + a2 - (€ - vt) + (vt) ] .
[ J(vt
~
(C)Id=€OdE=
dt
- €)2 + S2
-
I
2
{
v(vt-€)
J(vt - f)2
+ a2
-
v(vt)
+2V
J(vt)2 + a2
'
}
Asf -t 0, vt < f also --*0, so Id --* ~(2v) = AV = I. With an infinitesimal gap we attribute the magnetic field
to displacement current, instead of real current, but we get the same answer. qed
Problem 7.57
(a) V2V
= !~
s as
(
sO(Zf)
as
Ads
s = df:::}f = Aln(sjso)
= =-~
= O:::}~
OZ2
s ds ( ds )
ds ( ds ) = O:::}sdf
ds = A (a constant)
) + 02(zf)
sdf
sdf
(soanotherconstant).But (ii) :::}f(b) = 0, so In(bjso)= 0, so So= b, and
:::}
144
CHAPTER 7. ELECTRODYNAMICS
2
Ip
1
Ipz In(sjb)
.
But (1)::} Azln(ajb) = -(Ipz)j(7ra ), SOA = - 7ra2In(ajb)j V(s,z) = - 7ra2In(ajbf
= Azln(sjb).
V(s,z)
(b)E=-VV=_8Vs_8Vz=lpz
8s
1
8z
7ra2s In(ajb)
s+lpln(sjb)z=
Problem
= €o [Es(a+) -
= €o [ 7ra2In(ajb)
Es(a-)]
(::s+ln
(~b') z)
s
€olpz
(a ) - a
= 7ra3In(ajb).
]
7.58
-H
tg
.
= JloK = Jlo-;w1
= €oaj
= Bhl = Jlol
-hi
w
fl'B
I
1
(a) Parallel-plate capacItor: E
(b) B
7ra2In(ajb)
Z
Ip
(c) a(z)
Ip
7ra2In(ajb)
1 Q
V
Q
= Eh = €o wlh::}
=V
C
~
Jloh
= Ll::} L = -l::}
w
~
fowl
=h
CIJ
::}
oh
£ = -.w
(c) C£ ==Jlo€o = (47rx 10-7)(8.85 X 10-12) = 11.112X 10-17 s2jm2 .1
(Propagation speed l/m
= 1/ v'Jlo€o= 2.999 X 108 mjs = c.)
I
I
(d) D=a, E=Dj€=a/€,
so~ustreplace€o PY€j
H = K, B = JlH = JlK, so Just replace Jloby Jl. } I£C=€ Jl,.1 Iv=I/J€Jl.1Jl
Problem 7.59
(a) J = a(E + v X B)j J finite, a = oo::} E + (v X B) = O. Take the curl: VxE + Vx(v X B) = O. But
8B
8B
Faraday's law says VxE = -7Ft. So 7Ft = Vx(v X B). qed
(b) V.B = a::} § B. da = a for any closed surface. Apply this at time (t + at) to the surface consistingof
S, S', and n:
f B(t + at) . da + f B(t
is'
in
+ at) . da - f B(t + at) . da
is
=a
(the sign change in the third term comes from switching outward da to inward da).
= is'f
d
B(t + at) . da - isf B(t) . da
= isf
lB(t + at) - B(t)j . da - in
f B(t + at) . da
~
~~ dt (for infinitesimal dt)
d=
{l ~~ .
da} dt -In
B(t + at) . [(dl X v) at]
(Figure 7.13).
Since the second term is already first order in at, we can replace B(t + at) by B(t) (the distinction would be
second order):
= dt
d
l a: .da-dt i~(v XB).dl = dt{l
(~~) .da-l Vx(vXB).da}.
145
=
dip
dt
r
aB -VX(v
is [ at
X B)
.da=O.
]
Qed
Problem 7.60
(a)
V. E'
=
(V E) cosa + c(V. B) sin a = ~Pe cosa + c/1oPmsin a
0
fO
1
0
= .fO
- (Pecosa+c/1ofoPmsma
V. B'
V x E'
)
1
= /10Pm cosa
=
(V. B) cosa - ~(V' E) sina
=
/10(Pm cosa
=
.
(V x E) cosa + c(V x B) sma
C
-
~Pe
sin a)
C/1ofo
1,/
-Pm sma ) = fO
-Peo"
C
- 2-Pe sin a
CEO
= /10(Pm cosa -
CPesin a)
= /10P'm..(
aB
(-/1oJm - {it )
(
a
,aB'
-/10 Jm cosa - cJesma ) - at (B cosa - ~E sma ) = -/1oJm - 7ft.
aE
=(
)
(
=
=
=
(V x B) cosa
=
1
~(V
0
x E) sma
{LoJe+ /10f°{it
/1o(JeCosa+~Jmsina)+/1ofo~
C
ut (Ecosa+cBsina)
1
cosa - ~
aB
-/1oJm - {it
=/10J~+/10fOa~'0.(
ut
=
1
q' (E' + v x B' ) + q' (B' - -v
e
m
C2
=
(qe cosa + ~qmSina) [(ECOSa+ cBsina) + v x (Bcosa - ~Esina)]
x E')
+ (qmCOg
a - cqesina) [ ( B COga - ~E sina) - c~v x (E COga + cB sina) ]
=
qe [(ECOS2a + cBsina cosa
+v x (Bcos2a-
- cB sinacosa + Esin2 a)
~Esinacosa+
+qm[(~Esinacos~
+v x
.
sma
.(
(b)
F'
)
10
(
,
aE
cosa + C /1oJe + /10f°{it
0
V xB
1.
= fO
- (Pecosa+
c
( ~Bsinacosa-
~Esinacosa+Bsin2a)]
+ B sin2a + BCOS2a - ~Esinacosa)
c12Esin2a- c12Ecos2a- ~Bsinacosa
c
= qe(E + v x B) + qm( B - :2v x E) = Fo Qed
)]
)
0
sma
Chapter 8
Conservation
Laws
Problem 8.1
Example 7.13.
E- - --8
A 1
211"1008
B
Pol!.
~
}
211" 8 cp
p=
I
b
S. da=
I
/
E.dl=-
7.58.
E
211"100
a
Problem
= -~
z; a
= ~z
= J.LoKx = -J.Lol
w x
Q
= ---:i";
Q(t) = It
~
BE
1
1 2
(b) Uem = 2" €oE + J.Lo
B
S
-
~(E
- J.Lo
100
x B)
2
2
E(t)
=>
111"82
= J.Lo€o~
1I"€oa
1
= 2" [ 100
=
211"a2
1I"€oa2
-
/
butV=
~
a1
= -Y;
€oW
E'dl=~h,100
solp=IV.1
t
~
z.
1I"~a
~
J.Lo18
A
=> B(8, t) = ~
211"acpo
It
J.LOI8
- J.Lo 1I"€oa2
x B)
J.Lo
}
2
( )
)
- ~ ~
( )( ) (-8) - (
1
= -(E
~
S.da=SWh=alh,
B 211"8= J.LO€O-;:;-1I"8
ut
solp=IV.1
a
8.2
a~
(a) E
In(b/a).
11"100
l
~
/
8
211"100 ~d8=~ln(b/a),
8
211"100
S
p=
I
-d8=
_2>.I
b
100
B
I
S211"8d8
=_
A
a
Problem
b
>.I
a
b
ButV=
s =~ (E x B) =~411"2100821 .Zj
1
J.Lo18
( )] =
+ J.Lo 211"a2
12t 88
211"2€oa4
146
2
.
J.Lo12
211"2a4 [(ct)2 + (8/2)2]
.
147
QUem
Qt
= 27r2a4
/l0[2 2c2t = ~j
7r2€oa4
(C) Uem
=
I
-V. S
= 27r2€oa4
[2t
V.
/l [2
(8 S)
=~
= 8Uem.
./
7r2€oa2
8t
{b
UemW27r8dS
= 27rW2:2a4
Jo [(ct)2+ (8/2)2]sds= :a4
w[2b2
I
b2
= IJL027ra4 [ (ct)2 +
16 ]
-dUem = /low[2b24
= ~7r€oa = .Rn../
2
27ra 2c t
Problem 8.3
dt
.
Over a surface at radius b: .Rn =
[2wtb2
F
82
/l W[2
-
[ (ct)2"2
b
184
+ 44
]
I
0
[2wtb2
[2t
S. da = 27r2€oa4[bs. (27rbws)] = 7r€oa4 .
(Set b = a for total.)
=fT'
da
-
/lo€o
I
~
S dr.
The fields are constant, so the second term is zero. The force is clearly in the z direction, so we need
1
++
(T' da)z
=
1
BzBx dax + BzBy day + BzBz daz -"2B
(
Tzx dax + Tzy day + Tzz daz ="/l0
2
daz
)
= :0 [Bz(B'da)-~B2daz].
NowB = ~/loaRc.vz (inside) and B = 4/lon;(2cosOr + sinO8) (outside), where m = 3 7rR3(awR). (From
3
7rr
Eq. 5.68, Frob. 5.36, and Eq. 5.86.) We want a surface that encloses the entire upper hemisphere-say
a
~
hemispherical cap just outside r
=R
plus the equatorial
circular disk.
Hemisphere:
Bz
=
da
=
B2
=
H
(T' da)z
=
/lom
. n
47rR3 2cosO (r )z + smu
[
=
(
]
.
-
2 n
sm u ]
::;
/lom
= 47rR3 (3eos
2
0
- 1) .
(
)
)
2
( )
2 (~ )
/lo
3"
/lo
aWR2
= 2
=
/lom
2 n
z = 47rR3 [2cos u
)
(
/lo
(Fhemi)z
A
((} )
R2 sin 0 dOdcPr; B. da =
(2 cos O)R2sin 0 dOdcP;daz = R2 sin 0 dOdcPeos OJ
2
/lom
/lom 2
47rR3
(4COS20 + sin2B) = 47rR3 (3COS20+1).
1
1 /lom 2
/lo 47rR3 [ (3COS20- 1) 2eosOR2 sinO dOdcP- "2(3cos2 0 + 1) R2 sinO cosO dOdcP]
aWR
=
A
awR2
(~)
1.
[ "2R2 smOcosOdOdcP ] (12cos20
2
- 4 - 3eos20 -1)
.
2
(geos 0 - 5) smOeosOdOdcP.
2
1r/2
27r
I
0
(9cos3 0
/l07r(aw3R2r (0 + ~ - ~)
-
.
5eosO) smOdO
= _/l~7r(a~R2r
awR2
= /l07r
.
(~)
2
1r/2
9
5
[-4eos4 0 + "2eos2 0]
1
0
148
CHAPTER 8. CONSERVATION LAWS
Disk:
2
=
Bz
~
-Po(JRI.JJ;
da = rdrd
CHAPTER 8. CONSERVATION LAWS
150
=
Therefore the angular momentum of the cylinders is L
so
N dt
= -~J.tonI(R2
- a2) Z
/
J I'dt = Q,
I'dt. But
L = -~J.tonIQ(R2- a2)ZI (in agreement with Eq. 8.35).
I
Problem
(a)
/
8.8
0,
(r
< R)
~J.toMZ,
E-
- { ~~~ [2cosOi'+sinOO], (r > R) } (Ex. 6.1)
- { 4:€O~ i', (r > R) } ,
4
(where m
(r < R)
. B3
= "371"RM);
p
J.to Qm
= €o(E
x B)
A
= (471")2-;:s(i'
A
A
x 0) sinO, and (i' x 0) =
The force on a patch of surface (da) is dF
IE
=r
x dF
only the z component (Oz = - sinO):
N
1f
Here r = Rj
/
0
= -3&
= _J.t~0" d:: z
/
(r
dt
)
(0"
=
~J.todM
A
fx(Oxi)
sinB
- cos B (f x (fJ)]R2 sin B dB d4>
'---v---'
-9
~
O(f . i) - i(f . 0)
=
poM
127r
dq
dt
()
2
A
(1 + cosB)[2(0 x i) + 2cosB(0 x f)].
(l+cosB)2[
247r dt
1
PoM
2"
=
~[2i+2cosBf+sinBO];
]
A
dq
dt
PoM R2
A
(1 + cos B)R [cosB0 + cos B0] dBd4>=
()
67r
A
(1 + cos B) cos BdBd4>O.
211"
The x and y components
integrate
() /
dq
dt
PoMR2
Nz
=
-
-
_poMR2
3
67r
=-
to zero; (O)z
11"
(27r)
0
sin B, so (using
.
(l+cosB)cosBsmBdB=-
( )(~) = - 2poMR2dq.
dq
dt
3
9
dt
/
0
d4>
= 27r):
poMR2
3
dq
dt
( )(
Si~2
B-
CO~3
B)
[
9 MR2dqdt i,l
IN = - 2po
Therefore
0
L
=
/
N dt
=-
2~0 MR2 i
/
Q
dq = ~MR2Qi
I
I
(same as (a)).
(1used the average field at the discontinuity-which
is the correct thing to do-but
sameanswer using either the inside field or the outside field.)
Problem 8.9
dIs
2
(
)
(a) E = -di;
dif> if>= 7ra2 B; B = ponIs; E = IrR. 80 Ir = -Ii 1 po7ra n yt'
in this case you'd get the
I
(b)
dif>
f E. dl = -- dt
8 = ~(E
Po
x B)
2 dIs =>
=> E(27ra) = -PO7ran-
d
=~
Po
_poan dIs
2 dt
(
)(
t
E = -- 1 Poan-dIs cPoB
2
d
t
A
poIr
b2
2
(b2 + z2)3/2
)
((fJ X i)
=
_
= paIr
2
b2
A
3/2 z (Eq. 5.38).
(b2+ z2)
-~poIr dIs
ab2n
f.
4
dt (b2 + Z2)3/2
152
CHAPTER 8. CONSERVATION LAWS
Power:
!
=
P
=
!
S.da=
-=
.
The mtegralls
'
(
Z
.
b2yz2
dls
2
dIs
2
t
(S)(27ra)dz=--7rpoa2b2nIn-d
.
= - 7rpoandI
1
)
100
1 -
=
+ b2 -=
1
-
( )
b2
I
b?
~
=
1
!
3/2dx
-= (b2 + Z2)
2
= b2
'
2
= (RIr)Ir = Ir R. Qed
Ir
Problem 8.10
According to Eqs. 3.104, 4.14, 5.87, and 6.16, the fields are
-- 1 P,
3£0
E=
1
(r
-11 0 M
31""'
B=
1
(r>R),
{
}
(r
,
pom
~~
--[3(p.r)r-p],
47r£0 r3
{
2
< R),
~~
< R),
(r > R), }
--[3(m.r)r-m],
47r r3
where P = (4/3)7rR3P, and m = (4/3)7rR3M. Now P = £0f(E x B) dT, and there are two contributions, one
from inside the sphere and one from outside.
Inside:
1
Pin = £0
2
! (- ) ( )
3£0 P
X
Outside:
3fLoM
2
dT
= -gfLo(P
1 Po
Pout = £0-4
7r£047r
=
!
X M)
!
2
dT
4
= -gfLO(P
8
3
3
= 27fLO7rR (M
X M)37rR
X F).
r16{[3(p. f) f - p] X [3(m. f) f - ill]} dT.
Now f X(p x m)
p(f.m) -m(f.p), so f x [f x (p x ill)] = (f.m)(f x p) - (f.p)(f x ill), whereasusing the BACCAB rule directly gives f x [f x (p x ill)] = f[f.(p x ill)] - (p x m)(f.f). So {[3(p. f) f - p] x [3(m. f) f - mn =
-3(p.f)(fxm)+3(m.f)(fxp)+(pxm)
= 3 {f[f. (p x ill)] - (p x m)}+(pxm) = -2(pxm)+3f[f.(pxm)].
Pout = 1~;2
!
r16{-2(p x m) + 3 f[f
. (p
x ill)]} r2 sin 0 dr dO dl/>.
To evaluate the integral, set the z axis along (p x ill); then f. (p x m) = Ip x ml cosO. Meanwhile,f =
sin 0 cos I/>x + sin 0 sin I/>y + cos 0 z. But sin I/>and cos I/>integrate to zero, so the x and y terms drop out, leaving
Pout
=
1~;2
Po
=
-
=
-~
Ptot =
167r2
(100 r14
dr) { -2(p x m)
1
=
( -- 3r3 ) 1R
127rR3
(
~7rR3p
3
[
!
sin 0 dOdl/>+ 31p x ml z
47r
-2(p x m)47r+3(p x m)-
) (
x
~7rR3M
3
)=
3 ]
4po
27 R3(M x F).
(287+ 2~) POR3(Mx P) = ~POR3(M x p).1
I
= --(p
!
Po
127rR3
COS2
0 sin 0 dO dl/>}
x m)
153
Problem 8.11
(a) From Eq. 5.68 and Frob. 5.36,
r < R: E = 0, B =
~J.Loo"Rc.vZ,
e
4 R 2;
with 0"=
3
~
4
r > R: E = - 1 ~ r, B = J.LO
~0"c.vR4.
4~ nr ; (2cosBr + sinBO), with m = -3
4~fO r
{
The energy stored in the electric field is (Ex. 2.8):
1 e2
= 8~€0
WE
The energy density of the internal
1
uB
=
2
1
= 2J.Lo
2J.Lo B
magnetic
2
field is:
e
(
R'
2
)
3"J.LoRc.v 4~R2
J.Loc.v2e2
J.Loc.v2e24
3
J.Loe2c.v2R
= 72~2R2' so WE;" = 72~2R2 3"~R =
54~
.
Theenergy density in the external magnetic field is:
1
J.L5 m2
00
WBout =
J.Loe2c.v2R4
(18)(16)~2
.
2
= 2J.Lo16~2 -;:6 (4cos
UB
=
WB;"
e2c.v2R4 J.Lo 1
=
!
ROO
!
+ Wbout=
2~
.
2
(3 cos B + 1) sm BdB
!
J.Loe2c.v2R4
d 8d8dt/J; [(x-a)2+y2]=82+a2-28acost/J,
a)
8
+ a2 - 28a cas t/J)
cost/Jdt/J
-
a2
+ 82
21!'
d 8 d ,!.
'f'
(
1-
A
).
{27r
-
dt/J
21!'
Jo (A + Bcost/J) - B
VA2_B2'
Jo (A+Bcost/J) - VA2_B2'
A2 - B2 = (82 + a2)2 - 482a2 = 84 + 282a2 + a4 - 482a2 = (82 - a2)2; viA2 - B2 = a2 - 82.
~
J [1 - ( a2 -
)+
82
2a2
(a2 - 82)]
8 d8
/loqnI
= -y a
R
~
10
8
d8
=
J..LoqnIR2
2a
~
y,
156
CHAPTER
8. CONSERVATION
LAWS
Angular Momentum.
l
=
r x p
=
The x and y terms are odd in z, and integrate to zero, so
X2 + y2 - xa
..
J-LoqnI
- - 411" z [(x - a)2 + y2 + z2]3/2 dx dy dz. The z mtegralls the same as before.
=
- J-LoqnIZ
= J-Loq~I
r x [yx 4~
A
L
211"
=
-J-Loqnlz
!
!
!
X2
+ y2
-
(x - a) y]
xa dx dy
[(x - a)2 + y2]
[ a -8
!
= - J-LoqnI
z
211"
- a) x + zy Y- [x(x - a) + y2]z} .
8 - a cos
T[ik2AT
-
~
IT (¥Zlo+ - ¥ZIJ
iki (AI
ki
= (ki + k2 -
l+i{3
AR = ---=-
h
1
2-
A -
tan.
1 - i{3
~
..
1 ~fJ"
(
i{3
1 + i{3 - 1 ~
Thus ARe'"
2
-
)
IP2
{3_mw2_m(kIvI)2_mkIT
k IT k IT
( )A I, were =
1 + i{3
( )( ) 1 - Z{3
ki - k2
SImIlarly, 1 - i{3 - Ae
A-I,
i
~
A -
iT
-
) AT,
+ imw2lT
-
or AT =
-
~
II OR
2
)( + )
1 - i{3
(ki + k2 - imw21T ) AI.
= 00, then kd ki = 0, and we have AT =
Ai.
1
i{3
= 01+tan-'
-
-
AI.
and e - (1 - i{3)(l+ i{3)-
(
2kI
(1 ~ i{3) AI,
l+i{3
-- A e , WIt
.h
.
Z{3
or -- m-.ki N ow _1
- - T -,
PI
PI
(1 + i{3)2 - 1 + 2i{3- {32
~ e" Aie;" => I AR ~
2
2 -
i
mw2
(ki + k2 - imw2IT )
+ k2 - imw2IT
If the second string is massless,so V2= JT
-
m~lo'
-
- -- AT
- - AI- -- 2kl.- (ki + k2- imw2lT) AI
- --
AR
~
- AR)] = m( -W2 AT), or ki (AI - AR) = (k2 - im;2) AT.
Multiplyfirst equation by ki and add: 2kIAI
-
I
V2 ). Now gI(Z, t), gT(Z, t), and hR(Z, t) are each functions of a single variable u (in the
VI + V2
- VI t, in the second u
Z - v2t, and in the third u = Z + VIt). Thus
2V2
(
VI
(
Adding these equations, we get 2gI( -VI t) =
t
VI
1 8gT( -V2t)
-- V2
8t
~ gI( -VIt) - hR(VIt)
(
1 + (32
~
(~)
.1
4
2
A 1 + {32~
- J1+732'
)
i
CHAPTER 9. ELECTROMAGNETIC
160
Aei
=
I
2
= J1+i32AI;
AT
I
1
2
= 2(1 + i{3)
(1 - ~(3)(1+ i{3)
I
I
2\1 + i{3)
(1 + (J2) =} tan 4>= {3.
So ATeiOT =
"A.
J1+i32
e"" A Ie
WAVES
ior.
,
liT = iiI + tan-1 {3.1
Problem 9.7
a2f
af
a2f
a2f
= T-az-?6.z -'- at .6.z= JL.6.zat- ' or
(a) F
a2f
af
a = JL at2 + '- at
T z-
?
'J
-
?
'
(b) Let ](z, t) = F(z)e-iwt; then Te-iwt~~~ = JL(-r.,})Fe-iwt + ,(-iw)Fe-iwt =}
d2F
- d2F
-2 -2 W
- 1,
- 1,
T dz z = -w(JLW+ if)F, -dz 2 = -k F, where k ==-T (JLW+ if). Solution: F(z) = Ae"z + Be-' z.
Resolve k into its real and imaginary parts:
2kK,
= w,
- =} K, = -'W,
T
2kT '
k2 - K,2
W,
2
= k2 - (-2T )
k = k + iK, =} P = k2 - K,2 + 2ikK, = f(JLW+ if).
1
JLW2
- = -'
or k4 - k2 (JLW2 / T ) (w, / 2T) 2 = 0 =}
k2
T'
= !!Jfl-[1:J: VI
k2 = ~ [(JLw2IT):J:v(JLw2IT)2 + 4(w,/2T)2]
+ bIJLw)2]. But k is real, so k2 is positive,so
-1/2
]
[
iJLj
we need the plus sign: k = wyIT
w, = ..j2TJL
I
1 + VI + bIJLw)2. K,= 2kT
1 + VI + blJLw)2
.
Plugging this in, F = Aei(k+iK)Z
+ Be-i(k+iK)Z= Ae-Kzeikz + BeKze-ikz. But the B term gives an expo-
nentially increasing function, which we don't want (1 assume the waves are propagating in the +z direction),
so B = 0, and the solution is ](z, t) = Ae-Kzei(kz-wt) .1 (The actual displacement of the string is the real part
of this, of course.)
(c) The wave is attenuated by the factor e-Kz, which becomes lie when
I
z
=~
K,=
..j2T
, JL
j 1 + 0-+(,1
JLw)2;Ithis is the characteristic
penetration
depth.
-
(d) This is the same as before, except that k2 --*k + iK,. From Eq. 9.29, AR
AR
=
k1-k+i"'
k1-k-i"'
k1+k+i",
)(
( ) (
2
AI
(where k1
= WIV1
k1 - k - i"'
k1 + k + i",
(
)
=
k1+k-i",
= wVJLl/T,
(k1
-
)=
(k1-k)2+",2.
(k1+k)2+K,2
AR=
)
(k1-k)2+",2AI
(k1+k)2+",2
k - i",)(k1 + k + iK,) = (k1)2 - k2 - ",2 - 2i",k1;:=} UR
(k1 + k)2 + K,2
(k1 + k)2 + K,2
sum f = fv + fh lies on a circle of radius
A.
At time t
=
0, f = Acos(kz)x - Asin(kz)y.
At time t = n)2w, f =
A cos(kz-900) x- A sin(kz-900) y = A sin(kz) x+A cos(kz) y.
Evidently it circles counterclockwiseI. To make a wavecircling
I
the other way, use lih = -90°.
;J:
z
y
(
-
while k and K,are defined in part b). Meanwhile
Problem 9.8
(a) fv(z,t) = Acos(kz - wt)x; fh(z,t) = Acos(kz - wt +
90°) y = -A sin(kz - wi) y. Since i?; + f~ = A2, the vector
(b)
=
k1-k-iK,
k 1 + k + 'I'"
"AI;
= tan -1
-2k1K,
(
;J:
at
t ='Tr/
2W"---z
/
/
V
)
(k1)2 - k2 - ",2
\
,
" -
/ /
/
.
161
(c) Shake it around in a circle, instead of up and down.
Problem 9.9
w
k
(a) k =-~x;
n=z.
.r=
-~x
. (xx+yy+zz
w
~
I
~
~
(
I
~
)
~
~
~
w
k
) =-~x;
~
~
~
~
xn=-xxz=y.
E(x, t) = Eocos(~x + wt) z; B(x, t) = ~o cos(~x + wt) y.
:/;
x
z
z
,
"
,
"
,
y
"
,
',I
y
(b)
(a)
(b) k
since ft. k
I
~
I
I
1
I
I
I
I
~
~ (X + + z) ; fi ~
= 0,/3= -0:; and since
~.I
(moce fi;, p""allel to the x z plane, it mu,t have the fo,m ax
x y z
k'r=
%
v3c
(x+y+z).(xx+yy+zz)=
E(x, y, z, t)
% (x+y+z);
v3c
kxft=
~
v6
1
I
1
1 0
1
-1
1
= j6(-x+2y-z).
= Eocos[~c(x+y+z) -wt] (x;/);
B(x,y,z,t) =
Eo
ecos
w
[ V3c(x+y+z)-wt ]
(
-X+2Y-Z
j6
).
Problem 9.10
P
+ ~z;
it is a unit vector, 0: = 1/V2.)
=£c = 3.0
1.3 x 10:
10 = 14.3 x 10-6 N/m2 .1 For a perfect
reflector the pressure is twice as great:
18.6X 10-6 N/m2 .1 Atmospheric pressure is 1.03 x 105N/m2, so the pressure of light on a reflector is
(8.6x 10-6)/(1.03 x 105) = 18.3X 1O-11 atmospheres.!
-----..
162
CHAPTER
Problem
1 (T
=
If Jo
acos(k.
r
- LVt+
8a)bcos(k.
r
- LVt+
8b)
dt
ab (T
2T
Meanwhile,
in
1
the
.
(k
2f!J* = 2ae~
Problem
ab
-
[cos(2k. r - 2LVt + 8a + 8b) + cos(8a
Jo
complex
-
or-",tb*e-~
1
)
(k
1
-
<
0
«
or-",t = 2ab*
=
dt
2T
1
=
2abe~ va-vb,
Re
=
8b)T
a = aeioa,
b
-
2abcos(8a
=
beiob.
Ob).
So
1
1-
)
-
c~s(8a
( 219* )
= 2abcoS(Oa
- Ob) = (fg).
Qed
9.12
Tij
=
2
fa( EiEj-28ijE )
1
1
+ /-La(BiBj
the fields in Eq. 9.48, E has only an x component,
(i f:.j) terms
and
B
- 28ijB
2
)
.
only a y component.
So all the "off-diagonal"
are zero. As for the "diagonal" elements:
-
Txx
-
Tyy
1
Tzz
ExEx
1
--E
fa
1
fa
So Tzz = -foE5 cos2(kz- LVt+ 8)
I
momentum
2
1
1
/-La
2
1
2
1
( - -E ) + - (--B ) = - ( - -B ) =
( ) + - (B B - -B ) = - (-foE + -B ) = 00
(-2E ) + /-La(-2B ) = -u.
fa
2
I
1
2
.
2
The
8b)]
j = aeikor-"'t),9 = beikor-",t),where
notation:
.
)
1
With
WA YES
9.11
(fg)
1-
9. ELECTROMAGNETIC
2
1
/-La
Y Y
1
2
1
foE
2
1
2
2
2
O.
/-La
1
2
2
2
/-La
2
(allother elements zero).
of these fields is in the z direction, and
transportedin the z direction, so yes, it does make
it is being
sense that Tzz should be the only nonzero element in Tij. According to Sect. 8.2.3, - .da is the rate at which momentum
crosses an area da. Here we have no momentum crossing areas
oriented in the x or y direction; the momentum per unit time
0-'edt
1:
per unit q,reaflowing across a surface oriented in the z direction is -Tzz
= u = pc
(Eq. 9.59), so Llp = peALlt, and hence
Llpl Llt = peA = momentum per unit time crossing area A.
Evidently momentum fl4X density
Problem 9.13
I
2
R
=
Eo
(E )
R
o[
= energy density.l"'
(Eq.9.86)=} R= _11 I
(3
2
( (3)
+
2
/-LIVI
f2V2
(Eq. 9.82), where (3==-.
T= -
/-L2V2
1
=> IT
=
fi
(1 + (3)'
1
T+ R
= (1 +
[
Vl
I
101VI
(Eq. 9.82). [Note that "V,
2
(3)2 4(3 + (1 - (3)
]
1
= (1 + (3)2(4(3+
~
/-L2fl/-Ll
/-L2
p,
"p, VI
V, p,
~
2
1 - 2(3+ (3 )
()
1
V2
(E )
T
101VI
~
Eo
2 V2
VI
o[
(Eq.9.87)
= /-L2v2
/-LIVI = (3.]
2
= (1 + (3)2(1 + 2(3+ (3 ) =
1. "'
163
Problem
9.14
Equation 9,78 is replaced by Eolx + EORllR = EOTllT,and Eq, 9.80 becomesEotY - EoR(zx fiR) =
(3EOT(zx llT), The y component of the first equation is EoR sin()R
second is EoR sin()R
= -{3EoT
sinDT,
Comparing
= EoT sin()T;
these two, we conclude
that sin()R
the x component of the
= sin()T = 0, and hence
eR= OT= 0, qed
Problem 9.15
= Ceicx for all x,
Aeiax + Beibx
so (using x = 0), A + B = C,
Differentiate: iaAeiax + ibBeibx = icCeicx, so (using x = 0), aA + bB = ca.
Differentiate again: -a2 Aeiax - b2Beibx = -c2Ceicx, so (using x = 0), a2A + b2B = c2C.
a2A + b2B = c(cC) = c(aA + bE); (A + B)(a2 A + b2B) = (A + B)c(aA + bE) = cC(aA + bE);
a2A2+ b2AB + a2AB + b2B2 = (aA + bB)2 = a2A2 + 2abAB + b2B2, or (a2 + b2 - 2ab)AB = 0, or
(a - b)2AB = 0, But A and B are nonzero, so a b. Therefore (A + B)eiax = Ceicx,
a(A + B) = cO, or aC = cO, so (sinceC f:.0) a = c. Conclusion:a = b= c. qed
=
Problem 9.16
E- I
-- E- Otei(kt.r-""t) y,A
:/:
}
{ HI =
E- R --
VIIEolei(kl"r-""t)(-COSOI X+sin()1 z);
A
E- ORei(kR'r-""t)y,
{ HR
E- T
=
--
:1 EoRei(kRor-""t)(cos 01 X + sin 01 z);
E- OTe i(kTor-""t)
y,
{ HT
=
:2 EoTei(kTor-""t) (- COSO2X + sin ()1z); }
'
U'
.L
.L
ll
ll
( 111
) E 1 = E 2'
(1) £1E 1
£2E 2'
kR
z
}
A
=
Boundary co~ditions:
{ (ii) B.L1
0
sin O2
=
= B.L2 ,
(iv ) .!..BII
1'1 1
V2
=
Law of refract~on: --=--, [Note: kI 'r - VJt kR
sm 01
VI
exponential factors in applying the boundary conditions.]
BJ
= .!..BII,
1'2 2
'r
-
= kT . r -
VJt
VJt,at z = 0, so we can drop all
Boundary condition (i): 0 = 0 (trivial), Boundary condition (iii): Eot + EoR = EoT.1
1 1 1 "
'
'
. () E
VI sin
B d
d
E
()
E
()
E
E
( )
oun ary con ItlOn 11: -VI Ot sm 1 + VI OR sm 1 = -V2 OT sm 2 => Ot + OR = V2 sm.
But the term in parentheses is 1, by the law of refraction, so this is the same as (ii).
11 1 1 Boundary condition (iv): -EOt(-COS()I) + -EoR COS()1 -EoT(-COS()2)
=>
J.ll [ VI
VI
J.l2V2
]
- -
I
U
-
(
()2
() 1
)
-
E
OT'
=
-
EOt
-
-
-
EoR =
J.lIVI COS()2
(
J.l2V2cos ()1
)-
EoT'
Let
Solving for EoR and EoT: 2Eot
-
-
a == COS()2
-COS() 1 ; {3==-,J.lIVI
J.l2V2
= (1 + a{3)EoT =>
-
-
EoT
Then 1Eot - EoR
-
= a{3EoT.
- I
= (1 +2a{3) Eol;
-
1 - a{3
Eon = EoT - Eot = 1 + a{3 - 1 + a{3 Eot => EoR = 1 + a{3 Eol'
Since a and {3are positive, it follows that 2/(1 + a{3) is positive, and hence the transmitted wave is in phase
(
2
1 + a{3
)
(
)
with the incident wave, and the (real) amplitudes are related by EoT
I
=
(~)
EOt,1 The reflected wave is
164
CHAPTER
9. ELECTROMAGNETIC
in phase if 0:(3< 1 and 1800 out of phase if 0:(3< 1; the (real) amplitudes are related by EOR =:
These are the Fresnel equations for polarization perpendicular to the plane of incide~ce.
I
VI To construct the graphs, note that 0:(3=: (3
sin2 ()/(32
cos e
=:
J (32-
I
WAVES
~
Eo/.
I
sin2e
cos e'
where ()is the angle of incidence,
\12.25 - sin2e
.
so, for (3=: 1.5, 0:(3=:
cos e'
1
-!i
-/\
-7
-0
-5
--1
-3
-2
-1
O' 10 20 30 40 50 60 70 /\0!i0
81
Is there a Brewster's angle? Well, EOR =: 0 would mean that 0:(3=: 1, and hence that
0:
=:
1 =:
Vl-
(v2/vt)2sin2()
1
P2V2
=: - =: -,
or 1-
cos ()
(~:) 2 [sin2
(3
PI VI
V2
2
()
-
VI
.
P2V2
2
2
( )
sm () =: -
PI VI
() + (p2f PI)2 COS2e]. Since PI ~ P2, this means
2
COS(), so
1 ~ (V2/Vt)2, which is only true for optically
indistinguishable media, in which case there is of course no reflection-but that would be true at any angle,
not just at a special "Brewster's angle". [If P2 were substantially different from PI, and the relative velocities
were just right, it would be possible to get a Brewster's angle for this case, at
VI
()
2
=: 1- cos2e
+
V2
P2
( )
PI
2 cos2e => cos2e =: (VdV2)2 -1
(p2/pt)2-1
=: (P2f2fPIfI) -1
(p2fpd2-1
=: (f2/ft) - (PdP2) .
(p2/PI)-(pdp2)
But the media would be very peculiar.]
By the same token, OR is either always 0, or always 7[',for a given interface-it does not switch over as you
change (), the way it does for polarization in the plane of incidence. In particular, if (3 =: 3/2, then 0:(3> 1, for
(3
0:
.)2.25 - sin2e
=:
cos ()
.f
. 2 ()
2
. 2e
2
e
> 1 1 2.25 - sm > cos , or 2.25 > sm + cos e =:
1. ,(
In general, for (3 > 1, 0:(3 > 1, and hence OR =: 7['. For (3 < 1, 0:(3 < 1, and iSH =: O.
At normal inc'idence, 0: =: 1, so Fresnel's equations reduce to EoT =: (1 ~ (3) Eo/; EOR =:
consistent
with Eq. 9.82.
I
~
~ ~ Eon
,
I
Reflection and Transmission coefficients: R =:
(~~:)
2
=:
(~ ~ :~) 2.1 Referringto Eq. 9.116,
I
165
2
T
= E2V2a
EIVI
2
( ) = a{3(~ )
EOr
Eol
1 + a{3
I
=
R+T
.
(1 - a{3)2 + 4a{3 = 1 - 2a{3 + a2{32
(1 + a{3)2
+ 40:{3 = (1 + a{3)2 = 1 ./
(1 + a{3)2
(1 + 0:{3)2
.
Problem 9.17
Equation 9.106 ~ {3= 2.42; Eq. 9.110 ~
a= VI - (sinB/2.42)2 .
cosB
(a) B = 0
- 2.42
1+
~
=-
=
0:
1.42
3.42
=
1.0
1. Eq. 9.109 ~
I
-0.415;
0.8
Eol
0:+{3
0.6
( )
0.4
I
= 10.585.1
= tan-l
9.112 ~ BB
0
(2.42)
= !67.5°.1
91
--0.2
(c) EoR =Eor ~0:-{3=2;0:={3+2=4.42;
(4.42)2 COS2B = 1 - sin2 B/(2.42)2;
(4.42)2(1- sin2 B) = (4.42)2- (4.42)2sin2B
= 1 - 0.171 sin2 B; 19.5 - 1 = (19.5 - 0.17) sin2 B;
18.5 = 19.3 sin2 B; sin2 B = 18.5/19.3 = 0.959;
sine
-------------------
0.2
(~:~) = a: {3 = 3.~2
(b) Equation
EOR
0:-{3
--0.4
-0.6
= 0.979; Ie = 78.3°.1
Problem
9.18
(a) Equation 9.120 ~ r
= E/a.
Now f = fofr (Eq. 4.34), lOr~ n2 (Eq. 9.70), and for glass the index of
= 2 X 1O-11C2/N m2, while a = 1/ P ~ 10-12 n m
(Table 7.1). Then r = (2 x 1O-11)/1O-12 =
(But the resistivity of glass varies enormously from one
type to another, so this answer could be off by a factor of 100 in either direction.)
(b) For silver, p = 1.59 X 10-8 (Table 7.1), and f ~ EO,so c..Jf= 21TX 1010 x 8.85 X 10-12 = 0.56.
Since a = 1/ P = 6.25 X 107 » c..JE,
the skin depth (Eq. 9.128) is
refraction is typically
d
around
1.5, so 10~ (1.5)2 x 8.85 X 10-12
~
= ~ ~ V c..J~fL= V 21TX 1010 x
6.25 ~ 107 x 41TX 10-7
= 6.4 X 10-7
m
= 6.4 x 10-4 mm.
silver to a depth of about 0.001 mm; there's no point in making it any thicker, since the fields don't
penetrate much beyond this anyway.
(c) For copper, Table 7.1 givesa = 1/(1.68 x 10-8) = 6 X107, c..JEO= (21TX 106) x (8.85 X 10-12) = 6 X 10-5.
I'd plate
I
1
Since a »c..JE, Eq. 9.126 ~ k
A = 21f
~
~
-
c..JafLo
Vc..J~fL,
= 21f
so (Eq. 9.129)
V
= 4 X 10-4
m
=
21f x 106 X 6 X 107 X 41T X 10-7
i,From Eq. 9.129, the propagation speed is v
;
2
= ~ = ~ A = AV = (4 X 10-4)
0.4 mm.
I
X 106
I
= 1400m/s.!
In vacuum,
,\ = = 3 ;0~08 = 1300 m; Iv = c = 13 X 108m/s.1 (But really, in a good conductor the skin depth is so small,
compared to the wavelength, that the notions of "wavelength" and "propagation speed" lose their meaning.)
166
CHAPTER
ELECTROMAGNETIC
9.
WAVES
Problem 9.19
(a) Use the binomial expansion for the square root in Eq. 9.126:
~ ~ w f€i, 1 +
V2:
1
2
So (Eq. 9.128) d:;:: - ~ ~
a
ff
-.
P
[
~ (!!-.)2 2
1
:;::W f€i,~!!-.:;:: ~ ~.
1/2
V2: y2EW
]
EW
Qed
E:;:: ErEO :;::
80.1 EO (Table 4.2),
:;::po(l + Xm) :;::po(1 - 9.0 X 10-6) ~ PO
a:;:: 1/(2.5 x 105) (Table 7.1).
For pure water,p
{
2V~
(Table 6.1),
So d
:;:: (2)(2.5 X 105)
(80.1~~.~51~-~0-12):;::
11.19 X 104 m.1
(b) In this case (aIEw)2 dominates, so (Eq. 9.126) k ~~, and hence (Eqs. 9.128 and 9.129)
21f
21f
,\
,\
:;::
-k
~
-~
:;:: 21fd,
Meanwhile
1.3 X 10-8
:;::
-.
21f
Qed
2:
W Vf€i, Vfa
-;;:; :;:: Vwpa:;::
2
~ ~
(1O15)(41f x2 10-7)(107)
:;::
113nm.1 So the fields do not penetrate far into a metal-which
(c) Since k ~~,
.
or d :;::
as we found in (b), Eq. 9.134 says q;:;::tan-l(l)
Bo
Meanwhile, Eq. 9.137 says -E ~
0
H
8 X 107; d
is what accounts for their opacity.
:;::45°.
Qed
For a tYPIcal metal, then, E
Pff-." I.
Bo
EP-:;::
EW
:;::
:;:: .!.
~ :;:: ---2
8 X 107
W
0
:;::
!1O-7 s/m.! (In vacuum, the ratio is lie:;:: 1/(3 x 108) :;::3 X 10-9 slm, so the magnetic
about 100 times larger in a metal.)
Problem
9.20
V
(1O7)(41fX 10-7)
1015:;::
field is comparatively
(a) u:;:: ~ (EE2 + tB2)
:;:: ~e-2I. 1
z; (8)
:;::
qed
_21p EoBoe-2KZ cosq;z.
average of the product of the cosines is (1/21f) J027r
cos ecos(e+q;)
de :;:: (1/2) cos q;.]So I :;:: -kEoBoe-2KZ
1
K
-EJe-2I<
~ 1 =1=2.
1:f: --1-
",,2
W2
9. ELECTROMAGNETIC
(
)(
",,2+""6-2""6=F""01'-2(l=1=I'/2""0)-2
So Q ~ ~Qrnax at ""1 and ""2.
""0
!:><
1 =1=
2""0 -2
~
!:><
2""0 -2.
Qed
Problem 9.25
""
k=
~
dk =~
d""
1+ Nq2"
IV 2
<
Co
I
Problem
Ij
2mfo ~
Vg = c [ 1 + ~2m Co
I v,
(""; - ",,2)] .
[ 1 + 2mfo ~
c [
m,;
h
N q2"
(""J
?
~L
= ~ = C [1 +
~
("'I
~
(v x
(
~
v x
y -
DEz- BEy =
ay
a
Z
ax
E ) = BEy - aEx=
Z
-
~
( v x B )x
-
(
ax
ay
= aiJz - aBy =
aY
az
~
( v x B )y
= aBx - aBz =
(V x B)z
= oByaBx=
ax
ay
aZ
kE-
'
-
aX
r'
is
("";+,,,,2)
2mco ~
(
Z
Oz
(k
-
ax
aye.
)
-
( aY
'
(
- -
ZkB Oz
e
,0
J (""J
S (11) aEz
a
00
y
.
i(kz-",t)
than
i",,2
'
,
zk""By + -;;2E", => Z
Co depending
i""-
(
",,2
)
z-UJt.
S ( ) kE
III Z x
-
aEz
ax
- . B
- z"" y.
'
ay
- .
. S0 (v ) aBz
aY ZkB y
B
z"" z.
=-
i""
c2
E ""
aBoz e i(kz-UJt) . S0 (VI' ) ZkB '" - aBz = - i""E
ax
ax
c 2 y'
'
)
aBo. - OBoz
oy )
( ox
)
(k
- ' kE ='
z
y
z""B x.
ei(kz-UJt)
,
So (iv) oBy
ox
- oB",
= - i""Ez,
ay
C2
-
k aa~z
-"" a~z +i""kBy=
i
.
aEz
aBz
aEz
aBz
()
k - -;? Ex = k ax + "" ay , or 1 Ex = (",,/c)2- k2 k ax + "" ay
2
on "',
'
= 2'c at = -2EoeJ
c
This confirms Eq. 9.179. Now multiply (Hi) by k, (v) by"", and subtract: ik2 Ex
.
",,2)2 ]
-
1 aE
-
V x B
0 1 ax
e
c odess
S ( ) BEy - aEx ='
i(kz-UJt)
)
)
;
'00
i(kz-",t)
aBoz - Z.kB o.
t)
'
e i(kz-UJt) ,0
oEoz
aEo. - OEoz
(
than
Z-UJ
)
ax
great'"
at = i""BoeJ
-
aEoz - z'kE 0
0y'
aE", - oEz oz
"")
aB
x E = --
In the terminology of Eq, 9,178:
E) -
[
.1 Since the second term in square brackets is positive, it follows that
""
-
-
C
("".,+ "" )
(a) From Eqs. 9.176 and 9.177,V
~
( v x E ) '" =
- ",,2)2 ]
J (""J
-I
2
=~ 1+ Nq2 "I'
-(-2",,)
~
L h (.J.""j- 2)2]
w he,eas v
9.26
= dk = (dk/m,;).
Vg
+",,"1'
- ",,2)
1
(
).
.
Multiply (ii) by k, (vi) by"", and add: k a~z -ik2 Ey+i""kBx-""a:xz = i""kBx- i~2Ey =>i (~: - k2) Ey=
169
8Ez
-k 8y
8Bz..
+w
.
"
M u I tIp Iy (II )
.
(
2
)
W2
z k - 7!i
Multiply
.
(
2
W2
z k - 7!i
i
8x ' or
8Bz
- w 8x
).
.wk
.
. 2
8Bz - .w2B
W 8Ez
.wk
b Y w I c, (VI) bY k , an d a dd '. 2Ey
- Z2 E y + Zk B x - k _8
Z2
x - Z2
8
c
y
c
x
c
c
8Bz
w 8Ez
i
"'
( )
Bx = k 8x
(
c
)
W
8Ez
8Bz.
By = c2 8x + k 8y
.
i
USIngEq. 9.180, (w/c)2 - k2
82 Ez
82 Ez
(
(
, or
(IV) By = (W/C)2 - k2
of Eq. 9.180.
8Eo>
(
8x
82Ez
+ 8Eoy + ikEo
8y
82Bz
w 8Ez
z
)
i
x
8Bz
y
w 8Ez
k 8y
+ c2 8x
:}
)
c
iwk
+ ~Ex
c
:}
)
.
ei(kz-",t) = 0 => 8Ex + 8Ey + ikEz = o.
8x
8y
)
(
82Ez
8Ez
)+
+ (W/C)2- k2
82Bz
k 82y - w8x8y
)
.
+ zkEz = 0,
2
[(WI c) - k ] Ez = O.
+ 8~y + ikBz
W 82Ez +
i
Likewise, V . B = 0 => 8:Xx
(
(w/c)2 - k2
82Bz
82Bz
c
i
k 8X2 + w8x8y
2
+ 82y +
i
8Bz
- C2 8y , or 11l Bx = (w/c)2 - k2 k 8x - C2 8y
.
2
.wk
w 8Ez - k-8Bz
. 2
.w2
(iii) by w/c , (v) by k, and subtract: z2Ex - 2- 8
8 + zk By = z2By
(b) V. E = 8Ex + 8Ey + 8Ez =
8x
8y
8z
8x2
8Ez
k 8y
2
This completes the confirmation
or 8X2
(
Ey = (w/c)2 - k2
(ll)
k 82Bz 8X2
c2 8x8y
2
[
2
+ 82y + (w/c) - k
)
= 0 =>
(w/c)2 - k2
] Bz =
(
k 82Bz
8y2
+
W
C28x8y
ikBz = 0 =>
O.
This confirms Eqs. 9.181. [You can also do it by putting Eq. 9.180 into Eq. 9.179 (i) and (iv).]
Problem
9.27
Here Ez = 0 (TE) and w/c = k (n = m = 0), so Eq. 9.179(ii) => Ey = -cBx, Eq. 9.179(iii) =>Ex = cBy,
Eq. 9.179(v) => 8~z = i (kBy - ~Ex)
i (kBx - ~Bx)
= i (k_3y- ~By) = 0, Eq. 9.179(vi) => 8:Xz = i (kBx
+ ~Ey) =
= O. So 8:Xz = 8~z = 0, and since Bz is a function only of x and y, this says Bz is in fact
a constant (as Eq. 9.186 also suggests). Now Faraday's law (in integral form) says
f E . dl = - ! ~~ . da,
and Eq. 9.176 => ~~ = -iwB, so f E. dl = iwf B. da. Applied to a cross-sectionof the waveguidethis gives
f E . dl = iwei(kz-",t) !
Bz da = iwBzei(kz-",t) (ab) (since Bz is constant, it comes outside the integral). But
if the boundary is just inside the metal, where E = 0, it follows that
which we already know cannot exist for this guide.
Problem 9.28
Here a
= 2.28
em and b
= 1.01
em, so VlO
c
V30
= 32a = 1.97
c
x 1010 Hz; VOl
=
2b
=
= 1.49 x
-.!...WlO
211"
10
=
c
2a
I
Bz = 0.1So this would be a TEM mode,
= 0.66 X 1010 Hz;
C
10 Hz; V02 =
<
V
< 1.32
X 1010 Hz.!
A
= ~,
V
so AlO
= 2a;
A20
= 2 2ca = 1.32 X 1010Hz;
10
C
~
22{;= 2.97x 10 Hz; Vll = "2V;;2+ b2 =
1.62 X 1010 Hz. Evidently just four modes occur: 110, 20, 01, and 11.1
To get only one mode you must drive the waveguide at a frequency
10.66x 1010
V20
= a.
between VlO and V20:
12.28em < A < 4.56 em.!
170
CHAPTER 9. ELECTROMAGNETIC
Problem
WAVES
9.29
FromProb. 9.11,(S) = -2 1 (E x B*). Here (Eq. 9.176) E = Eoei(kz-wt), B* = B~e-i(kz-wt), and, for the
J-Lo
TEmn mode (Eqs. 9.180 and 9.186)
-ik
.
-m1l"
m1l"x
n1l"Y
( );
(c.vjC)2- k2 (~
Bosm (~ ) CDSb
)
-n1l"
-ik
. n1l"Y
(
)
(c.vjc)2_k2 (b ) Bocos ~
sm (b ) ;
=
B;
m1l"x
B; =
B; =
Bo CDS(m:x) CDS(n;y) ;
ic.v
-n1l"
m1l"x
. n1l"Y
Ex =
(c.vjC)2- k2 b
Bo CDS~
sm b
;
-zc.v
-m1l"
. m1l"X
n1l"Y
Bo sm ~
CDS b
;
Ey = (c.vjc)2- k2 ~
Ez = O.
( )
( )
(
(
) ( )
) ( )
So
1
m . m1l"X
m1l"x 2 n1l"Y
i1l"c.vBg
(S) = -2 {
2
2
sm
CDS
(
)
)
( a
(- a ) CDS ( - b ) x
J-Lo (c.vc
j ) - k
a
A
n
i1l"c.vBg
m1l"X
2
+ (c.vjc)2- k2 (b ) CDS(~
+
!
n1l"Y
n1l"Y
n
2 m1l"X
. 2 -n1l"Y +
- 2 CDS
sm
( b )
)
(
)
[(
[(c.vjC)2- k2]2
b
a
c.vk1l"2 B5
1
(S) . da
.
m 2
c.vk1l"2
B2
= -8J-Lo[(c.vjc)2-
0
2 ab
k2]
-
[( a )
Joasin2(m1l"xja) dx = Joacos2(m1l"xja) dx = aj2;
A
) sm (b ) CDSb
( )Y
n 2
+ (-b ) ] . I
m 2. 2
SIll
a
( )
m1l"X
(-
a
A
2 n1l"Y z.
)] }
) CDS (- b
[In the last step I used
J: sin2(n1l"yjb) dy
= J: cos2(n1l"yjb)
dy
=
bj2.]
Similarly,
(u)
=
1
-1-- foE.E*+-B.B*
4
(
J-Lo
= -fO
c.v211"2
Bg
n
)
m1l"x
- )2 CDS2 (-
4 [(c.vjc)2- p]2 [( b
+
~
4J-Lo{
+
!
ab
(u) da ="4
l
B2 COS2
0
m1l"x
(
a
)
COS2
a
m 2. 2 m1l"x
.
) sm (- b ) + (-a ) sm (- a )
2
n1l"Y
2
CDS
n1l"Y
(- b
)]
n1l"Y
(b )
n 2 2 m1l"X . 2 n1l"Y . m 2 . 2 m1l"X
2 n1l"Y
CDS
- CDS SIll ( - ) T (- ) sm ( )
( - b )] } .
)
(
)
[(
[(c.vjc)2- k2]2 b
a
b
a
a
k211"2B5
fo
c.v211"2
Bg
n 2
m 2
) (
{ "4 [(c.vjc)2- k2]2 [( b + ~
Bg
1
k211"2
Bg'
n 2
) ] + 4J-Lo
+ 4J-Lo
[(c.vjc)2
- k2]2[( b )
m 2
+ (~ ) ] } .
171
These results can be simplified, using Eq, 9.190 to write [(U.l/C)2- k2] = (U.lmn/c)2,follo = l/c2 to eliminate €o,
and Eq. 9.188 to write [(m/a)2 + (n/b)2] = (U.lmn/7rc)2:
/
/
= 8/l-0U.lmn
U.lkab~2B5i
(8) . da
U.l2ab
(u)da
2
= 8lloU.lmn
2 Bo'
Evidently
energy per unit time
energy per unit length
- J (8) . da
- J(u)da
kc2
c
= -::; = ~";U.l2-U.l~n = Vg (Eq. 9.192). qed
Problem 9.30
Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with Ez =F0, Bz = 0, subject to the boundary
conditions 9.175, Let Ez(x,y) = X(x)Y(y); as before, we obtain X(x) = Asin(kxx) + Bcos(kxx). But the
boundary condition
requires
Ez
But this time m = 1,2,3,...
Ez
= Eosin (m:x)
=0
= 0)
(and hence X
=0
when x
and x
= a,
so B
=0
and kx
= m7r/a.
, but not zero, since m = 0 would kill X entirely, The same goes for Y(y), Thus
sin (n;y)
with n, m = 1,2,3,... .
The rest is the same as for TE waves: U.lmn
= c7rv(m/a)2 + (n/b)21 is the cutoff frequency, the wave
velocity is v = c/Vl - (U.lmn/U.l)2,
and the group velocity is Vg = cVl- (U.lmn/U.l)2.The lowest TM mode is
I
11, with cutoff frequency
U.l1l = c7rv(l/a)2
TE frequency is C7rV(I{:;~~
(l/b)2
=
+ (l/b)2.
So the ratio of the lowest TM frequency
to the lowest
VI + (a/b)2.1
I
Problem 9.31
= ~~(sEs) = 0./; V.B = ~~(B",) = 0./; V xE = oEs (b-~ oEsz = - Eoksin(kz - wt) (b:b
(a) V.E
aB
s as
s o
EoU.lsin(kz - U.lt);.
- -at = - - c
.
(sInce k
OZ
s o
a
Eok sin(kz
s + - - ( sB q,) z = -
oB",
= U.l/ C) ; V x B
A
1
'/J"
= - -oz
s
s as
c
. .
1 aE EoU.lsin(kz - U.lt)A
II
2~
s ./. Boundary condItions: E = Ez= 0 ./;Bl. = B8= 0./.
C ut = ~c
s
/
s
- U.lt)
A
A
s
s
?
=
(b) To determine A, use Gauss's law for a cylinder of radius s and length dz:
= Eo cos(kz - U.lt)
(27rs)dz = -Qenc
= -Adz
=>II
A= 27rfoEocos(kz
-U.lt).
1
1
f E.da
To determine I, use Ampere's law for a circle of radius s (note that the displacement
S
.
.
fO
"
A
'
.
Ioop ISzero, SInce E ISIn t h e S d IrectlOn: )
fO
current through this
Eo cos(kz - U.lt)
f B .dl = -
c
27'0Eo
cos(k z
(27rS ) = Ilo1enc => I = -
S
lloC
- wt) .
The charge and current on the outer conductor are precisely the opposite of these, since E = B = 0 inside
the metal, and hence the total enclosed charge and current must be zero.
Problem 9.32
I
j(z,O) =
-00
1
i:
1
00
j(z,O)
i:
A(k)eikz dk => i(z, 0). =
00
A(-WeilZ(-dl)
=
A(-Weilz dl =
-00
= Re [i(z,O)]
1
I
A(k).e-ikz dk. Let I ==-k; then i(z, °t =
00
A(-k).eikZ dk (renaming the dummy variable 1-+ k).
-00
= 2~ [i(z,O) + !(z,o).] = 1-00
roo _21 [A(k)+ A(-k).]
eikz
dk. Therefore
172
CHAPTER
1 r-
I
-
oo
::)IA(k)+A(-kr
] =-27r j
'" L
-00
Meanwhile,
(Note
that
=
Iklv, here,
j(z,O)*
j(z,O)
-~w
i:
fez, t) =
w
not
come
outside
=
Re [fez,0)] = ~ [f(z,O) + fez,0)*] =
-00 [ilklvA(k)*]e-ikz dk =
[ilklvA(-k)*]eikz dk =
i:
[iwA(
-k)*]eikz
i:
=
Adding these two results,we get A(k)
I
= 2; j-00
1
=
~[-iwA(k)
+ iwA( -k)*]eikz dk.
= -f; i:
i,
[
f(z,O) +
[illlvA(-l)*]eilz(-dl)
dk.
j(z,O)e-ikZ dz, or ~ [A(k) - A(-k)*]
I
dk.
_=
=
i:
i:
[-iwA(k)]eikz
integral.)
=
-oc [iwA(k)*]e-ikz dk =
-
Problem
j
the
j
2~
i:
A(k)( -iw)ei(kz-"'t) dk => f(z,O) =
so it does
=
=
WAVES
'k
f(z,O)e-,'zdz.
=
[A(k) - A(-k)*]
9. ELECTROMAGNETIC
[~j(Z,O)]
e-ikz dz.
,
;:;f(z,O)
]
e-.kz
dz.1 Qed
9.33
(a)
Gauss's
(i)
(ii)Faraday's
aB
at
law:
V .E = --J:-o a;:: = O.
rsm
a.
10
A
(
k .
.
a.
)
A
(
)]
-
0.
B
k
ut or casu = - smu; or smu = casu.
I .
I E0 '
A I
.
.
I.
I
---=-- - 2 sm 0 cas casu - -k smu r - - E osm - k smu + - 2 smu - - casu.
kr
r sm 0 r
r
r
r
o(
o(
)0
)
Integrating with respect ta t, and noting that!
B
A
V x E = ---=---(smOEq,)
r - --(rEq,)
0
r sm 0 00
r or
sin20
I a
.
I.
I a
I.
---=--0casu - -smu
r - -osm 0 casu - - smu
r sm 0 00 [E r
kr
kr
] r or [E
a
=
.I
law:
I
=
u= -2 0 27r
/LoC
0
1
=
47r E2
-~.
3 /LoC
)
174
CHAPTER
Problem
9. ELECTROMAGNETIC
WAVES
9.34
x
z
CD
y
{
Z < 0:
BI(Z, t) = .lE1ei(kIZ-UJt)
y
VI
EI(Z, t) = Elei(kIZ-UJt)X,
En(z, t) = Enei(-kIZ-UJt) X, Bn(z, t) = - ;1 Enei(-kIZ-UJt) y.
Br,(z t ) = .lE
ei(k2z-wt)Y
V2 r
~
Er(z, t) = Erei(k2Z-UJt) x,
EI(Z, t) = Elei(-k2z-wt) X,
Bl(Z, t) = - vl2Elei(-k2z-wtj y.
O d:
{ ET(Z,t) = ETei(k3Z-UJt)
X, BT(Z,t) = ;3ETei(k3Z-UJt)
y.
= J.L2 = J.L3 = J.La):
Boundary conditions: E~ = E~, B~ = B~, at each boundary (assuming J.LI
EI + En = Er + El;
Z --. O.
1-
Ereik2d
Z -
-.
1-
1-
- -En
{ -EI
VI
VI
1-
= -Er - -El
V2
+ Ele-ik2d
V2
-
-
--
=>EI - En = {3(Er- El), where {3==VdV2.
== ETeik3d;
d.
]...Ele-ik2d = ]...ETeik3d=>Ereik2d- Ele-ik2d = aETeik3d, where a ==v2/v3.
{ ]...Ereik2dV2
V2
V3
We have here four equations; the problem is to eliminate En, Er, and El, to obtain a single equation for
ET in terms of EI.
Add the first two to eliminate En :
2E1 = (1 + {3)Er + (1 - {3)El;
Add the last two to eliminate El :
2Ereik2d = (1 + a)ETeik3dj
Subtract the last two to eliminate Er : 2E1e-ik2d = (1 - a)ET eik3d.
Plug the last two of these into the first:
2EI
=
(1 + {3)~e-ik2d(1
+ a)ETeik3d + (1 - {3)~eik2d(1
- a)ETeik3d
2
2
4EI
=
[(1+ a)(1 + {3)e-ik2d+ (1 - a)(1 -
=
=
[(1 + a{3) (e-ik2d + eik2d) + (a + {3)(e-ik2d - eik2d)] ETeik3d
{3)eik2d] ETeik3d
2 [(1+ a{3)cos(k2d) - i(a + {3)sin(k2d)] ETeik3d.
175
Now the transmission coefficient is T
l1El12
;:--a /3 IETI2
=
T-I
VlflElo
VI
_I [(1 + a/3) cos(k2d)
= -I
-
= vafaEto = Va
-
a /3 2
J.LOfa
I~TI2
J.Lofl
IEl12
( )
= VI
I~TI2
=a
Va IEl12
..
IETI2
/3IEI12'
so
2
'
z(a + /3) sm(k2d)] etkad
I
1
=
4~/3 [(1 + a/3)2cos2(k2d)+ (a + /3)2sin2(k2d)]. But cos2(k2d) = 1 - sin2(k2d).
=
4~/3 [(1 + a{3)2+ (a2 + 2a{3+ (32- 1- 2a{3- a2{32)sin2(k2d)]
1
4a {3 [(1 + a{3)2 - (1 - a2)(1 - (32)sin2(k2d)] .
c
c
c
na
n2
But ni = -, n2 = -, na = -, so a
-,
{3 -.
-
VI
= - 1
4nlna
[
V2
(n2n2
1-
(nl + na)2 +
= na
Va
)( n2
-
2 2 a
n2
= nl
n2 )
2 sin2(k2d)
]
.
Problem 9.35
=
T = 1 => sinkd = 0 => kd = 0,11',211'
The minimum (nonzero) thickness is d = 1I'/k. But k w/v =
27rv/v = 211'vn/c, and n = VfJ.L/fOJ.LO
(Eq, 9.69), where (presumably) J.L~ J.Lo.So n
Vf/fO = Fr, and hence
=
1I'C
c
d = 211'vFr = -2vFr
Problem 9.36
From Eq, 9.199,
T-l
108
= 2(103x X109)
~2.5 = 9.49 x
.
=
1
-a
I
m, or 9.5 mm.
I
[(4/3) + 1]2+ [(16/9) - (9/4)][1 - (9/4)] Sin2(3wd/2C)
(9/4)
}
4(4/3)(1) {
3 49
(-17/36)(-5/4).
=
16 [ "9
+
=
T
10
49
2
(9/4)
sm (3wd/2c)]
48
85
= 48 +
.
2
(48)(36) sm (3wd/2c).
.
49 + (85/36) sin2(3wd/2c)
.
. 2(3wd/2c) ranges from 0 to 1, Tmin = 49 + 48
48 = rnnon-l
Sillcesm
(85/36) = 0.935; Tmax= 49
~
Not much
I
I
variation, and the transmission is good (over 90%) for all frequencies. Since Eq. 9.199 is unchanged when you
switch 1 and 3, the transmission is the same either direction, and the fish sees you just as well as you see it.
Problem 9.37
I
(a) Equation
9.91 => ET(r,t)
kr(x sin OT + Z cos OT)
= xkT
= EoTei(kT,r-I4It);
sin OT + izkTV sin2 OT - 1
.
Wn2
k == kTsmOT= (K, == kTvsin2
ET(r, t)
=
kT'
c
nl
.
) -smOI
n2
r = kT(sinOTx
+ yy + zi)
= kx + iK,z, where
,..lnl .
c
= -smOl,
OT - 1 = W;2 v(nl/n2)2
EOTe-ltzei(kx-l4It). Qed
+ COSOTZ)' (xx
sin2 01 -1
= ~/n~ sin201- n~. So
I
=
176
CHAPTER 9. ELECTROMAGNETIC WAVES
(b) R
.
2
2
= ~:: = I: ~~I.
1
1
with a real: R
ia-/3
-:--w + /3
(
=
Here /3is real (Eq. 9.106) and a is purely imaginary (Eq. 9.108); write a = ia,
-ia-(3
a2+(32
-w'/3+
= a2 + (32
2
1-a(3
1-a(3
1-ia(3 2
(l-ia(3)(l+ia(3)
f1l
)(
)
=[I]
(c) From Prob, 9,16, EOR= 1 + a(3 EO1,so R = 1 + a(3 = 1 + ia(3 = (1 + ia/3)(l - ia(3) = L!:J
(d) From the solution to Prob. 9.16, the transmitted wave is
1
1
1
1
/
1
E(r , t ) = E OT ei(kT'r-uJt)Y, :B(r,t) = ~EoTei(kT.r-"'t)(-cos(hx+sin(hz).
V2
A
Using the results in (a): kT.r=kx+iK,z-wt,
sinOT=
-
B (r,t ) =
E(r, t) = EoTe-l 0, as here). For Ixl > ct, S = o.
So the energy per unit time entering the box in this time interval is
dW
dt
=p =
/
S(d) . da
=
(ct Lf!:.oo:2lw
4c
I
Note that no energy flows out the top, since S(d + h) = o.
179
d+h
_(d+h-x)3
d)2.
]
d
_
-
foJ.L~o:2lwh3
I
12
CHAPTER
180
t2
it
(ct - d)3
(ct - d)2 dt
dlx
= /LoEo, this
Since l/c2
Problem
= /LO~:lW [
=
(d+h)le
]
3c
die
agrees with the answer to (a).
10.3
E=-VV_aA
B=VXA=@J
at=
This is a funny set of potentials for a stationary point charge q at the origin. (V
I
I
course, be the customary choice.) Evidently
Problem lOA
E
= - VV
B
=
I
p
= qJ3(r);
~~ = -Ao cos(kx-
-
VXA=Z:X
J
= 0.1
wt)y(-w)
=
I
Aowcos(kx - wt)y,
= --aB
at
= - Aowk sin(kx
=
/LoEo
- wt) Z,
..(.
~~ = Aow2sin(kx - wt) y.
VxB = -y :x [Aokcos(kx- wi)]= Aok2sin(kx- wt)y,
Problem
I
[Aosin(kx-wt)] =IAokcos(kx-wt)z.1
VxE = z :x [Aowcos(kx - wi)] = -Aowksin(kx - wt) Z, - ~~
So V x B
=~
41TEO~,
r A = 0 would,of
= 0,,(, V.B = 0..(.
Hence V.E
so VxE
AND FIELDS
(d+h)le
f Pdt = /LO~:lW f
(c) W =
10. POTENTIALS
~
provided
I
= /LoEOW2,lor, since C2 = 1/ /LoEo, w
k2
I
= ck.1
10.5
a>..
V I =V--=O-
at
(
---
1 q
41TEo r
)
~
= --;
q
41TEOr
I
A =A+V>"=---r+
1 qt
41TEOr2
~
(
1
--qt
47r£0
1
)( )=
--r
r2
fI)l
~
LQ:J
.
This gauge function transforms the "funny" potentials of Frob. 10.3 into the "ordinary" potentials of a stationary point charge.
Problem 10.6
Ex. 10.1: V.A
= 0; ~~ = o.
I
Both Coulomb
and Lorentz.!
r
qt 3 (r); ~
aV
Frob. 10.3: V.A = --41TEOV. 2"
= --15
= O. I'NeIther. I
r
EO
vt
qt
aV
Frob.
()
~
10.4:V.A = 0; at = o. ~
181
Problem
10.7
8V
Suppose V.A i: -Jl.OE°Ft,
8V
.
(Let V.A + Jl.QE°Ft = cI>-some known functiOn.) We want to pick A such
8V'
that A' and V' (Eq. 10.7) do obey V.A'
-Jl.oEo8t .
=
I
.
W
W
2
V.A +j.toEO 8t =V.A+V
~A
A+Jl.QE°Ft-Jl.QEO8t2
2
=cI>+0
A,
This will be zero provided we pick for A the solution to 02 A = -cI>,which by hypothesis (and in fact) we know
how to solve,
We could always find a gauge in which V' = 0, simply by picking A = J; V dt'. We cannot in general pick
A = O-this would make B
= 0. [Finding such a gauge function would amount to expressing A as - V A,and
we know that vector functions cannot in general be written as gradients-only
zero, which A (ordinarily) does not.]
Problem 10.8
l.From the product rule:
V. (~) = ~(V .J) + J
ButV~1- = -V/~,1- since4=r-r/,
'
(
V ~)
if they happen to have curl
, V'. (~) = ~(V/.J)+J, (V/~).
So
- J . (VI ~) = ~ (V .J) + ~ (V'
V. (~) = ~(V .J)
- V'. (~) ,
.J)
But
V.J
= 8J%+
8Jy
8x
8y
and
+ 8Jz
8z
8tr
1&
-.=
8x ---,
c8x
= 8J%8tr
8tr 8x
+ 8Jy 8tr + 8Jz 8tr,
8tr 8y
-- ---,
1 81-8tr
8y
c8y
8tr 8z
8tr 1 818z - -~8z'
so
V.J = -~
c
8J%& + 8Jy & + 8Jz &
[
8tr 8x
8tr 8y
8tr 8z ]
= -~c 8tr
8J
. (V1-).
Similarly,
- 8p
- ~c 8tr
8J
8t
V/.J =
. (V/1-).
[Thefirst term arises when we differentiate with respect to the explicit r/, and use the continuity equation.]
thus
V.
(~) = ~ -~ 8tr
1-
1-
[
C
+
8J ,(V/1-)
]
~ - 8p
-~
8J
c 8tr
[ 8t
1-
. (V/1-)
- V.
]
(~) = -~
1-
8p
1- 8t
- V'.
(~)
1-
(the other two terms cancel, since V1-= -V/1-), Therefore:
.
8
Jl.o
V.A=- 411" --8t
[
f
p
-dr..,.
f
V.I
J
()
8
1
f
p
Jl.o
f
J
-..,. dr =-Jl.oEo-411"EO -dr
-- 411" -.da.
8t
11-
]
[
]
Th~last term is over the suface at "infinity", where J = 0, so it's zero. Therefore V.A = -Jl.oEo~~. ./
CHAPTER 10. POTENTIALS
182
Problem
10.9
< riG, A = OJfor t > riG,
(a) As in Ex. 10.2, for t
V(ct)2-r2
A(r,t)
=
=
I
J.t°z 2
)
(411"
J.tok
211"z
k(t-vr2+z2/c)dz=J.tokz
tin
211"
- r2
ct + y(ct)2
r
(
)
8t
-
211"
+ z2
C
( ct +
dz
°
}
- r2 ) (;:) ( c +
1
2c2t
=
1-211" In (
{
J.tok
(
)
r
1
2c2t
2 y(ct)2 - r2 ) - 2c y(ct)2 - r2 }
_J.tok z In ct + y(ct)2 - r2 +
211"
8Az
)
1
y(ct)2
-
ct
-
y(ct)2 - r2
ct
y(ct)2 - r2 }
- r2
ct + y(ct)2
r
)
zl
(or zero, for t
8r
J.tok
(-2r)
[ r2V(ct)2-r2
r
-c
- 211" t (ct + y(ct)2 - r2 )
-ct2
211"{ ry(ct)2
+
- r2
t
-
y ( t )2
c
2
-r
r2
{
-- --J.tok
r
cy(ct)2 - r2 }
1
(-2r)
- 2c y(ct)2 - r2
- --J.tok (-c2t2
;.. 'P
]
+ r2) 'P
;.. 211"rcyl(ct)2 - r2
A(r, t) = (J.t~;o z) 2100 c5(t~ IJ-Ic)
1
2IJ-ch
211"rc
A(r, t) =
dz.
IJ-ch
= yIJ-2 - r2 =>dz = -2 VIJ-2- r2 = VIJ-2- r2 , and z = 0 => IJ-=
z
J.toqo roo ~c5
211" ir
IJ-
r, Z = 00 => IJ-= 00. So:
IJ-dJz,
(t - ~C) VIJ-2-
r2 .
A
}
-J.tok y (ct)2
(b) A(r, t) = 4J.toz ('X> qoc5(t- IJ-Ic) dz. But IJ-= vr2 + Z2, so the integrand is even in z:
11" i-oo
IJ-
Now z
I
- r2 ] . Accordingly,
r
(
{
1
-
vr2
= - 8A = _J.tokz In ct + y(ct)2 - r2 +
t
-
1
- ~y(ct)2
-~
dz
0
{
r
=
I
t
+ z2
vr2
"
v(ct)2-r2
v(ct)2-r2
0
( )[
E(r, t)
B(r, t)
AND FIELDS
cf>i
2
A
- r cf>
.
183
Now8 (t - ,z.jc)= c8(,z.- ct) (Ex. 1.15); therefore A
JLoqoc
A(r, t) =
E(r, t)
Problem
-
00
8(1.- ct)
ch, so
V,z.2- r2
Z (or zero, if ct < r);
r2
- aA = - JLoqoc - !
at
B(r,t) =
A
1
2n: V(ct)2
=
= -JLoqo
2n: z c 1r
2n:
( )
(_!)
2
- aAz (fJ = _JLoqoc
at
2n:
2c2t
-
[(ct)2
2
Z
r2J3/2
-
JLoqoc3t
- r2J3/2 z
A
-
-2r
2n:[(ct)2
A
-
I
-JLoqoCT
[(ct)2 - r2J3/2 4J -
(or zero, for t < Tic);
A
2n:[(ct)2 - r2J3/2 4J
I
(or zero, for t < Tic).
10.10
J (t -lJ,jc) dl = JLok { t J
A = JLo I(tr) dl = JLok
4n:
J
IJ,
4n:
IJ,
dl
4n:
IJ,
_!c
J dl } .
I
1 {
{b dx
f
4n: { a 1dl + b J2 dl + 2 x Ja -;; } . Here 1dl = 2ax (mner
!
JLokt
J
But for the complete loop, dl = 0, so A =
A'
A
circle),J2dl = -2bx (outer circle), so
JLokt 1
A=
4n:
[
1
a(2a)+b(-2b)+2In(bja)
]
JLokt
A
x=?
I
A
A=2;""ln(bja)x,
E
aA =
= - at
1
-JLOkln(bja)x.
2n:
The changing magnetic field induces the electric field. Since we only know A at one point (the center), we
can't compute V x A to get B.
Problem 10.11
In this case p(r,t) = p(r,O) and j(r,t) = 0, so Eq. 10.29 =?
E(r,t) =
=
~
~dT', but tr = t - ~ (Eq.10.18),so
p(r"O) +!(r',O)tr
+ p(r"O)
4n:fO J [
IJ,
Q'l. ]
C
~
p(r"O) + p(r',O)t - p(r',O)(lJ,jc) + p(r"O) ~dT' = ~
p(r', t) ~dT'.
4n:fO
Problem
J[
1J,2
CIJ, ]
4n:fo
J
1J,2
qed
10.12
In this approximation
JLO
B(r,t)
=
=
Problem
1J,2
we're dropping
1
the higher derivatives
.
J 1J,2J(r',t)+(tr-t)J(r',t)+cJ(r',t)
JLo
_4n: J J(r',t)IJ,2 x~ dT.' qed
4n:
[
of J, so j(tr)
IJ,.
= j(t),
and Eq. 10.31 =?
1-
]
x~dT',buttr-t=-c(Eq.1O.18),so
10.13
x
At time t the charge is at r(t) = a[cos(wt) + sin(wt)y], so v(t) = wa[- sin(wt)x + cos(wt)y].
4 = zz - a[cos(wtr) x + sin(wtr) y], and hence 1J,2
= z2 + a2 (of course), and IJ,= VZ2 + a2.
Therefore
~. v = IJ,
!(Ji' v) = IJ,
! {-wa2[- sin(wtr) cos(wtr) + sin(wtr) cos(wtr)]} = 0, so ( 1 - ~.Cv ) = 1.
--,
184
CHAPTER 10. POTENTIALS
AND FIELDS
Therefore
1
q
.
V(z, t) = 1411"1':0
VZ2 + a2'
(z, t) =
I411"I':OC2VZ2
qc.va.+ a2 [-
x
sm(c.vtr) + cos(c.vtr)y), where tr
I
I
=t -
~.I
Problem 10.14
Term under square root in (Eq. 9.98) is:
I
=
=
c4t2
-
2c2t(r . v) + (r . v)2 + c2r2 - c4t2 - v2r2 + v2c2tZ
(r. v)2 + (c2 - v2)r2 + c2(vt)2 - 2c2(r. vi).
(r. V)2
+
(C2
- v2)r2
r2v2
=
=
put in vt
=r
- R2.
+ c2(r2 + R2 - 2r. R) - 2c2(r2 - r. R)
= (r.
V)2 - r2v2
+ C2R2.
but
(r. v)2
-
=
=
((R + vi) . V)2 - (R + vt)2V2
(R. v)2 + v4e + 2(R. V)V2t - R2V2 - 2(R. V)tv2- v2t2v2
(R. v)2 - R2V2 = R2V2 COS2 () - R2V2 = _R2v2 (1 - COS2())
_R2v2 sin2 ().
Therefore
1= _R2v2 sin2 () + c2R2 = c2R2 (1
-
~: sin2 ()) .
Hence
t)
VCr,
1
. qed
q
- -411"1':0
R V 1- V2.
~sm
2 ()
Problem 10.15
Once seen, from a given point
x, the particle will forever remain
in view-to
disappear it would
have to travel faster than light.
ILight rays in + x direction
A person at point
:r;first sees the
particle when this point is reached
i.e. at x = _oct,or
,,
t = -x/c
;;
Region below wavy line represents space-time
points from which the particle is invisible
I
,,
,
,
185
Problem
10.16
First calculate tr: tr = t -Ir - w(tr)l/c ~
-c(tr - t) = x - Jb2 + c2t~ ~ c(tr - t) + x
= ylb2
+ c2t~;
4
l
c2t; - 2c2trt + c2t2 + 2xctr - 2xct + X2 = b2 + c2t;;
..
w(t,J
(
2ctr(x - ct) + (X2 - 2xct + c2t2) = b2j
2
2
x
b2
- (x - ct)2
2cx-ct
(
2ctr(x - ct) = b - (x - ct) , or tr =
)'
1
qc
41I"€0~
( - Ji. V)' and~c-Ji'v=~(c-v);
NowV(x,t)=-
p
~=c(t-tr).
~
1
2c2tr =
c2tr
=
c2tr
; C- v = c2tr + c(x - ct) - c2tr = c(x - ct) ;
2 Jb2 + c2t~
c(tr - t) + x
ctr + (x - ct) ()
ctr + (x - ct)
ctr + (x - ct)
V=
b2 - (x - ct)2
b2+ (x - ct)2
IJ,C
- Ji.V = c(t - tr)c(x - ct) = C2(t - tr)(x - ct) ; ctr + (x - ct) =
+ (x - ct) =
;
ctr + (x - ct)
ctr + (x - ct)
2(x - ct)
2(x - ct)
t - tr = 2ct(x - ct) - b2+ (x - ct)2
2c(x - ct)
1
IJ,C-
b2
Ii'
v
=
+ (x - ct)2
2(x
[
-
ct)
= (x -
ct)(x + ct) - b2 = (x2 - c2t2- b2),ereTh
2c(x - ct)
2c(x - ct)
1
] C2(X
-
-
ct) [2ct(x
2c(x
ct) -
-
ct)
b2 + (x
£ore
b2 + (x
-
ct)2]
= c(x -
ct) [2ct(x
-
- ct)2
ct)
-
b2
+ (x - ct)2] ,
The term in square brackets simplifiesto (2ct + x- ct)(x - ct) - b2 = (x + ct)(x - ct) - b2 = X2 So IV(x,
t) =
c2e
-
b2.
-
!L
b2+ (x ct)2
411"€0
(x - ct)(X2 - c2t2 - b2) ,
Meanwhile
A
=
Vv
C2
=
c2tr
ctr + (x
- q
b2
V X=
- ct) c2
[
b2- (x - ct)2
411"€oC
(x - Ct)(X2 - C2t2 - b2)
Problem
-
-
(x ct)2
2(x - ct)
q
b2 + (x - ct)2
X
2c(x ct) ] b2 + (x - ct)2 411"€0
(x - ct)(x2 - c2t2 - b2)
A
X
,
10.17
From Eq. 10,33, c(t - tr) = ~ ~ C2(t - tr)2 = ~2 = Ji' Ji. Differentiate with respect to t:
aJi
aJi
atr
atr
2
2c(t-tr)
l~at
=2Ji, at' or 02- I-at
=Ji. at' NowJi=r-w(tr),
so
(
( )
atr
atr
ali
ow
.ow atr
atr
- = -= --= -v-'
02- 1- = -Ji' v-'
at
at
atr at
at ' ( at )
at '
atr
and hence ~ut
)
02-
= -,
Ji' U
Qed
atr
atr
02-= -(02- -Ji' v) = -(Ji' u) (Eq. 10.64)
at
at
'
-,
I
I
CHAPTER 10. POTENTIALS AND FIELDS
186
v
Now Eq. 10.40 says A(r, t) = 2'
c VCr, t), so
oA
ot
=
-!.. ov V
ot
(
1
[
v OV
+
C2
otr
ot
) = -!..(otrot
ov otr V
+
c2
lqc
1
-qc
+ v 41TfO(".
v OV
ot
)
0
u)2 ot (1-C
-". v)
=
c2 aFt
=
&, 0"
c241TfO" . u Ft - (". u)2 Cot - ot .V-".
1 qc
41TfO'" u
[
a
otr
(
But 1-= c(t - tr) =>ot = c (1- Ft ) ,"
OV
01-
OV otr
otr
a-
{
". {
41TfOC('"u)2
=
41TfOC~
U )2
0"
otr
= r - w(tr) =>ot = -Vat (asabove),and
otr
ot - otr ot - ot'
otr
q
a""u
(
) --v
=
]
ov
ot )] .
v
c2
[(
ot
1--
otr
ot
)
+v 20tr
--""aot
-c2v+[(",.u)a+(c2-v2+".a)v]0;r
" . u )2 {-c2v+[(".u)a+(c2-v2+".a)v]~
41TfOC
r". )3 [-c2V('" u) + ~(",. u)a +
qc
1
(-z.c-".v) (-v+-a ) +-(c
41Tfo(-z.c-",.v)3 [
c
c
=
U
=
]}
". u }
v2
~(c2
1-
ot
}
ut
41TfOCr
=
otr
1-
2
+ ",.
-v
2
a)v]
+".a)v
].
Qed
Problem 10.18
E = 41TfO
q (",. 1-U )3 [(c2_v2)u+"x(uxa)].
Here v =
v X, a ==aX, and, for points to the right,.i = X. So u =
(c-v)x,
uxa=O,
and""u=1-(c-v).
E
=
-.!L
B
=
~.i x E = O. Qed
c
41TfO 1-3(c
1-
- v)3
For field points to the left,
E=--
q
(c2- v2)(C- v) X = -.!L~
.i =
1-
2
41TfO1-3(C +
v )3(C
(c + v)(c
2
10.19
R
dx
A
2 2
(a) E = -(1
- v /C )
2"
3 2'
41TfO
R [1 - (v/C)2 sin2 8] /
The horizontal components cancel; the vertical component of R is sin 8 (seediagram). Here d = Rsin8, so
1
sin2 8
x
2
d
-
(c - v)3
+ v) x, so",.
.
-v )(c+v)x=
V)2
X = -.!L~
c
,
+v
( ) x;
41TfO1-2 C- v
u = 1-(c + v), and
-q
I
p
x
41TfO 1-2
-x and u = -(c
..x
-R04
1
2'
41TfO1_
c
-
V
( )
-+ v
C
.
x; B=O.
Problem
/
- -d = cot 8, so dx = -de - csc 8) d8 = -:-r
R 2 = --.:n-;
usm 8 d8j
y
xi,
...x
dq=>.dx
187
1
R 2 dx
d
= ~sm
=
E
=
sin20
dO
d
0 d2 dO = _
'
>.
y
-(1
( )1
- v2je2) -d
411"£0
>'(1 - V2je2)
y
411"£od
1
411"£od
.
sinO
11"
0 [1 - (vje)2 sin2 0]
1
1-1 [1-
>'(1 - V2j e2) y
=
Thus
2
3/2 dO. Let z ==cosO, so sm 0 = 1 - z
2
.
3 2 dz
(vje)2 + (vje)2z2] /
1
+1
z
[ (vje)3 (e2jv2 - 1)v(ejv)2
- 1 + Z2] -1
1
~
>'(1- v2je2) e
1
2
.
-2>'Y (same as for a hue
charge at rest).
4
d
d
(
2j
2)
Y
=
411"£0 v 1 - e v v(ejv)2 - 1 + 1
11"£0
A
A
I
(b) B"; e~ (v x E) for each segment dq = >'dx. Since v is constant, it comes outside the integral, and the
same formula holds for the total field:
B
But >.v= 1, so B =
I
= ~(v
e2
4;"d
J-to 21 (jJ
A
x E)
I
= ~v~
2>'(x x y) = J-to£ov~ 2>'z =
e2 411"£0d
(the same as we got in magnetostatics,
411"£0d
J-to2>'vz.
411" d
Eq. 5.36 and Ex. 5.7).
Problem 10.20
w(t)
= R[cos((,)t)x + sin((,)t) Y]j
y
v(t) = &[- sin((,)t) x + cos((,)t) Y]j
a(t) = -R(,)2[COS((,)t)x + sin((,)t) y] = _(,)2w(t)j
'"= -w(tr);
x
IJ-=R;
tr = t - Rj e;
4 = -[cos((,)tr) x + sin((,)tr)y];
u
'" x (u x a)
=
=
=
",. U =
c4 - v(tr) = -e[cos((,)tr)X+ sin((,)tr)y] - (,)R[- sin((,)tr)x + cos((,)tr)y]
- ([e cos((,)tr)- (,)Rsin((,)tr)]x + [esin((,)tr)+ (,)Rcos((,)tr)]y} j
('". a)u - (",. u)a; ",. a = -w . (_(,)2W) = (,)2R2j
R [ecos2((,)tr) - (,)Rsin((,)tr) cos((,)tr) + esin2((,)tr) + (,)Rsin((,)tr) cos((,)tr)] = Rej
V2= ((,)R)2. So (Eq. 10.65):
E
=
q
R
2
2 2
2
q eu - Ra
[ (
)
]
411"£0
(Re)3 u e - (,) R + u((,)R) - a(Re) = 411"£0(Re)2
= ~
1
411"£0(Re )
2 {-[e2 cos((,)tr) - (,)Re sin((,)tr)]x - [e2sin((,)tr) + (,)Recos((,)tr)] y
+ R2(,)2cos((,)tr) x + R2(,)2sin((,)tr) Y}
1
411"£0(R e )2 {[ ((,)2R2 - e2) cos((,)tr) + (,)Rcsin((,)tr)] x + [((,)2R2 - e2) sin((,)tr) - (,)Re COS((,)tr)]Y} .
= I~
188
CHAPTER
B
L
1
"
( E y-~y
-~xE=-~x
C
C
=
.
Ex )
.
C 411"/::0
(R C) 2 {COS(wtr) [(W2R2
- sin(wtr)
AND FIELDS
Z
l
= - ~---L
10. POTENTIALS
-
C2) sin(wtr)
-
wRccos(wtr)]
[(W2R2 - C2) cos(wtr) + wRc sin(wtr)] } Z
= -- q
_q
4w~ R :~-~ [-wRccos2(wtr) - wRcsin2(wtr)] Z = 4w~ R :c3wRd
=
--zq
w.
4WEORc2
.
Notice that B is constant in time.
I,
To obtain the field at the center of a circular ring of charge, let q -t A(2wR) j for this ring to carry current
2wI 1 .
we need I = AV = AwR, so A= I/wR, and hence q -t (I/wR)(2wR) = 2wI/w. Thus B = R 2 Z, or,
4WEO c
since l/c2 = EoJ.Lo, B
I
Problem
=
~ z,
I
the same as Eq. 5.38, in the case z = O.
10.21
A(t/J,t) = Aolsin(O/2) I, where 0 = t/J- wt. So the (retarded) scalar potential at the center is (Eq. 10.19)
V(t)
~dl'
= -24WEOJ 1-
=
_
A
0
4WEO
= -2-
(21. From Eq. 10.19 (again)
A(t)
=
J.Lo
4w
J 1-! dl' = 4w 10
J.Lo {21< AOwa Isin[(t/J - wtr)/2]
a
1 4> adt/J.
But tr = t - a/cis again constant, for the t/Jintegration, and 4>= - sin t/Jx + cos t/Jy.
=
J.Lo~~wa121~-/
Gb-
Q'..
~V:
.::tJ
~~
~~
~G
~
.~\(
..f
.G b-
~':':'-1
~x
~,
Yt~.
~ v.
(>
~
/,,0
I;
Q
x,€:,
"c>\
.0
~(>
~~
",'-=
y
(
I>~
,/;.
r~,
"..x..."
~~,
~/~"
y"..g.
/ v
I; ~.
~~
"' -!(~
o '
~
;"<~
[ I ~valuate at the retarded
time.t.
E=-VV-
ilA:B=VXA
ilt .
:..r~
,,~,
. x.<;."
,,~,
~
190
CHAPTER 10. POTENTIALS
AND FIELDS
Problem 10.23
Using Product Rule #5, Eq. 10.43 =}
v .A
=
{Loqev
411"
. V [(e2t- r. V)2+ (e2- v2)(r2 - (h2)] -1/2
=
-~ [(e2t - r. V)2 + (e2- v2)(r2 - e2t2)]-3/2V [(e2t411"
{ 2
--{LOqe [(e2t-r'v. )2 + (e2 -v 2 )( r 2 -e 22t )] -3/2 v. { -2 (e2t-r.v
{LOqeV'
=
r. V)2+ (e2- v2)(r2 - e2t2)]
}
2
) V (r.v ) + (e -v
811"
2
2
) V (r )} .
Product Rule #4 =}
V(r. v)
(
V)
v.
r
V(r2)
V.A
=
v x (V x r) + (v. V)r, but V x r = 0,
=
(vxax +vYay +vzaz ) xx+yy+zz
=
V(r.r)=2rx(Vxr)+2(r.V)r=2r.So
a
a
a
(
~
~
~
)
~
~
~
=vxx+Vyy+vzz=v,
an
d
=
2
-- {Loqe [(et-r'v
811"
=
{Loqe[(e2tr. v)2 + (e2- v2)(r2 - e2t2)]-3/2 {(e2t - r. v)v2 - (e2- v2)(r . v)} .
411"
2
2
2
2
2 2 -3/2
2
v. [-2et-r.vv+
) + (e -v )(r -et )]
(
2
2
(e -v )2r ]
)
But the term in curly brackets is : e2tv2 - v2(r. v) - e2(r. v) + v2(r. v)
-
{Loqe3
-
~
= e2(v2t -
r. v).
(V2t - r. v)
[(e2t- r. v)2 + (e2 - v2)(r2 - e2t2)]3/2'
Meanwhile, from Eq. 10.42,
aV
-{Lofo-
=
at
1
-{Lofo-qe
411"fo
a
(
--
1
2
)[
2
(e t-r.v
2
2
2
2
22
) + (e -v )( r -e t )]
-3/2
x
-at [(e2t- r. v)2 + (e2- v2)(r2- e2t2)]
=
- {Loqe[(e2t
- r. v)2 + (e2 - v2)(r2 - e2t2)] -3/2 [2(e2t - r. v)e2 + (e2 - v2)( -2e2t)]
811"
= Problem
(a)
I
(e2t - r . v - e2t + v2t)
32
411" [(e2t - r. v)2 + (e2 - v2)(r2 - e2t2)] /
{Loqe3
= V . A.
./
10.24
F2
a;(t)
_
= 4q1q2
1
1I"fO (b2 + e 2 t 2) X.
~
~
,
(This is just Coulomb's law, since q1 is at rest.)
00
OO
1
q1q2 -1
q1q2
tan -1 (et
(b ) 12 4
b
4
[
/b)] -00
1I"fO -00 (b2 + e 22)
t dt
1I"fO e
j
=_
q1q2
= 4nobe
11"
=_
11"
[2" - (- 2")] =
qz
q1q2
11"
411"fObe'
1
= -4q1q2b [tan -1 ( 00 ) -tan
1I"fOe
-1 -00
(
)]
..a;
191
q2
c-
1
V
( )
~
(c) From Prob. 10.18, E = -- 4
7TfO---"2x
C+ v x. Here x
t ,.
and v are to be evaluated at the retarded time tr, which is
given by c(t - tr) = x(tr) = Vb2 + c2t~ :::}c2t2 - 2cttr +c2t; =
2
2 2
t,.(t)
.
c2t2 - b2
b + c tr :::} tr = 2c2 t . Note: As we found In Prob. 10.15,
q2first "comes into view" (for qI) at time t = O. Before that it
can exert no force on q1, and there is no retarded time. From
the graph of tr versus t we see that tr ranges all the way from
-00 to 00 while t > O.
2c2t2
x(tr) = c(t - tr) =
c2t2
(
-
b2
c2t2 + b2
2ct
=
c- v
c+ v
= (C2t2+ b2) +
b2 + c2t2
=
2ct
b2 + c2t2
)(
v(tr)
2t
-
c2t2
)= (
-
2b2
b2
)
(for t > 0).
q2
(for t > 0).
0,
q1q2
{ -
(d) h
= - 47TfO
q1q2 4b2 Joroo
1 roo
C4 Jo
80
11
1
[(b/c)2 + t2J2 dt
=-
c2t
47TfO (b2
4b2
+ c2t2)2 x,
so
Therefore
b2
= 2c2t2 = c2t2
F1 =
2c2t
(for t > 0). vet) = -2 ylb2 + c2t2 = -,x
C c2t2 + b2
(c2t2 + b2) - (c2t2 - b2)
(c2t2 - b2)
1
2ct
E
4c2t2
= - 47TfO(b2 +
b2
~
c2t2)2 c2t2 x:::}
t < O'
'
t > O.
(b2 + 1c 2t 2)2 dt. The integral is
1
= c4
C2
( )[
2b2
t
(b/c)2
00
roo
+ t2 0 + Jo
1
1
[(b/C)2+ t2)] dt]
1
7TC
7T
= 2C2b2 ( 2b ) = 4cb3'
q1q2 7T
47TfO bc'
(e) F1 "I -F2, so Newton's third law is not obeyed. On the other hand, h = -12 in this instance, which
suggests that the net momentum delivered from (1) to (2) is equal and opposite to the net momentum delivered
from (2) to (1), and hence that the total mechanical momentum is conserved. (In general, the fields might
carry off some momentum, leaving the mechanical momentum altered; but that doesn't happen in the present
case.)
Problem 10.25
1
1
S = -(E x B); B = 2'(v x E) (Eq. 10.69).
Po
c
1
So8 = ~
poc [E x (v x E)] = fO [E2v - (v. E)E].
The power crossing the plane is P = J S . dR,
~E
~x
192
CHAPTER 10. POTENTIALS
= 27rrdr x
and da
=
p
(see diagram). 80
-
fO I(E2V
E;v)27rrdr;
= EcosB, so E2- E; = E2sin2B.
Ex
it
27rfOV E 2,2sm Br dr. From Eq. 10.68, E _q1
- _4
2"
7rfO'R2[1-(vje)2sin2B]
1
q
27rfOV
2
(4uo ) ,
-
1
10
,,/2
q2 1
;j-2"
2, 47rfOa
vq2
rsin2B
00
2"
v
-
3 dr.
= atanB
Now r
R4 [1- (vje)2sin2BJ
3/2 where,
-
1
= -
V1-v2jcZ
1
1
coso
=}dr = a-Z B dB; R =-.
cos
.
a
sin3BcosB
10
3 dB.
[1 - (vje)2 sin2 BJ
{1
U
167rfoa2,4Jo [1- (vje)2u]3 du =
Problem
AND FIELDS
Let u ==sin2B, so du = 2sinBcosBdB.
vq2
167rfoa2,4
~
,4
( ) = ~a2'
2"
I
10.26
(a) F12(t)
I
_
= 47rfO
1 q1q22 Z.
(vt )
z
A
,
(b) From Eq. 10.68, with B = 180~,R = vi, and it = -z:
F 21 ()t -- - - 1 q1q2(1-v2je2)
47rfO
,
(vt)2
Z.
A
y
:1:
No, Newton's third law does not hold: F12 f:. F21,
because of the extra factor (1 - v2je2).
I
I
(c) From Eq. 8.29, p = fOJ(ExB) dT. Here E = E1 +E2, whereas B = B2, so ExB = (E1 xB2)+(E2 xBz).
But the latter, when integrated over all space, is independent of time. We want only the time-dependent part:
p(t)
= fO
E2
q2
=_
4
E2
= -4 q2
I
(E1 x B2) dT. Now E1
2
2
_
= 47rfO
1 rq~ r,
'
(1 - v j e )
R
. 2 Blje 2)3/2 R2' But R
7rfO(1 - V2sm
(1-v2je2)
7rfO [1-
(vrsinBjRe)2]
3/2
(r-vt)
R3
while, from Eq. 10.69, B2 = e~ (v X E2), and (Eq. 10.68)
. 1 r sin B . 80
= r - vi; R = r + v t - 2rvtcosB; sma = - R
2
2
2 2
A
. Finally,
noting that v x (r
-
vi)
=v
x r
= vr sin B 4>,we get
I
1
rsinB(r x cb)
vrsinB
J.. 80 P (t -- fO-q1 q2(1-v2je2)v
B 2 -- q2(1-v2je2)
'1-"
.
47rfoe2
[R2 - (vrsinB/e)2]3/2
)
47rfO
47rfoe2
r2 [R2 - (vr sin Bje)2]3/2
But r x cb = - {j = - (cos Bcos .= 103m, R = 790(5 X 10-5)2 = 2 X10-60,
which is negligible compared to the Ohmic resistance.
Problem
11.4
By the superposition principle, we can add the potentials of the two dipoles. Let's first express V (Eq. 11.14)
POW
(
)
in Cartesian coordinates: V(x, y, z, t) = - 4
2
z2
2 sin[w(t-r Ic)]. That's for an oscillatingdipole
1fEOC x + y + z
along the z axis. For one along x or y, we just change z to x or y. In the present case,
.
P = Po[cos(wt) x + cos(wt - 1f12) y], so the one along y is delayed by a phase angle 1f12:
= -cos[w(t
sin[w(t - TIc)] -+ sin[w(t - TIc) -1f/2]
V
X.
-- PoW
-
41fEOC
{ x2 + y2 + z2
=
- TIc)] (just let wt -+ wt -1f/2). Thus
sm[w(t
- PoW sinO {coscPsin[w(t
41fEOCr
-
-
y
TIc)] -
TIc)]
-
x 2 + y 2 + z 2 cos[w(t - TIc)] }
sincPcos[w(t - rlc)]}.
Similarly,
-/-L0POW{sin[w(t - TIc)] x - cos[w(t - TIc)] y}.
41fr
A =
We could get the fields by differentiating these potentials, but I prefer to work with Eqs. 11.18 and 11.19,
using superposition. Since z = cosOf - sinOn, and cosO = ziT, Eq. 11.18 can be written
2
E
= /-LO::: cos[w(t -
TIc)]
E =
B
-
(z-
~ f). In the case of the rotating dipole, therefore,
2
. /-L~ {cos[w(t- TIc)](x - ~ f) + sin[w(t- TIc)](y - ; f)} ,
1~(fXE).1
s = ~(E
x B) = ~
[E x (f x E)] = ~
[E2f - (E .f)E] = /-LoC
E2 f (noticethat E. f = 0). Now
/-Lo
/-LoC
/-Loc
E2
=
( 41fr )
/-LOPOw2
2 {a2 COS2[W(t
-
TIc)] + b2sin2[w(t
- TIc)]+ 2(a. b) sin[w(t- TIc)]cos[w(t- TIc)]} ,
where a ==x - (xlr)f and b ==Y- (ylr)f. Noting that x. r
=x
and
y. r = y, we have
197
a
2
= 1+
E2
X2
-
r2
-
X2
- 2-
r2
X2
=1-
-j
r2
(/Lo::;2) 2 {(1-
b2
y2
=1-
YX
XY
xy
xy
-j
a. b = ---r r - -+ -r2 = --.r2
r2
r r
::) COS2[W(t
-TIc)] + (1- ~:) sin2[w(t-TIc)]
- 2:; sin[w(t-rle)]cos[w(t-rle)]}
(/Lo::;2) 2 { 1 - r12(x2 COS2[w(t
- TIc)]+ 2xysin[w(t - TIc)]cos[w(t- TIc)] + y2sin2[w(t- TIc)])}
(/Lo::;2r
{1-
r12(xcos[w(t - TIc)]+ ysin[w(t - rle)])2}
= r sin B cos "b2(7Ty+OX)=7rb2>..oY.
>..r dl
= J (>"0 sin
4»(b sin 4>y + b cos 4>x)b d4>=
>"ob2
(y
127r
sin2
4>d4> + x 121r
)
sin 4>cos 4>d4>
As it rotates (counterclockwise,say) pet) = Po[cos(wt)Y- sin(wt) x], so p = -W2p, and hence (p)2 = w4p6.
_
= 6fi7TC
0 W4(7Tb2 >"0)2 =
Therefore (Eq. 11.60) P
Problem
7Tfi w4b4 >..2
0
6C
O.
11.10
= -eyy, y = ~gt2, so P = -~get2y; p = -gey. Therefore(Eq. 11.60):P = 67rc
fio (ge)2. Now, the time
it takes to fall a distance h is given by h = ~gt2 ::} t = V2h/ g, so the energy radiated in falling a distance h
p
is Urad = Pt = fio(ge)2
67TC V2h/g.
f
= Urad =
Upot
fiog2e2
(2h ~
67rc V
9 mgh
Meanwhile, the potential energy lost is Upot = mgh. So the fraction is
fioe2
=
I
{2g
67rmc V h'
= (47T X 10-7)(1.6
1
X 10-19)2
67r(9.11 X 10-31)(3
X 108)
(2) (9.8)
(0.02)
Evidently almost all the energy goes into kinetic form (as indeed I assumed in saying y
= 12.76 X 10-22.1
= ~gt2).
200
CHAPTER
Problem
11.11
PoW
(a) Vi: =
=F4
r:f: =
1
vr2
1
r:f:
r
-
11. RADIATION
~
COSO:f:
1f€OC
-
(
-
r:f:
+ (d/2)2
=F2r(d/2)
=F(d/r) cosO ~ r (1 =F 2~ COSO) .
cosO ~ rv1
d
2r
(l:1:-cosO
).
rcosO=F(d/2)
cosO:f: =
sin[w(t - r:f:/c)]
V:f: =
=
=
Vtot
)
.
sm[w(t- r:f:/c)]. Vtot= V++ V_.
r:f:
d
(
COSO=F2r
1
r
d
d
1:1:- cosO =cosO:l:-coS20=F2r
2r
) (
-
d
2r
)
=
cosO =F2~ (1- cos2 0) = cosO =F~ sin2 O.
=
sin {w [t - ~ (1 =F2~ COSO)]}
=
sin(wto) COB
= sin (wto:l: ~: COSO)
(~: COB 0):I:cos(wto) sin (~:
, where to ==t
- Tic.
COSO)!:!!sin(wto) :I: ~: COB 0 cos(wto).
=F4:~:er { (1:1: 2~ COSO)(COSO=F2~ sin2 0) [sin(wto):I: ~: COSOCOS(wto)]}
=F
4::er
{ (cos 0 =F
POW
=F
4 1f€oer [
POW
= - 41f€oer
=
=r
I
[
COB 0
~ sin2 0 :I: ~ COS2 0)
sin(wto) :I: Wd COS20 cos(wto) :I: d (COS2
0
2c
2r
Wd COS2Ocos(wto)
+
c
COS20 cos(wto)
- :OW2~
1f€OCr [
In the radiation zone (r»
Meanwhile
A:f:
Atot
[sin(wto):I: ~: cas 0 COS(wto)]}
+
~r (COS20- sin2 0) Sin(wto)
~
wr
(COS2
0 - sin2 0) sin(wto
-
sin2 0) Sin(wto
]
]
].
I = - :OW2~
1f€OCr COS2Ocos[w(t
w/c) the second term is negligible, so V
-
Z
=
=FJ140POWsin[w(t
1fr:f:
=
=FJ1-~:w { (1 :I: 2~ COB0) [sin(wto):I:
~: COB0 COS(wto)] } Z
=
=FJ1-~~~W
[sin(wto):I:
:I: ~ cos 0 Sin(wto)] Z.
=
°Pow
A+ + A= - J141fr [ Wd
c cosOcos(wto)
= I-
r:f:/c)]
~: cosOcos(wto)
+
~r cos 0 Sin(wto)] Z
c.
A
J1-opow2d 0
( ) - sm (wto )] z.
4 1fer cas
COBwto + wr
[
.
- Tic)].
201
In the radiation zone, I A = _J.LO:°UJ2d
rrer cosOCOS[UJ(t
- Tic)] z.
(b) To simplify the notation, let 0:==_J.LOP;;2d.Then
=
V
COS20
0:- r
cos[UJ(t- Tic)];
= ~~ f+ ~~~ O=O:COS20{- :2 cos[UJ(t-rle)] +
VV
- 2 COg0 sin 0
[ (
+ a
COgUJt
r2
E
B
= -VV -
I-O:UJsinOcosOsin[UJ(t-rlc)]O.
er
=
V x A
1 8
= -r
co; 0 sin[UJ(t - Tic)] (cosOf
.
8
8
COS20
A
=
'::( - sin 0 COg0) ~e sin[UJ(t- r le)];P (in the radiation zone)
er
f..Lo
= ~
=
~e
= ~E
J.LoC
x (f x E)
-
sinOcosOsin[UJ(t
{ O:UJ
re
The integral is:
1
=~
[E2 f J.LoC
rlc)]
} f.
= 1- cr
0::
4>
sin 0 COg0 sin[UJ(t
-
r le)];P.
= f..LoC
E2 f
(E. f)E]
1=-
2
1
(O:UJsinOcosO
)2.
re
2~e
O:UJ 2
1
2f..Loc
UJ2
J.L~
- ~
2
4
COBS 0
'/1"
1
0
+
4
7? 16rr2(Pod) UJ2rr 15 =
S-
2
'/1"
1
0
= 0 (since the
2
O:UJ 2
'/1"
(1- COS20) cos2OsinOdO.
4
= 3 - 5 = 15'
J.Lo
2
6
60rrc3 (Pod) UJ .
Notice that it goes like UJ6,whereas dipole radiation
Problem 11.12
Here V
] }
2J.Loc(-c ) 2rr 10
J (8) . da = -J.Loe(-c ) J sin2 OCOS2OsinOdOdcp = _
1
COS30
=
80 [ -;:-cos[UJ(t-rlc)]
= !e (f x E) and E . f = O.
2-(E x B)
I
0)
A
er { 8r(cosOcos[UJ(t-rlc)](-smO»-
=
- sinO 0).
8Ar
-8r (rAo) - - 80 4>
[
]
=
Notice that B
P
=- 0:;
,.
(m t h e ra dlatlOn zone.)
A
sin[UJ(t - Tic)] (COS2Of - COS2
Of + sinO cosO
=
0:
8
~ = -~
.
( - r Ie)]r
sm [UJt
- r I e)](J = 0:-e - r
= ~ co;0 cos[UJ(t
- Tic)](cosOf - sinOO). ~~
A
~sin[UJ(t-rle)]}f
0 .
UJcos2
A
goes like UJ4.
ring is neutral), and the current depends only on t (not on position), so the retarded
f
I(t -IJ-Ic) dl'. But in this case it does not suffice to replace IJ-by
vectorpotential (Eq. 11.52) is A(r, t) = J.Lo
4rr
IJ-
~
cdKrb'lN:iT.11?-"~
.
I'1JJll'1TION
r in the denominator-that
~
~ ~
-2- r
(+~
1
r
sin () cas
would lead to Eq. 11.54, and hence to A = 0 (since p = 0). Instead, use Eq. 11.30:
4>'
).
Meanwhile,
dl' = bd4>'$ = b( - sin 4>'x + cas 4>'y) d4>', and
J (t - Iz-I c) ~ J (t - r I c + (b I c) sin () cas 4>')
= J (to
+ (b I c) sin () cas 4>') ~ J (to) + j (to )
(carrying all terms to first order in b). As always, to = t
j, etc. are all to be evaluated at to.) Then
= ~;
A(r, t)
f ~ (1 + ~ sin
J.tob {27r
~
4Mh
J
+j
[
() cas 4>')
~
sin
c
() cas 4>'
{27r
(J + j ~ sin
+ J
~
sin
r
-
(ji'
() cas 4>') b( - sin 4>'x + cas 4>'y) d4>'
]
(- sin 4>'x + cas 4>'y) d4>'.
{27r
{27r
But 10 sin 4>'d4>'= 10 cas 4>'d4>'= 10 sin 4>'cas 4>'d4>'= 0, while 10
r ' A
.b.
b.
J.tob A
J.tob2.
'Ir Y
- sm () + J - sm
=
(
2
sm
+
-c J ) y.
()(J
4'1rr
r
4r
)[J c
()]
=
() cas
rico (From now on I'll suppress the argument: I,
() cas 4>'
{27r
~
c sin
COs24>'d4>'= 'If.
-
.
In general (Le. for points not on the x z plane)
y -t
$; moreover, in the radiation zone we are not interested
.
sin ()
J.to
b2
in terms that go like l/r2, so A(r, t) = 4c
J(t - TIc) :;:- cpo
[
I
E(r,t)
aA
J.tob2
u
[
= --=at
]
-- 4c J(t-rlc)
A
]
sin ()
A
-cpo
r
a
VxA= ~
ar (rAsin ()) f - r~a
B(r, t) =
j
1.
= J.tob2
- 4c rsm()r
--:--2sm()cos()r
[
A
1
1
.
{j
-- c sm()(J = J.tob2
-J4c2
]
( )
) (-$ x 0) = ~
u
- -J
r
A
I
usin
() A
r
(J.
2
S
=
P
=
2..
J.to
(E x B)
= J.toc
~ (
J.tob2iSin()
4c
r
16c3
r.
(b2i )2 sin2()
r2
J S. da = 1~~3(b2i) 2 J Si~: ()r2 sin()d()d4> = 1~~3 (b2i) 2 (2'1r)(~) = ~~: (b2i) 2
~
~m
=
'2
6'1rC3'
u
(Note that m
= J'lrb2,so in = J'lrb2.)
Problem 11.13
2 2
(a) P
_J.toq2a2
6'lrC
= J.t°6q
a , and the time it takes to come to rest is t = vola, so the energy radiated is Urad = Pt
'lrC
Vo
. ..
.
.
.
.
.
-. The Initial kInetIc energy was Ukin = ~mv5, so the fractIOn radiated IS f
a
1 2 1 v5 v5
v5
(b) d= '2at = '2aa2 = 2a' soa= 2d'
f
J.tOq2a
= -Urad
ukin = -3'1rmvoc .
Then
= 3'1rmvoc2d
J.tOq2v5 = J.toq2vo
(4'1rx 10-7)(1.6X10-19)2(105)
6'1rmcd = 6'1r(9.11x
10-31 )(3 x 108)(3 x 10-9)
= 12 X 10-10
.I
=
203
80 radiative losses due to collisions in an ordinary wire are negligible.
Problem 11.14
1 q2
V2
= 411"€0r = ma = m-r
F
=>
-""2
v
_
1 q2
-.
41I"€0
mr
=
= 0.5A),
At the beginning (ro
-1/2
1
(1.6 X 10-19)2
~ - [ 411"(8.85x 10-12)(9.11 x 10-31)(5 X 10-11) ]
3 X 108
v -
= 0.0075,
and when the radius is one hundredth of this v/c is only 10 times greater (0.075), so for most of the trip the
velocity is safely nonrelativistic.
2
_ (- ) =_ (_
V2
J.Lq2
l,From the Larmor formula, P = 60
1I"C r
where U is the (total) energy of the electron:
U = Ukin + U ot
P
80--
dU
dt
1
J.L0 q2
q2
~r
61I"C 41I"€0
m
(
211"€omc
(
2
q
2
=P= -
)
2
r
dr
=>
= - 3C
t
2
)
(
( q )1
211"€omc
2
dr
1
dt
3c
, andhence- = -0
ro
= (3
2
r
211"€omc
=c(
dr
2
108 ) 211"(8.85x 10-12)(9.11 x 10-31)(3 x 108) (5 10-11 ) 3
x
(1.6 x 10-19)2
x
[
]
Problem 11.15
According to Eq. 11.74, the maximum
2 sin 9 cos 9
=0
- 5 sin2 9({3sin 9)
(since a
2
q
=
.
{3
= 3i3
6{3
( VI+ ~i2 -
1
) =
=
=
{3
r2
3
ro
I
13
10-11
)
(N
. X
s.
ot very ong.
1
~ [(I-s~::S9)5] = O.Thus
R . 29 = R
2
5 (1 9)
9(1 - R
9) = 5
2
cos
'
,
.
the plus SIgn, SInce 9m ~ 900(cos9m
= 0) when
.../1+ 15{32-1
3{3
(
).
-t 0 (FIg. 11.12): 9max= cos
For v ~ c, {3 ~ 1; write
2
)
1
-, or
occurs at
(:1:.../1 + 15{32 - 1). We want
-1
I
(
2
,
1
=
=
q2
211"€omc
)
(1-{3cos9)5
(1-{3cos9)6
=> cos
,..,cos
,..,sm
,..,
2cas0 - 2{3COS2 9 = 5{3-5{3 COS2 9, or 3{3COS29 + 2 cos9 - 5{3 O. So
.
- 2 :1:.../4 + 60{32 1
cosO
= v2/r), and P = -dU/dt,
)
2
dt = - 3C
)
q2 - ~
q2 = -~
q2.
= ~mv2
- 411"€0r
~ q2 = ~2 ~
2
411"€0r
411"€0r
811"€0r
q2
--1 q2
611"€oc3 411"€0mr2
1 q2 dr
811"€0r2 dt
=
2
=1-
€ (where € « 1), an'd expand
3(1 ~ €) [.../1 + 15(1
- €)2 - 1] ~
~(1+€)[v'16-30€-1]
15
1
-(1
+
€)
3
-€
3
4
(
Evidently 9max ~ 0, so cos9max ~ 1 -
to first order in €:
~ (1 + €) [.../1 + 15(1
=~(1+€)[4.../1-(15€/8)-1]
5
~
4
) = (1 + €)(1 - -f)
!9~ax= 1 -
1+€
5
- -€
4
- 2€) -
1]
=~(1+€) [4(1-
~~€) -1]
1
= 1 - -€
4'
t€ => 9~ax = !€, or 9max~ J€T2 = 1.../(1 -
{3)/2.1
204
CHAPTER
(1
-
(dP l dOI
=
Let f
-
lem
(dPldO
)
ur
)rest
(3cas Bmax) ~ 1 - (1
1
"(=
em
=
~
-
--
.
B
max
. Bmax ~ €/2, and
5
. Now sm2
1
cosB
(3
[(
max ) ] ur
-
€)(1
t€) ~ 1 - (1 - € - i€)
c::"
(1-
€)2 -
1
.)1-
f =
Problem
2
Sin
1
.)1-
11. RADIATION
4
_1
(1-
2€) -
= 2€.
=
So f
4
€/2
=
(5€/4)5
1
(i ) 25
5
But
2€4'
y2€ => € = 2"(2' Therefore
(~r ~(2"(2)4
(~r
=1 ~
= 2.62"(8.
"(8
11.16
.
EquatiOn
11.72 says dP
do
H
- _6q2
I~x
-
(u x a)12
. Let (3 = vie.
C~.u )5
111"€o
2
v = ci - vz => ~ . u = e - v(~. z) = e - v casB = e (1- ~ cas B) = e(1 - (3cas B);
a. u = ae(x .~) - av(x. z) = aesin Bcas 4>; u2 = u. U = e2 - 2ev(~' z) + V2 = e2 +V2 - 2ev cas B.
U
= ci -
~ x (u x a)
=
(~. a)u - (~. u)aj
I~x (u x aW
=
=
(~. a)2u2 - 2(u. a)(~' a)(~' u) + (~. u)2a2
=
=
a2 [e2(1 - (3cas B)2 + (sin2 BCOS24>)(e2 + V2 ,--2cv cas B - 2e2 + 2cv cas B]
dP
dO
(e2 + V2 - 2cv cas B) (a sin B cas 4»2
a2e2
4>
)(a sin B cas
4>
)(e - v cas B) + a2e2(1 - f3COS8)2
[(1- (3coSB)2- (1- (32)(sinBcos4»2].
/Loq2a2 [(1-
=
- 2(ae sin Bcas
(3coSB)2 - (1-
1611"2e
(32) sin2 BCOS24>]
(1 - (3cas B)5
The total power radiated (in all directions) is:
P
=
j
dP dO=
dO
{21r
-
j
jj
dP.
BdBdA..=/LOq2a2
dO sm
'I'
1611"2e
.
sm BdBd4>.
{21r
But Jo
d4>= 211" and Jo
/Loq2a2
r [2(1
- (3cas B)2 -
1611"2e
11"
Jo
COS24>d4>= 11".
(1 - (32) sin2 B]
(1- (3cas B)5
Let w ==(1 - (3cas B). Then (1 - w)1 (3 = cas Bj
2w2
[(I-(3COSB)2-(I-(32)Sin2Bcos24>]
(1 - (3cas B)5
.
smBdB.
sin2 B = [(32- (1
-
W)2] 1(32,and the numerator becomes
- (1 ~f2) ((32- 1 + 2w - W2) = ;2 [2w2(32+ (1- (32)2- 2(1 - (32)W+ w2(1 - (32)]
=
;2 [(1- (32)2- 2(1 - (32)W+ (1 + (32)w2];
205
dw
= {3sin0 dO=} sin 0 dO =
~ dw.
2a2 1
= ~167rc
P
fJ
(1+{3) 1
j
a35
When 0 = 0, w = (1 - (3); when 0 = 7r, W = (1 + (3),
(1-{3)
W
[(1 - {32)2 - 2(1
-
{32)W + (1 + {32)w2]dw,
The integral is
Int = (1- {32)2! :5dW - 2(1- (32)! :4 dw + (1+ (32)! :3 dw
=
~
1+{3
W2 1-{3
1
~
1+{3
w3 1-{3
1
~
1+{3
W4 1-{3
1
-
[(1- {32)2(- 4~4) - 2"(1-(32)(- 3~3) + (1+ (32)(- 2~2)] I:::.
1
1
- (1 - 2{3+ (32)- (1 + 2{3+ (32)- -~
(1 + {3)2 (1 - {3)2(1 + {3)2(1- {3)2
- (1 - {32)2'
1
1
- (1 - 3{3+ 3{32- (33)- (1 + 3{3+ 3{32+ (33)- - 2{3(3+ (32)
(1 + {3)3 (1 - {3)3(1 + {3)3(1- {3)3
(1 - {32)3.
1
1
- (1 - 4{3+ 6{32- 4{33+ (34)- (1 + 4{3+ 6{32+ 4{33+ (34) - 8{3(1+ (32)
(1 + {3)4- (1 - {3)4(1 + {3)4(1- {3)4
- - (1 - {32)4.
2 2
Int
P
1
(-4 )
-8{3(1 + (32)
2
(-3 )
=
(1-{3)
=
2{3
2
2
2
2
(1 - {32)2 [ (1 + (3 ) - 3(3 + (3 ) + (1 + (3) ]
~~
=
p,oq2a2
167rc {333 (1
(1-{32)4
{33
- {32)2
-2(1-{3)
=
a2 -
2
(-c
Problem
xa
)
= a2
(
1-
v2
2"
C
)
= (1 -
-2{3(3 + (32)
(1-{32)3
8
=
/
2
+(1+{3)
(-2 ) (1-{32)2
.
1
VI -
{32
Here v x a
= va(z x x) = vay, so
1
{32)a2 = 2"a2,
so the Lh~nardformula says P
/
p,q2"f6a2
= ~ 67rC
11.17
2"'
/
,(
2
For circular motion, r(t)
= V = -&2
-4{3
{32)2'
(a) To counteract the radiation reaction (Eq, 11.80), you must exert a force Fe = -
aCt)
1
{33
= 3 (1 -
where
Is this consistent with the LiE~nardformula (Eq. 11.73)?
V
1
= R [cos(wt) x + sin(wt)y],
[cos(wt)x + sin(wt)y] = -w2r;
0Q a,
f..l
67rC
vet) = r = & [- sin(wt)x + cos(wt)y];
0q w2v.
a = -w2r = -w2V. So Fe = f..l
67rC2 I
2
Pe= Fe' V = ~w2V2
,I
This is the power you must supply.
Meanwhile, the power radiated is (Eq. 11.70) Prad
2
Prad= f..l0q w2V2, and the two expressions agree.
6
2 2
= f..loq
a , and
67rC
a2
= w4r2 = W4R2
= w2V2,so
7rC
(b) For simple harmonic motion, r(t) = A cos(wt)z; v = r = -Aw sin(wt)z; a = v = _Aw2 cos(wt)Z =
-w2r; a = -w2r
= -W2V. SO IFe = f..lOq2w2v;
Pe = f..lOq2
67rc
67rcW2V2.1 But this time a2
=
w4r2
= w4A2COS2(wt),
!
206
CHAPTER 11. RADIATION
whereas
c.v2V2
= c.v4A2sin2(c.vt),
so
2
Prad = 1-t0q c.v4A2 COS2(c.vt)
2
=I-
6 ~c
Pe
=
I-toq c.v4A2sin2(c.vt)j
6~c
the power you deliver is not equal to the power radiated. However, since the time averages of sin2(wt) and
COS2(c.vt)are equal (to wit: 1/2), over a full cycle the energy radiated is the same as the energy input. (In the
mean time energy is evidently being stored temporarily in the nearby fields.)
(c) In free fall, v(t) = ~gt2 yj V = gt y; a = 9 y; Ii = O. So IF e = 0; Ithe radiation reaction is zero, and
2
hence Pe = 0.1But there is radiation: Prad = I-toqg2.
Evidently energy is being continuously extracted from
6~c
I
the nearby fields. This paradox persists even in the exact solution (where we do not assume v « c, as in the
Larmor formula and the Abraham-Lorentz
formula)-see
Prob. 11.31.
Problem
11.18
(a) 'Y = c.v2T, and T = 6 X 10-24 s (for electrons). Is 'Y« c.v(i.e. is T « 1/c.v)? If c.vis in the optical region,
c.v= 2~v = 2~(5 X 1014) = 3 X 1015j l/w = (1/3) X 10-15 = 3 X 10-16, which is much greater than T, so the
damping is indeed "small". ,f
(b) Problem
9.24 gave
~w
2:! 'Y
=
=
W5T
= 11
[2~(7 x 1O15W(6 x 10-24)
X 1010 rad/s.!
Since
we're
in the
region of c.vo~ 4 X 1016rad/s, the width of the anomalous dispersion zone is very narrow.
Problem
11.19
F
(a) a
= To' + -m
=}
dv
-
da
= T- dt +
dt
F
-
m
/
=}
dv
-dt dt
=T
/
da
1
-dt dt + -m
/
Fdt.
210
[v(to + E) - v(to - E)] = T [a(to + E) - a(to - E)]+ -Fave,
where Fave is the average force during the interm
val. But v is continuous, so as long as F is not a delta function, we are left (in the limit 10-+ 0) with
[a(to + E) - a(to - E)] = O. Thus a, too, is continuous. qed
da
1
.
.
da
da
1
/
(b) (1) a = Ta = T dt =} ~ = -:;.dt =}
~ = -:;. dt =} lna = -:;.t + constant =} a(t) = Ael T, where A
I
is a constant.
..
(11)a
Bel/T
=}
I
a(t)
.
F
= Ta + -
m
da
=}
= ~ + Bel/T,
T-
I
dt
F
=a-
-
m
/
da
=}
I
1
= -T dt
a - F/m
=}
In(a - F/m)
t
= -T
I
+ constant
F
=} a - m
=
where B is some other constant.
(iii) Same as (i): a(t) = Cel/T, where C is a third constant.
I
I
(c) At t
= 0, A = F/m + B; at t = T, F/m + BeT/T = CeT/T=}C = (F/m)e-T/T + B.
a(t)
=
<
::;OJ
[(F/m) +B]el/T,
t
[(F/m) + Bel/T] ,
a ::; t ::; T;
[(F/m)e-T/T
t? T.
+ B] el/T,
So
To eliminate the runaway in region (iii), we'd need B = -(F/m)e-T/T;
(i), we'd need B = -(F/m).
Obviously, we cannot do both at once.
to avoid pre acceleration in region
I
207
(d) If we choose to eliminate the runaway, then
(F/m) [I - e-T/T] et/T, t:5 OJ
a(t) =
(F/m) [I-e(t-T)/T] ,
0 :5t :5T;
0,
t ~ T.
!
(i)v = (F/m) [I - e-T/T] et/Tdt = (Fr /m) [I - e-T/T] et/T + D, where D is a constant determined
by the condition v( -00)
(ii) v = (F/m)
[I -
(Fr/m)
= 0 =>D = O.
[t- re(t-T)/T
l + E,
where E is a constant determined by the continuity of v at t
e-T/T] = (F/m) [-re-T/T] + E =>E = (Fr/m).
(Hi) v is a constant determined by the continuity of vat t
(Fr /m) [1 - e-T/T] et/r,
v(t) = { (F/m) [t+r-re(t-T)/T]
(F/m)T,
= T: v = (F/m)[T
- r] = (F/m)T.
+r
t :50;
, O:5t:5Tj
t ~ T.
(e)
ullchar!(.d
particle:
vet)
a(f)
chargcd particle
(N'?;"" w"";'",'
F/m
£(1m
e-'/T)
chargoo particle
(with radiation rcaction)
T
()
Problem
T
11.20
(a ) F rom E q. II .80 ,
(b) Frad = ~a
Ihc
{L
10
end
J1.o(q/2)2 .
F fad
a,
=
61I'C
{101112AdY2
F.
so fad =
1
1
J1.0q2.
2
F int 2 r:>end J1.oq2.
fad+ rrad = 611'C
a [2' +
4 ] = 611'ca. ./
2AdYl' (Running the Y2
}
integral up to Yl insures that Yl ~ Y2, so we don't count the
()
Y
dYI
same pair twice. Alternatively, run bothintegrals from 0 to Lintentionally double-counting-and divide the result by 2.)
L
)dY2
v
= 0:
208
CHAPTER
.
F,fad
Problem
= 12
f..Loa (4A2)
-rrc
L
10
Yl
-rrc
2
= f..Lo
AL 2'a = f..Loqa.
. ./
6-rrc( )
6-rrc
11.21
(a) This is an oscillating
electric dipole, with amplitude
2
Poynting vector is given by Eq. 11.21: (8)
I, = (8) . z =
=
dI,
dR
.
L2
= 12
f..Loa (4A2) 2
dYl
11. RADIATION
32-rr2C
)
(f..L;::~)
.
Po = qd and frequency W = vk/m.
r, so the
SI;2
.
sin2 (Jcos(J
R
power per unit area of floor is
h
= -,r
But sm(J
r2'
The (averaged)
2(J
cos(J = -,
r and r2
= R2
+ h2.
R2h
f..LOq2d2W4
( 32-rr2c ) (R2 + h2)5/2.
2R
R2
d
= 0 =>
(
f..LOP5W4
=
4
dR [ (R2 + h2)5/2 ]
(R2 + h2) - ~R2
= 0 =>h2 =
= 0 =>
5
R2
(R2 + h2)5/2 - 2 (R2 + h2)1/2 2R
~R2 => R =
I
= 0 =>
V2/3h, for maximum intensity.
I
(b)
=
p
f..Lo(qd)2w4
(
J
J I,(R)
)
rx>
R3
da = I,(R) 2-rrRdR= 2-rr 32-rr2c h 10 (R2 + h2)5/2dR.
roo
R3
1 roo
x
1 r(2)r(1/2)
2
10 (R2 + h2)5/2dR = 210 (x + h2)5/2dx = 2h r(5/2) = 3h'
Let x ==R2 :
which should be (and is) half the total radiated power (Eq. 11.22)-the rest hits the ceiling, of course.
(c) The amplitude is xo(t), so U = ~kx5 is the energy, at time t, and dU/dt = -2P is the power radiated:
1 d x2 = f..LoW42x2 => d X2 = - f..Low4q2X 2
2
2
2 /
-kq
( )
( ) = -/'i,X => X = dent or xo (t ) = de-II R = h.1
209
.
.
..
.
h
A h I
l(h)
t t IS ocatlOn t e IntensIty IS
3P
f3Pl
h2
= 811"(2h2)2 = ~
(c) Imax= 3~::(~0~~:) = 0.026W/m2 = 12.6/-LW
/cm2.1
Problem
!
Yes, KRUD is in compliance.!
11.23
= M cos1/Jz +
M sin 1/J[cos(""t)x + sin(""t) y]. As in Prob. 11.4, the power radiated will be twice
that of an oscillating magnetic d~pole with dipole moment of amplitude mo = M sin 1/J. Therefore (quoting
(a) met)
/-L M2",,4 sin21/J
Eq. 11.40): P
I
=
0
611"c3
.
.1 (Alternatively, you can get this from the answer to Prob. 11.12.)
/-LoM
(b) From Eq. 5.86, wIth r ~ R, m ~ M, and ()= 11"/2:B = 411"
R3' so
M= -B
(411"x 10-7)(1.3
(c) P
=
-
10-7 x 10-5)
1- 411"(6.4x411"X
106)3(5
/-Lo
411"R3
1
x 1O23)2sin2(11°)
611"(3X 108)3
(
211"
24 x 60 x 60
= 11.3 X 1023 Am2.1
) =14 x 1O-5WI (not much).
4
- /-Lo(411"R3B611"c3
/ /-LO)2",,4
sin21/J= ~
(",,2R3B sin 1/J)2. Using the average value (1/2)
3/-Loc3
(d) P -
~
811".
P - 3(411"x 10-7)(3 X 108)3
Problem
!
!
J-LOZ
411"
=
2(1O4)3(108)
]
2~=12X1O36wl(alot).
2
11.24
(a) A(x, t) = ~;
=
[( 10-3 )
3;
!
K~r) da
K(tr) 211"rdr
vr2 + X2
vr2 + x2/e) r dr.
vr2 + X2
The maximum r is given by t - v'r2 + x2/e = 0;
Tmax= Je2t2 - x2 (since K(t) = 0 for t < 0).
J-LOZ
2
for sin21/J,
K(t-
z
"
y
(i)
A( x, t )
=
l
rm
/-LoKoz
2
0
r
v'r2 + X2
E(x,t) = -7ft
aA = -~z,
/-LoKoe
I
A
B(x,t) = V x A = - aa~zy =
dr
= /-LoKoz
J r 2 + x 2 rm =
2
1
0
/-LoKoz
2
(J r
for et > x, and 0, for et < x.
I
I
J-LO;Oy, I for et > x, and 0, for et < x.
2
m
-
X
2
-
- x)
) = /-LoKo(et
2
z.
A
X
210
CHAPTER
11. RADIATION
(ii)
/-Loaz
= -
A(x t )
,
('
~
yr2 + X2
E(x, t)
=
-~ ut =
B(x,t)
=
vxA=_aa~Zy=I-~(x-ct)y,lforct>X,
yr2 + X2
c
aA
yr2 + X2
and 0, forct x, and 0, for ct < x.
2
1 1
(Vr2 +
= -z2
/-Loa(x- ct)
r
rm
/-Loa
z t(ct - x) - ..!..(c2t2- X2) = /-Loaz (X2 2 [
2c
4c
]
(b) Let u ==-c
E(x, t)
10
= -/-Loaz t 1
2 [ 0
r dr
=
1
t-
2
(t-yr2+x2/c)
rm
) du=-z 2
[if K(-oo)
But at K (t [K (t---uco2 )]
X
.
~
a
X
~
00
1
u ) = - auK (t -
~
X
)
~-
u .
/-Loc
A
=--[K(t-x/c)-K(-oo)]z
=0].
Note that (i) and (ii) are consistent with this result. Meanwhile
B(x, t) = --
S
A
y
ax
/-Lo
=
=
aAz
~
--y2
I
~K(t
= --/-Loc
00
a
~
y
00
1
coax
a
-K
X
I
X
a
)
(t -
- - u au. But -K
c
ax
/-Lo
X
00
) du=--y 2 [K (t---uco2
1oau-K (t---uc
- x/c)y,
(t -
[if K(-oo)
A
)]
X
- - u
c
=-
/-Lo
1
a
X
t) = --K
c au (
- - u .
c
)
A
[K(t-x/c)-K(-oo)]y
1
= 0].
= :0 (E x B) = :0 (/-Lt) (~O) K(t - x/c) [-z x y] = /-L~C
[K(t -
X/c)]2X.
This is the power per unit area that reaches x at time tj it left the surface at time (t - x/c). Moreover,an
equal amount of energy is radiated downward, so the total power leaving the surface at time t is
Problem
11.25
p(t)=2qz(t)iP=2qziF=mz=--
..
.
.
UsmgEq.11.60,thepowerradlatedlsP=
1
.
q2..
2iZ=-47rEo (2)
Z
..
1
- q2
2 3
/-Loc q
2
7rmz
2iP=--
3 3 6
/-Loc q
_67rC= -67rC(--87rmz ) = 6(4)37r m 24=Z
/-LoP
/-Lo
2
Problem
/-Loc2q2..
47rEO4mz 2=-16
11.26
With a = 90°, Eq. 7.68 gives E' = cB, B' = -~E,
c
y
[K(t)]2 .
/-Loc2q3
87rmz 2'
W)
2
3
/-Locq
1
- 47r
6m
~. Z
q~ = -cqe. Use this to "translate" Eqs. 10.65,10.66,
211
and 11.70:
E'
= e(~~XE) =~x(-eB')=-e(~xB').
B'
= --E=--e
1
1 qe
12
[(e -V2)U+Ji.x(uxa)]
e 411"£0(Ji..u)3
1 ( q' /e)
P
1-
=
11"£0 (Ji. . U )3 [(e2-v2)u+Ji.x(uxa)]
-- e -4m
-
Jloa 2 - Jloa
-
2
2
-q
611"c e
-
2
1 '-611"c --qC m
( )
=
Jl0mq'
1-
411" (Ji. . U ) 3
[(e2-v2)u+Ji.x(uxa)].
2
, 2
( )
-q
611"C3 m .
Jloa
-
Or, dropping the primes,
B(r, t)
=
Jl0qm
411" (Ji.. U )3
E(r, t)
=
-c(~ x B).
P
= Jloq~a2
611"C3
.
1-
Problem
[(c2
- V2)U
+ Ji. x (U x a)] .
11.27
(a) Wext =
/
F dx
= F IT vet)dt. FromProb. 11.19,vet) = ~ [t+ r
F2
Wext
=
;;:
fT
fT
So
T
t2
F2
fT
- re(t-T)/T].
[ 10 t dt + r 10 dt - re-T/T 10 et/T dt] =;;:
[ 2"
+ rt - re-T/T ret/T ] 0
I
1
F2 1
= ;;: [ 2T2+ rT - r2e-T/T (eT/T- 1)] = IF2
m (2T2 + rT - r2 + r2e-T/T) .
(b) From Prob. 11.19, the final velocity is vI
(c) Wrad=
J Pdt.
= (F/m)T, so Wkin= -mvJ
22m = -m2T2
1
1
F2
= -.2m
I
F2T21
JLoq2a2
According to the Larmor formula, P = -, 611"c and (again from Prob. 11.19)
"
(F/m) [1- e-T/T] et/T,
(t::; 0);
aCt)=
{ (F/m) [1- e(t-T)/T],
(0::; t ::;T).
"v
CHAPTER
212
2
2 F2
Wrad
JLaqJ2
67fc m { (1 - e-Tlr)
F2
e-Tlr)
2
T
0
("2e2tlr)
1
{T
-00
dt }
{T
{T
+ Jo dt - 2e-Tlr Jo etlr dt + e-2Tlr Jo e2tlr dt }
[~ (1 - e-Tlr) 2 + T - 2e-Tlr (Tetlr) I~ + e-2TIT (~e2tlr) I~}
T~2
=
(1 -
i
T
-00 e2tlr dt + a [1 - e(t-T)lr]
2
T-:;;;:{
-
j
o
11. RADIATION
T~2 [~(1TF2
2e-Tlr + e-2Tlr) +T - 2Te-Tlr (eTlr -1) + ~e-2Tlr (e2Tlr -1)]
T
---;:;:;: [ "2
Tlr
T
2T l r
+ "2e-
- Te-
+
T
Tl r
2
- T + 2Te-
T
2T l r
T
+ "2- "2e-
]
=
I
---;:;:;:
TF2
(T -
T lr
T
+ Te-
).
I
Energy conservation requires that the work done by the external force equal the final kinetic energy plus
the energy radiated:
F2T2
Problem
TF2
+ Wrad = ~
Wkin
+
---;:;:;:
1
F2
(T -
T
+ Te-Tlr) =
-:;;;:
(
"2T2+TT - T2+ T2e-Tlr
) = Wext. ./
11.28
k
f
(a) a = To' + -8(t) =>
m
j
-f
a(t) dt
= v(€) -
v( -E)
j
=T
f
-f
da
k
-dt dt +
If the velocity is continuous, so v(€) = v( -E), then a(€) - a( -E)
I
j
f
8(t) dt
m -f
=
I
={
k
a(-E)] + -.
m
/;;;.1
k
6.a = B - A = - mT
When t < 0, a = To'=>a(t) = Aetlr; when t > 0, a = To'=>a(t) = Betlr;
k
. .
=>B = A - mT' so the general solutlOnIS a(t)
= T[a(€) -
Aetlr,
[A - (k/mT)] etlr,
(t < 0);
(t > 0).
To eliminate the runaway we'd need A = k/mT; to eliminate preacceleration we'd need A = O. Obviously,
(k/mT)etlr
(t < 0)'
you can't do both. If you choose to eliminate the runaway, then a(t) = { 0,
, (t > 0):
I
v(t)
=
j
t
-00
k
a(t) dt
= mT
-
j
t
-00
k
etlr dt
= mT
-
t
(Tetlr
)
I
-00
for t > 0, v(t) = v(O)+ (t a(t) dt = v(O)= ~. So v(t)
Jo
m
For an uncharged particle we would have a(t)
The graphs:
k
= -etlr
(for t < 0);
m
={
(k/m)etlr,
(k/m),
= ~8(t),
v(t) =
m
j
(t < 0);
(t > 0).
t a(t) dt
-00
={
0,
(t < 0);
(k/m),
(t
> 0).
213
a(t)
v(t)
neutral
neutral
(b)
/
=
/
Wext
=
Wkin
= !mv~=!m
22m
Wext =
/
Fdx
/
=k 8(t)v(t)dt = kv(O) = ~.
~ =~.
2m
Fvdt
()
2
2
Praddt = J.toq
61rC
/
[a(tW dt = rm
k
2
( )1
-mr
0
k2
e2t/Tdt = -
mr
-00
r
)
(-e2t/T
2
0
1
k2
= --
r
mr 2
-00
= -.k2
2m
Clearly, Wext = Wkin + Wrad. ,f
Problem 11.29
=
+ 8(x - L)], subject to the boundary conditions
Our task is to solve the equation a ra + Uo[-8(x)
m
(1) x continuous at x 0 and x Lj
(2) v continuous at x = 0 and x = Lj
(3) ~a
-3:Uo/mrv (plus at x = 0, minusat x = L).
=
=
=
The third of these followsfrom integrating the equation of motion:
dv
da
Uo
dtdt = r dtdt+ m [-I5(x)+8(x-L)]dt,
dt
Uo
[-8(x) + 8(x - L)] dx dx = 0,
~v = r~a + m
Uo
Uo 1
~a = = -mr
-v [-8(x)+ 8(x- L)]dx = -3:-.
mrv
/
/
/
/
/
In each of the three regions the force is zero (it acts only at x
a(t)
= Aet/Tj
v(t)
= Aret/T
+ Bj
x(t)
= 0 and
x = L), and the general solution is
= Ar2et/T + Bt + C.
(I'll put subscripts on the constants A, B, and C, to distinguish the three regions.)
Region Hi (x > L): To avoid the runaway we pick A3 = OJthen a(t) = 0, v(t) = B3, x(t) = B3t+ C3.Let
the final velocity be vI (= B3), set the clockso that t = 0 whenthe particleis at x = 0, and let T be the time
it takes to traverse the barrier, so x(T) = L = vlT + C3, and hence C3 = L - vITo Then
!
a(t)
= OJ
v(t)
= vI> x(t) = L + vl(t - T), I (t < T).
>-
CHAPTER
214
11. RADIATION
< x < L): a = A2et/T, v = A2ret/T + B2, x = A2r2et/T + B2t + C2.
Region ii (0
(3)
::}
0-A2e
(2)
::}
vI
(1)
::}
L
Uo
--::}
T/ T =
Uo - T/
A2 = -e
T.
mrvl
=MA2re ~T
+ B2
mrvl
~ + B2
= -mvl
Uor
= A2r 2eT/T + B2T + C2 = -mvl
C2
Uo
= L - VIT + -(T
mvl
= ~e(t-T)/T
vet)
=
=
x(t)
= vI
~
-.
mvl
-
Uo
+ VIT - -UoT + C2 = VIT + -(r
mvl
mvl
- T) + C2 ::}
- r).
aCt)
.
B2
::}
j
mrvl
vI + Uo [e(t-T)/T - 1] ;
(0 < t < T).
mvl
L + vIet - T) + Uo [re(t-T)/T - t + T - r ] ;
mvl
[Note: if the barrier is sufficiently wide (or high) the particle may turn around before reaching L, but we're
interested here in the regime where it does tunnel through.]
In particular, for t = 0 (when x = 0):
-r
0 = L- vlT + Uo [re-T/T + T
mvl
Region i (x
]
L
::}
= vlT -
Uo [re-T/T
mvl
+ T - r . Qed
]
< 0): a = A1et/T, v = A1ret/T+ B1, x = A1r2et/T+ B1t + C1.Let Vibe the incidentvelocity
(at t -+ -00); then B1
= Vi. Condition
(3) says
Uo -T / T A - Uo
-e
- 1--,
mrvl
mrvo
where Vo is the speed of the particle as it passes x
e-T/T -1
Vo = vI + Uo
Therefore
(
mvl
Uo
Vi
) . But
= O.
From the solution in region (ii) it follows that
we can also express it in terms of the solution in region (i): Vo = A1r+vi.
Uo
Uo
= VI+-mvl (e-T / - 1)- A 1r=vl+- mvl (e-T / - 1) +---e
mvo
Uo
-
T
Uo
Uo
= VI-
T
Uo
VI
1- Vo = vI - mvl { 1- vI + (Uo/mvl) [e-T/T -1] }
(
VI
)
=
VI
-
V - 1Qed
I mvl {
1+ (Uo/mvJ)[e-T/T- 1] } .
mvl + mvo
Uo
mvl
Uo -T / T
mvl
I
If !mvJ = !Uo, then
L = vlT - vI [re-T/T + T - r] = vI [T - re-T/T - T+ r] = rvl (1 - e-T/T) ;
215
Vi =Vf -VI
[1-
1 +e-;/T
-1]
= vf (1-1
+eT/T) = vIeT/To
Putting these together,
~ =1-
e-T/T ::} e-T/T = 1 -
TV!
Vf
~
Tvf
::} eT/T =
4
KE-
.
1
::} Vi =
vf
(L/TVf)
1 - (L/vfT)
-
1
In particular, for L = VfT/4, Vi = 1- 1/4 = '3Vf' so KE; = ~mv} =
16
16 1
8
KEi = -KEf
= --Uo
= -Uo.
9
92
9
Problem 11.30
16
2
vo
lmv2
Qed
(v; ) = "'9 ::}
II-
= oi -
(a) From Eq. 10.65, El = (q/2) 3 [(c2 - V2)U + (Ji. a)u - (Ji. u)a]. Here u
41r€o (Ji. U )
V, Ji = l x + dy,
V = V x, a = a x, so Ji. V = lv, Ji. a = la, Ji. u = O'l-- Ji. V = O'l-- tv. We want only the x component.
(el- VII-)/II-,
we have:
that Ux = (c/II-)l V
Noting
- =
q
El.
1
II-
2
2
=
(
)
II- - VII-)C - V + la - a(O'l-- tv) ]
-81r€o(O'l-- lV)3 [ -(el
=
81r€om
~
( -II V)3 [(el- VII-)(C2
- V2)+ el2a - vII-La
- am2 + aim] . But 1I-2
= l2 + d2.
= _8
1r€om
q ( -II V)3 [(el- VII-)(C2
- V2)- acd2].
=
Fself
81r€o (O'l-- ll)3
V
L
[(el- VII-)(c2- V2) - acd2] x.
(This generalizes Eq. 11.90.)
Now xCi)- x(tr) = l = vT + !aT2 + !aT3 :+-"', where T = t - iT>and v, a, and a are all evaluated at the
retarded time tr.
(CT)2 = 1I-2= l2 + d2 = d2 + (vT + ~aT2 + ~aT3?
2
6
c2T2(1 'Yd
T
= -C
V2
/ C2) = c2T2/'l
= d2 + vaT3
c2 'Y2~
(1 + Ad + Bd2 +...)
::}
2~
"(C
1
= 2va
Comparing like powers of d: A
2B
=
=
= ~
{ 1+
"(3
"2 c3 d
+
Va
2C4
[ "'3
Va
3va"(3
2B + A2 = ~A
)
'Y4
(
(1 + 2Ad + 2Bd2 + A2d2) = d2+va~(1+3Ad)+
C
(
va
Va a2
-3 + 4
"(3d3
3va"(3~va 'Y3- ~v2a2 'Y6+ va "(4 + a2"(4 = va
C3
2 c3 4
c6
3 c4
4C4
3
"(4
v2
V2
V2
Va
a2'Y2
- - +1 - - - - + 6::} B =
C4 [ 3
4
C2
C2
C2 ]
"(d
T
+ (~va + ~a2) T4. Solvefor T as a power series in d:
"(3
C3;
= d2 + v2T2 + vaT3 + ~VaT4
+ ~a2T4;
3
4
"(2a2
+4
+
a2
("'3 + 4" ) c4'
+~
(~ ) V2
'Y4 Va
'Y2a2
2C4 [ 3
4
- +-
"(4
)
"(4
- V2
"(4 + 'Y6a2
c4
4c4 "(2 C2.
-
_
(
V2
( 1 + 4 c2 ) ] d2 } + ( ) d4 + ...
2
1 + 4-
c2
v2a2"(6
c6
)] .
(generalizing Eq. 11.93).
4d4.
C
216
CHAPTER 11. RADIATION
=
l
=
1 2 + -aT
1.
vT + -aT
2
6
va ')'3
WYd
-
=
c
{
3
+...
V2
1 + - -d + -')'4 -Va + -')'2a2
2 C3
v')'
2C4 [ 3
V2.
a ')'4
(
4
V2
C2)]
2
}
Va
v')' ')'4
2
1 ')'2d2
')'3
d2 + -a1 + va-d
1 + 4-
C2 [
v2
')'2 a2
1.')'33
+ -a-d
c3 ]
6
1 ')'2
(- c ) d + -2 -C2 (1 - -C2 + -C2) d + { -2c -C4 -3 + - 4 ( 1 + 4-c2 ) ] + -a-va2 C2
a
V2
v2
v')'4a2 1 V2
(-'-C ) d + (-2C2 ) d2 + -2c3 -3 (1 + ')'2-c2 ) + - c2 (-4 + -c2 + 1 - -c2 )] d3
a
5 v')'2a2
d3+()d4+...
(-c ) d+ (-2C2 ) d2+- 2c3 -+-3
4 c2 ]
.
[
.
=
v"'
a')'4
')'3
V')'
a')'4
')'5
C3
')'3 + -a1 . ')'3 d3
C3 6 C3}
[
-
[
It =
clt-lv
cl-
=
va ')'3
va')'4
2
c')'d+-d2c2
V2
(
c')'d 1-- C2
=
-d + -d3
C
')'5a2
')'
8C3
)
=
(
a
2c2 [ 3')'2
')'5
')'3
a
2c
2C2
3
')'6a2
1+-d2
av')'4
+...
2
2C2
5v2')'2a2
3
4
C2
]
a
')'5v
5v')'2a2
-- 2c3 -+-4
[3
3
d +...
C2 ]
d3+...
3
--q2 ')'3
=
-
87fEoc3d
)]
(
1-
q2
-+ 4
(
_')'3-
V')'2a2
c2
a
V2a
v')'2a2
C2 3
c2
5
1
V2
(4
4
c2
3
)] d
1
( 4-+4
+()d
V2
3
c2 )]
d +()d
1
V2
(
- + d3
4
c2 )]
4
+...
+...
V')'2a2
) d3+()d4+...
= (- ) 1-3-d2
(
) +...
C2
3
')'6a2
8c4
)
) {[( ) + L (~
a v')'2a2
) [--+- (-+- ) d] x
a')'2
3 ')'6a2d2
8C4
2
8 C4
4C2d
c2
cd
1---d3 ')'6a2
q2 ')'3 1
- - -ac
87fEoc3d 2 [
47fEO [
)
Va
V')'5
- - + ')'2a2
2C4 [ 3
v2a')'4
d3 - v')'d - --d2
2 c3
5 v')'2a2
')'
8c4
1
L
87fEO (cd )
-
4
v2')'
va
)
c2
a
')'5
a')'2
=
=
v2
2C2 [ 3
2c
[
3
-+4
C2
5v')'2a2
- + -3
4
2
2
(
(
a
')'5
+ 2c2
c2
a')'2
')'
v2
1
[
(1-- ) d +(- ) d +- -+(- ) d2+- (-+--3
-
2
)] d } + ( ) d
d --d--d
4
C2)]
C
( -+-
Va
')'5
V2
cd
=
1
2 2
(
1 V2
-4 + -C2
+ ( ) d4+ .. .
Ya')'4
-2c
=
Fself
2 2
+- 2c3 -+')'2a2
3
v')'d + -d2
2c
=
Va
')'5
+- 2C3 -+')'a
3
a')'4
=
=
(CIt-lv)-3
Va
[
=
Vlt
')'4
cT = ')'d 1 + --d
+ ~2C4 -3 + ')' a
2 C3
{
[
a
d2
2c
ac
')'
2
2
2C2
3
+ v')'2a2
C2
d3
C2
]
')'2
-
acd2
}
x
A
3
C2
v')'2a2
) d + ( ) d2 + ... ] x
')'4
a v')'2a2
+- ++ ()d+ ... x (generalizingEq. 11.95).
4C3 ( 3
C2 )
]
+ ')' - +
(3
~
C2
Switching to t: v(tr) = v(t) + v(t)(tr - t) + ... = v(t) - a(t)T = v(t) - a')'dfc. (When multiplied by d, it
doesn't matter-to
this order-whether
we evaluate at t or at tr.)
217
1- [V(~r)]2 = 1- [v(t)2 -C;Va"(dJC]= [1- V~~2] (1 + 2a~td),
2 -1/2
"(=
= "((t)(I-Va;:
[ 1-(v(:r)) ]
Evaluating everything now at time t:
= - q2
Fself
-"(3 (1-
411"€o[
"(3
"(3a
q2
"(4
(
4c3
[
v"(2a2
3
C2
(
4c3
)
+ ( ) d2 +...
(
. it
va2"(2
a+-+3-+3
c2
v"(2a2
~
)
)
c2
+()d+,..
~
X
]
"(4 it v"(2a2
it"(
va2"(2
-+3+-+- C2 +()d+,..
c
c3
4c3 3
va2"(2
+
3
it
+
3---z+ ( ) d +...
-411"€o
-~
d
4c
3c
c
[
(
"(3a
q2
it
+ -"(4 - + -
)
(
4C2
= -411"€o--+4c2d
=
.
d); a(tr)=a(t)-Tit=a(t)_a;d.
3va"(3dJc3) (a - it"(dJc)
4c2d
"(3a
q2
-411"€o --+4c2d
[
=
so
3
]
x
x
~
]
'
"(4
]x
)
(generalizing Eq. 11.96).
The first term is the electromagnetic mass; the radiation reaction itself is the second term:
2 2
2
F:~j
= ~~:c
"(4
( +3
it
)
vac2"(
(generalizing Eq. 11.99), so the generalization of Eq. 11.100 is
J-LOq2 4
-"(
611"c
Frad=
3"(2a2v
(
(b) Frad = A"(4 it +
~
)
l
t2
l
=-
FradV dt
t!
l
Rewrite the first term:
4
3 d"(
Nowdt("(v)=4"(
d
"(4itvdt
=
dt ("(4v)
t2
[
+ "(4a
t2
"(4itvdt
{12
it!
4
"(4
(
Problem
= "(4va
I
3"(2a2:!2
itv
+
d"(
dtv+"(aj
va"(3
4"(3v~
c2
-
l
t!
l (
or
h
"(4
v2a2"(2
itv
+ 3~
)
dt
t2
d
c
=
l
= "(6a
da
("(4v)
t2
-d t dt = "(4va t! I
t!
1
1
d
( y'1-v2Jc2 )
dt =dt
t2
t2,
V2
V2
C2
4 C2
l
t!
.
a =v
l
t2
~
.t!
dt
a2"(6
2va
va"(3
(-7 ) =~.So
=-2(I-v2Jc2)3/2
v2
( - + ) = (1 + 3 ) .
(1 + 32" ) dt, and hence
( )
+l
1
-d t ("(4v)a dt.
1
"(6a
C2
V2
"(6a2
) dt =
t2
"(4va
I
t!
c t2
h
v2
[ -"(6a2
a2v2
1 + 3-;;2 + 3"(67
dt
]
= "(4va
t2
I
t!
-
l
11.31
2 2 6
(a) P
=-
Sect. 11.2.2).
t2
tJ
d
)
P = Aa2"(6 (Eq. 11.75). What we must show is that
t2
Pdt,
t!
(except for boundary terms-see
va2"'2
+ 3---!....
c2'
J-Lq2
,where A == 6:c'
t2
(
it
= J-LO~1I"C
a
"(
c2
= yb2
2t
(Eq. 11.75).
c2t(C2t)
w
= y'b2 + c2t2 (Eq.
c2
10.45); v
'2
2 2
= tV = yb2 c+ c2t2 ;
2 2
b2c2
(
)
+ c2t2 - (b2+ c2t2)3/2 = (b2 + c2t2)3/2 b + c t - c t = (b2 +
c2t2)3/2 ;
h
t2
"(6a2 dt.
Qed
-,
CHAPTER 11. RADIATION
218
1
2
, = 1-
- J1.0q2
P -
1
v2jc2
= 1-
b4C4
(b2 + c2t2)3
611"c(b2 + c2t2)3
b6
24
(b) Frad
-
3b2c4t
3
(
3 (b2 + c2t2)
b2
1
= b2 (b2 + c2t2).
So
q2c
=~.
I
22
= J1.oq,
a+ ~
611"c
c2
(b2+ c2t2)5/2 + C2
b2 + c2t2
[c2t2j(b2+ c2t2)] = b2 + c2t2 - c2t2
Yes, it radiates
I
(in fact, at a constant rate).
22
)j
b4c4
a = -~
b c (2c2t)
-
3b2c4t
2 (b2+ c2t2)5/2 - - (b2+ c2t2)5/2j
c2t.
(b2+ c2t2)3../b2+ c2t2 = 0.1 Frad = 0.1
(. + ~ ) =
a
3,2a2v
C
I
No, the radiation reaction is zero.I
Chapter
12
Electrodynamics
and Relativity
Problem 12.1
Let u be the velocity of a particle in S, u its velocity in 5, and v the velocity of 5 with respect to S.
Galileo's velocity addition rule says that u = u + v. For a free particle, u is constant (that's Newton's first
law in S).
(a) If v is constant, then u = u-v is also constant, so Newton's first law holds in 5, and hence S is inertial.
(b) If 5 is inertial, then u is also constant, so v = u - u is constant.
Problem 12.2
(a) mAUA + mBUB = mcuc + mDuDi Ui = Ui+ v.
mA(uA + v) + mB(uB + v) = mc(uc + v) + mD(uD + v),
mAnA + mBUB + (mA + mB)v = mcuc + mDuD + (mc + mD)v.
Assuming mass is conserved, (mA + mE) = (mc + mD), it followsthat
mAUA+ mBuB = mcuc + mDuD, so momentum is conserved in 5.
1
2
1
2 - 1
2
1
2
(b) 2mAUA + 2mBUB - 2mcuC + 2mDuD =>
!mA(U~ + 2UA' v + V2) + !mB(U~ + 2UB' v + v2) = !mc(ub + 2uc' v + V2)+ ~mD(ub + 2UD' v + V2)
!mAU~ + ~mBU~ + 2v. (mAuA + mBuB) + !v2(mA + mE)
= ~mcub + !mDub + 2v. (mcuc + mDuD) + ~v2(mc + mD)'
But the middle terms are equal by conservation of momentum, and the last terms are equal by conservation
1
-2
1
-2
1
-2
1
-2
~
0f mass, so 2mAuA + 2mBuB = 2mcuC + 2mDuD' qed
Problem 12.3
VAB+VBC
~ Va (1 - ~ C
(a ) Va = VAB + VBC' , VE = 1+VABVBC/C2
) => ~ va
In mi/h, c = (186, 000 mi/s) x (3600sec/hr) = 6.7 x 1O8mi/hr.
... ~
va
= (6.7xlO
(5)(60J )2 = 6 . 7 X 10-16
=> 16.7 x 10-14% error
=
~. C
, I (p rett y small! )
[10].
(b) (!c + ic) / (1+ ~ . i) = (~c) / en = l.:!rJ (stllliess than c).
(c) To simplify the notation, let /3==VAC/C,/31==VAB/e, /32==vBc/e. Then Eq. 12.3 says: /3 = f~%~;2'
or:
/32=
/3f + 2/31/32+ /3~ =
(1 + 2/31/32
+ /3f/3~)
1 + 2/31/32+ /3f/3~ - (1 + /3f/3~- /3f - /3~) = 1
(1 + 2/31/32
+ /3f/3~)
(1 + 2/31/32
+ /3f/3~)
219
-
(1
-
/3f)(l - /3~) = 1 - A
(1 + /31/32)2
'
~
CHAPTER
220
ELECTRODYNAMICS
12.
AND RELATIVITY
where ~ ==(1-,8f)(1-,8~)/(1 + ,81,82)2is clearly a positive number. So ,82< 1, and hence IvAGI< c. qed
Problem
12.4
= ~c= ~gc.
(a) Velocity of bullet relative to ground: ~c + lc
Velocityof getawaycar: ~c = 192C.Since
Vb > vB'
:+ 1\
1c+1c
(b) Velocity of bullet relative to ground:
2 3
bullet does reach target.!
5
20
= T6 = 'fc= 28c.
Velocity of getaway car: ~c = ¥Bc. Since Vg >
Problem
I
~c
Vb,
I
bullet does not reach target.
I
12.5
(a) Light from the 90th clock took i~~~8°:'/S= 300 s
111:55 am.
= 5 min
to reach me, so the time I see on the clock is
I
(b) I observe 112 noon. I
Problem
12.6
light signal leaves a at time "7'
t~; arrives at earth at time ta = t~+ dalc,
{ light signal leaves b at time t~; arrives at earth at time tb = t~+ dblc.
:.~t=tb-ta
=t~-t~+
(db-da)
=~t'+
c
(-v~t'cosO)
c
=~t'
[1-
~cosO .
c
]
(Here da is the distance from a to earth, and db is the distance from b to earth.)
~s
,.
= v~t
vsinO..
vsinO ~t
SIll 0
= (1 - VI c cos 0); u = (1 -
v cos 0)
C
I
~cosO)(cosO)
- sinO(~ sinO)]
(1 - ~COSO)2
du
dO -- v[(I-
.
11sthe the apparent veloCity.
~C cosO) cosO = ~ sin2 0
= O:::}(1-
_V
cosO= -(SIll
0 + cos 0) -c
c
v.2
:::}
-1
( I)
max = COS V c.
A h
0
1
I
.
. al
t t 1Smax1m
1
vV1-v2/c2
ang e, u =
1-v2/c2
2
v
= V1-v2/c2'
As v ~ c, u ~ 00, because the denominator ~ 0, eventhoughv < c.
I
I
Problem 12.7
The student has not taken into account time dilation of the muon's "internal clock". In the laboratory, the
muon lasts 'YT =...;
1-v2/c2
T
, where T is the "proper" lifetime, 2 x 10-6 s. Thus
d
v=
TI../lT
2 2
(d)
V2
1
c2 = 1 + (Tcld)2;
v
v2
= 1-
e2;
d
= -vl-v2/c2,
v2/c2
T
2
1
r
2
v
[(d )
+
e2
whered=
] = 1;
Te - (2 x 10-6)(3 x 108) - ~ -~.
d 8.00
- 8 - 4'
2
v
=
V2
c2
800m.
1
(rld)2 + (1/e)2'
-
1
-
16.
- 1+ 9/16 - 25'
Iv = ~e.1
221
Problem
12.8
(a) Rocket clock runs slow; so earth clock reads 'Yt
= Vi-v
1 2
. 1 hr.
/c2
Here
= Vi-v
"1
1 2 /C2
=
~
5.
1-9/25
= 4_
.'. According to earth clocks signal was sent 11 hr and 15 min Iafter take-off.
(b) By earth observer, rocket is now a distance ac) (~) (1 hr) = ~c hr (three-quarters of a light hour) away.
Light signal will therefore take ~ hr to return to earth. Since it [eft 1 hr and 15 min after departure, light
signal reaches earth 12 hrs after takeoff.!
(c) Earth clocks run slow: trocket=
Problem
Lc
= 2LV"Ye
. h = h.
'Y.'
Problem
(2 hrs)
"1'
= ~ . (2 hrs) = 12.5hrs.1
12.9
(12 )
so 'Ye
2. =...L
'Y. = Vl-
2 = V4';:Y;
Ii. 1 = 1-
~V2
16'~
=..1..
12.10
= 1-..1.1616'=
v2
4' c
= V13
13.lv
I
.
Say length of mast (at rest) is 1. To an observer on the boat, height of mast is [sin 0, horizontal projection
is [cos O. To observer on dock, the former is unaffected, but the latter is Lorentz contracted to l[
'Y cos O.
Therefore:
-
tan 0
Problem
=
[sin 0
1
::y[cosO
-tan
.
=
"1
tan 0, or tan 0 =
I
0
vI
-
12.11
.
V2/C2
.
Naively, circumference/diameter = ~(27rR)/(2R) = trh = 7rVl- ((;JR/C)2 - but this is nonsense. Point
is: an accelerating object cannot remain rigid, in relativity. To decide what actually happens here, you need a
specific model for the internal forces holding the disk together.
Problem 12.12
(iv)
=>
t
= ~ + ~.
Put this into (i), and solve for x:
v2
l VX
1
x vt;
- x = 'Yex+ vt). ..(
x = 'YX- 'Yv ::y+ e2 = 'YX 1 - e2 - vt- = 'Yx"12vt- = ::y-
(
)
(
)
I
I
Similarly, (i) =>x = it
'Y + vt. Put this into (iv) and solve for t:
-
t
Problem
= 'Yt-
'YV
X
v2
(
-e2 "1- + vt
12.13
-
V
t
) = 'Yt(1 - -c2 ) - -xC2 = -"1 -
v -x;
C2
It='Y(l+~x).I..(
'-~
Let brother's accident occur at origin, time zero, in both frames. In system S (Sophie's), the coordinates
of Sophie's cry are x = 5 X 105 m, t = O. In system .5 (scientist's),l ='Y(t - ~x) = -'YVX/C2. Since
this is negative, Sophie's cry occurred before the accident,
f = - (l!) (He) (5 x 105)/c2
= -12
X 105/3
X 108
= -4
in.5.
X 10-3.
"1
53.
= V1-(12/13)2
1
14 X 10-3 s earlier.
So
- 144 = 1
= v'16~3
I
Problem 12.14
(a) In S it moves a distance dy in time dt. In.5, meanwhile, it moves a distance dy = dy in time dl =
"((dt
- ~dx).
dy
.'. dl
dy
= 'Y(dt -
~dx)
-
(dy/dt)
uy
= "1(1- ~~~) j or Uy = "1(1- 7)
1
-
Uz
; Uz = "1(1-
7)'
222
ELECTRODYNAMICS AND RELATIVITY
CHAPTER 12.
=-
(b) taniJ
=-
Uy
Ux
= -ccos{}'
In this case Ux
1
-
= -'Y
tan {}
b
Uy/
(1 - ~
(ux - v)/ (1-~)
, UY
sin {}
(
cos {}+vc /
)
)]
=.!.
.
(-Uy)
'Y(ux - v)
-esinO
= csin{}=}taniJ = 1.
'"Y -eeosO-v
(
-.
).
0
= 'Y~~n0
s in Prob.
. [Compare tan {}
12.10. The point is that velocities are sensitive
not only to the transformation of distances, but also of times. That's why there is no universal rule for
translating angles-you have to know whether it's an angle made by a velocity vector or a position vector.]
Problem 12.15
5
~c - lc
(1/4)c = -c.
2
Bullet relative to ground: -c, Outlaws relative to police: 4 3 2 1
8
5
)
(
/
7
1-4'2
5
2c - ~c
-(1/28)c
1
=
Bullet relative to outlaws: 17- ~ ~i = (13/28) = - 13c. [Velocity of A relative to B is minus the velocity
of B relative to A, so all entries below the diagonal are trivial. Note that in every case Vbullet< Voutlaws,so no
matter how you look at it, the bad guys get away.]
speed of -+
relative to .j.
Ground
Ground
Police
Outlaws
Bullet
Problem
Police
0
-e
-ie
-¥e
Bullet
¥e
!e
--he
0
Outlaws
e
0
-e
-!e
ie
e
0
-he
Do they escape?
Yes
Yes '!Yes
Yes
12.16
(a) Moving
V1-(4/5)2
1
= 32 .
clock runs slow, by a factor 'Y =
Since 18 years elapsed
~ X 18 = 30 years elapsed on the stationary clock. 151years old.
on the moving clock,
I
(b) By earth clock, it took 15 years to get there, at tc, so d = tC x 15 years = 112c years I (12 light years).
(c) It = 15 years, x = 12c years.
(d)
It= 9 years,
x = 0.1
I
[She got on at the origin in 5, and rode along with 5, so she's still at the origin. If
you doubt these values, use the Lorentz transformations,
(e) Lorentz transformations:
~= 'Y(x + vt)
= 'Y(t+ ~x)
(note that v is negative, since
{ t
:. x
with x and t from (c).]
S is going to
the left).
}
= ~(12c yrs + tc.
i=
15 yrs) = ~ . 24c yrs = 140c years. I
~(15 yrs + t2r .12c yrs) = ~ (15 + ~8) yrs = (25 + 16) yrs
= 141 years.
I
(f) Set her clock ahead 32 years, from 9 to 41 (t -+ i). Return trip takes 9 years (moving time), so her clock
will now read @QJyears at her arrival. Note that this is ~ .30 years-precisely what she would calculate if the
stay-at-home had been the traveler, for 30 years of his own time.
I
(g) (i)
t = 9 yrs, x
I
= O. What is t? t = ~x +
f=
~
.9
= 257 = 5.4 years,
and he started at age 21, so he's
126.4 years old. I (Younger than the traveler (!) because to the traveler it's the stay-at-home who's moving.)
(ii)
i = 41 yrs, x = O. What is t? t =
!45.6 years old. I
f=
~
. 41
=
1;3
=
24.6 years, and he started
at 21, so he's
223
(h) It will take another 15.4 years I of earth time for the return, so when she gets back, she will say her
twin's age is 45.6 + 5.4 =
years-which is what we found in (a). But note that to make it work from
~
traveler's point of view you must take into account the jump in perceivedage of stay-at-home when she changes
S to S.
coordinates from
Problem 12.17
-a,°rp + a,lfjI + a,2fj2
+ a,3fj3= -"l(aO - {3aI)(bO- {3bI)+ -l(aI - {3ao)(bI - {3aO)+ a2b2 + a3b3
= -"-?(aobo - {3/bI - {3jl'b°
+ {32aIbI
-
aIbI
+ {3jl'b° + {3/bI
- {32aobo)+ a2b2
+ a3b3
= -72aobo(1- {32)+ 72aIbI(1- {32)+ a2b2+ a~b3
[Note: 72(1 - /32) =
= -aobo + aIbI + a2b2+ a3b3. qed
Problem
1.]
12.18
1
ct
000
c~
(a) I
( g)
( -{
~
I
(b) A =
(
~ !
) ( ~ ) I(usmg
~
7
0
-7{3
0
0
-7{3
0
1
0
0
0
7
0
0
O'
1
the notation
of Eq. 12.24,
10' best
compadson).
)
'YO
.
.
(c) MultIply the matrIces: A =
-'Y~
0
(
1
0
-'Y~ 0 'Y
00010001
0
0
0
)(
7
-7{3
0
-7/3
7
0
7'Y
0 0
0 0
1 0
-7/3
)
=
(
-7'Y{3 -'Y~
7
-'Y7~
7'Y{3~
0
001
0
0
0
'Y
O'
)
Yes, the order does matter. In the other order, "bars" and "no-bars" would be switched, and this would give
a different matrix.
Problem 12.19 ~
I
I
(a) Since tanh 0
7=
= ~~~~~,and cosh20 - sinh20 = 1, we have:
1
y'1- v2je2
-
cosh 0
. A..1 (b) u
- sinh 0
0
( 0
u-v
=
1
y'1- tanh20
sinh 0 0
cosh0
0
0
fi
=
cosh 0
y'cosh2O- sinh20
0
0 0
1 0 .
0 1
.
= coshOj7{3= cosh0tanh 0 = smhO.
cas 1
41!. '/ 4/, 'If,t..
'(f,J/,,-'I -L
I il -----/ 'i / I It..I
/ -;--; I I '---71/ I /
II I II l--I/ tZ -1'/ 7/7
i
I
;
!
'
I
.i-
.
'I
I
,--LJ-----
'
'
-- -T+-
'
I I VJ PJ ./
fT v-: r It ./
/,
./
./
V
./
v
./
VJ ..) V; ...A I/f
r
r
I...--:'V:=lL---fT
? fT
t- r/
r
/
/
./
./
c,t
'?
c,t
'?
c,t
./
./
./
(b)
'}.
~
\.
"
./i-'
= slope = H
= ~:; c =
-
O.95c.1
= tC, so v = l+sos
~c:i:
- S!.fJ!k
<1-
I
\37725J
r35J
=@ = O.95c. .(
x
I
I
ii'
itll
:
U
i i
12.24
(a) (1- ~)ry' = u'; .'(1 +~) = ry'; u
I
=
(b)
1
1
Vl-u2/c2
Vl-tanh21J
Problem
12.25
= ,/1
+lry2/C'
11°1
= Vcosh21J-sinh21J
coshlJ
= coshOJ
1/= Vl-u2/c2
1
u=coshOctanhO=lcsinhO.1
(a) u. = u, = uco,45° = }, Joe = Ae.1
I
(b) Vl-~2/C2 =
R
= ~ = J5;
'I}
= Vl-:2/C2
~
l1]x
= 1]y = v'2col
(c) 1]0: ,c = J5 col
I
- -- 1":~
u -v -- .,fiTsc-fi75c
-- rQ"l
ux
1--5
~
02
(d) Eq. I2.45~
-
{
Uy
1
1
1
Vl-fj,2/c2 - ~Problem 12.26
-
° 2
~
(V2c -
2
~
2/5c
~- j;)'f.,
2
= y'iTo°=[lE]
~J5 c) = [QJ
ijx
(f)
1] 1]/-,--(1])
u
= " (,-~) =/1-.
(e) ijx = ,(1]x - (31]°)= VI -
/-'
v
(c) v'
./\.
./J
%
./
./ +9j-.LL
./LLLi--1
'?
./
/
./
~
I c,t \J
' '?
c,t
./
./
:c,t'?
/
./
&fTlld_i
Problem
./
./
/1 ....1
i././IA'11...J:1Ji/}
./
./
..) /}
C--r
/} A. VI
./ V 1...-1
i./V
,
!
'?
c,t'?
I
ijy
= 1]y = v'2col
= V3Ux = O. .(
- V3° - - V3 ii
,'1}~ { ijy=V3uy=V2c..(
2-
1
+1] - (I-u2jc2)(-c
2
2 -
+u )--c
2 (1 - u2jc2)
(I-u2jc2)-~
}
-
r:2l
226
CHAPTER
Problem
= 0 we
(b) VX2
11.31 we have
want T
-
b2
= 0: 0 = £c Inb + k, so k =
+ x = beeT/hi VX2
V - xVx 2 C
b2
--
-
b2
+ Vb2 + e2t2) + k; at
.
~(et + Vb2 +e2t2)
~e In [ b
C
_£ In bj IT =
Also from Frob.
Vb2 COS
h2 (CTj
bcDsh(CT/b)
t = cosh
= ~ In(et
]
= beCT/b- Xj X2- b2 = b2e2cT/b- 2xbeCT/b+ X2; 2xbeCT/b= b2(1 + e2CT/b);
C
(c) 1]1'= "((e, V, 0, O)j "( =
J ~dt = b J ~
= tvb2 + e2t2. :. T =
"(
x = b(ecr/b+2e-cr/b)
= Ibcosh(eTjb).1
Problem
AND RELATIVITY
12.27
(a) From Frob.
t
12. ELECTRODYNAMICS
b)
11.31: v
= e2tlvb2
+ e2t2.
eT
( )
sinh CTb --I etan h b'
- b2 - e cosh2(cT/b)-Icosh(cTb
- ecosh(cT/b
T' so 1]1'= cosh T (e,etanh T' 0,0) = Ie (coshT' sinhT' 0,0) .
12.28
Ui + v
.
'.
1 + (uivle2)
UA + v
UB + v
ue + v
UD + v
mA
0- .
2 + mB
2 = me
2 + mD
1 + (UAVI e )
1 + ( UBVI e )
1 + (uevle )
1 + (uDvle4)
This time, because the denominators are all different, we cannot conclude that
(a) mAUA + mBUB
= moUe
= moUe
mAfiA + mBUB
+ mDUD;
=
Ui
+ mDuD.
As an explicit counterexample, suppose all the masses are equal, and UA = -UB = Vj Ue = UD = O. This
is a symmetric "completely inelastic" collision in 5, and momentum is clearly conserved (0 = 0). But the
Einstein velocity addition rule gives UA = 0, UB = -2uj(1 + u2je2),ue = UD= -U, so in S the Olfcorrectly
defined) momentum is not conserved:
m
(b) mA1]A + mB1]B
= me1]e
+ mD1]Dj
-2u
(
)#
1 + u2/e2
1]i
-2mu.
= "((ih + !3ij?). (The
inverse Lorentz transformation.)
mA"((ijA+ !3ij~)+ mB"((ijB+ !3ij~) = me"((ije + !3ij~)+ mD"((ijD+ !3ijC];).
The gamma's cancel:
mAfiA + mBijB + !3(mAij~ + mBij~)
= meije
+ mDijD + /3(meij~ + mDijC];).
But mi1]? = p?= Ede, soif energyis conserved in S (EA + EB =
(correctly defined):
mAfiA + mBijB = meije + mDijD. qed
Problem 12.29
"(me2 - me2 = nme2 =>
=n + 1=
1
=> 1 - ~ =
1
I
I
"(
Vl-u2/c2
~
C
Ee + ED), then so too is the momentum
.
1
~.I
... ~u2 -- 1 - (n+l)2
-- n2+2n+l-l
(11,+1)2 -- Tfi+T)2'
U -- vn(n
n + +1 2) e.
Problem 12.30
Er = E1 + E2
+"';
pr
= PI + P2 +...;
fir
= "((pr -
/3Erle)
= 0 =>!3= vie = prelEr.
v = e2pr I Er = e2 (Pi + P2 + . . . )I (E1 + E2 + . . . ) .1
I
Problem
12.31
(m; + m~) 2
2
(m; + m~)
EJL=
e = "(mJLe => "( =
=
2m71"
2
~=1--=1-
e2
1
"(2
2m71"mJL
4m2m2
71" I'
(m~ + m~)2
m4 + 2m2m2 + m4
=
71"
71" I'
I'
1
VI - V2j e2
-
(m~ + m~)2
4m2m2
71" 1'=
V2
1
; 1 - 2" = 2j
e
"(
m2 ) 2
71"
I' jv=
(m2
-
(m~ + m~)2
m2 - m2
( m~ +m~ ) e.
227
Problem
12.32
Initial momentum:
E2
-
= m2e4
= 3me2.
p2e2
Initial energy: 2me2 + me2
-
=* p2e2 = (2me2)2
= 3m2e4 =* p =
m2e4
V3me.
Each is conserved, so final energy is 3me2, final momentum is V3 me.
E2 - p2e2 = (3me2)2
-
(V3me)2e2
= 6m2e4 = M2e4
=*
I
M
= V6m
I
~ 2.5m.
(In this process some kinetic energy was converted into rest energy, so M > 2m.)
E = V3me
3me2 C2= V3
v = pe2
e .1
I
Problem
12.33
First calculate p,ioQ.'senergy: E2 = p2e2 + m2e4 = I96m2e4+ m2e4 = ~~m2e4 =* E = ~me2.
Conservation of energy:
2me2 = EA + EB
2E - 2 2
Conservation of momentum:
tme2 = PA + PB =
=* tme2 = EA - EB }
A - me.
~ ~
=* EA
I
Problem
= me2j
I
I
EB
= ~me2.1
12.34
Classically, E = ~mv2. In a colliding beam experiment, the relative velocity (classically) is twice the
velocity of either one, so the relative energy is 4E.
1
CD-"-
~
r
s
.5
CD
,E
~S
Let be the system in which
E1 = poc2 -
2
= Pac - POCEA+
POCEA+ E~
1232
= [VP5C2 + m2c4 + mc2 -
= 2m2c4 + 2mc2vp5c2
EA(mc2 + Vp5C2 + m2c4
E A =mc 2
-
Problem
EA) + m2c4
- 2EAmc2 + E~. Or:
-
2EAmc2;
+ mc2Jpoc2 + m2c4;
. (mc2 (mc2 -
(mc2 + VP5C2 + m2c4)
-
EA] 2
2EAVP5C2 + m2c4
= m2c4
(mc2 + VP5C2 + m2c4
= mc2
-
+ m2c4
poc/2)
AND RELATIVITY
4"EA + 4"EA
= POC2+ m2c4 + 2VP5c2 + m2c4(mc2 -POCEA
12. ELECTRODYNAMICS
poc/2)
2
VP5C2-t-m2c4 - poc/2)
VP5C2 + m2c4
2
(,,?c' -(#c4
P6e' -- ,,?c'
- ..",/- #c4)
P6e'+ m'c')
pomc3- +lPome'
p~t - P5c2
-
Poc/2)
= I
me'
2
0
(me
+ 2Po(mc++vP'o!po)+ m'c')
12.36
F
dU
1
d
P mu
dt
- dt - dt Vl- u2/C2 - m { Vl- u2/C2 + U -2
m
u(u a)
=
a+
. qed
{
Vl-u2/c2
(c2-u2) }
-
d
( )
1
dU
2
-<:2" u. dt
(1- u2/c2)3/2 }
ay"
.
Problem 12.37
At constant force you go in "hyperbolic" motion. Photon A, which left the origin at t < 0,
catches up with you, but photon B, which
passes the origin at t > 0, never does.
Problem
12.38
(a)
0
a
d1]o
=
-
a=d1]=dtd1]=
dr
dr dt
=1
1
(1
-
u2/C2) [
dr
= [ dt ( Vl-
dr
1
U2/C2
1
)]
(-~)2u'a
(
)
VI - U2/C2 -2 (1 - u2/c2)3/2 C
1
d
VI - U2/C2dt
a+ u(u
c
d
d1]o dt
= ill
ct
. a)
(c2 - U2) ]
.
.
u
( VI
- u2/C2
)
=
~U2/C2
1
u.a
C(1 - u2/C2)2'
1
a
+u(-~)
-~2u'a
VI - U2/C2 { VI - U2/C2
2 (1 - U2/C2)3/2}
229
J.L-
(b)
o 2
--(a)
aJ.La
=
-
1
.a)2
(u
1
+a:.a:--e2(I-u2/e2)4
1
(1 - u2/e2)4
{
1
1
e2
.a) 2 + a2 (1 -
--(u
U2
2
{ (
= (l-u2/e2)4
a
[(
+ (l-u2/e2)4
2
2
2
U
-
e2
)
2
+ -
e2
+
)
2
U
(
1- -
e2
U2
e2
2
1
]
a 1- e2 + e2u(u.a)
(u.a)2
)
1- e2
U2
)
1
e4
2
(u. a) + -u
U2
2
(u. a)
2
}
}
(,-1+2-2;2+;2)
v
(1 - ~)
1
-I
-
(c)
'T]J.L'T]J.L
[
a
(1 - u2/e2)2
2
+
(u. a)2
(e2 - u2)
= -e2, so iT ('T]J.L'T]J.L)
= aJ.L'T]J.L+ 'T]J.LaJ.L=
(d) KJ.L=
¥r =
Problem
12.39
!r(ml}J.L)=
~
I
.
J
2aJ.L'T]J.L
KJ.L'T]J.L =
= 0, so
I
0.1
aJ.L'T]J.L=
maJ.L'T]J.L
= 0.1
KJ.LKJ.L= _(KO)2 + K. K. From Eq. 12.70, K. K = (1-~:/C2)' From Eq. 12.71:
KO_~dE-
1
me2
d
( viI
e dr - evil - u2/e2 dt
- u2/e2
)
m
But (Eq.12.73):
KO =
u.F=uFcosO=
uFcosO
evil - u2/C2;
viI - u2/e2
-
me
-
viI - u2/e2
[
-~
U2 (u
[
(u.a)+
F2
(-I/e2)
2 (1 - u2/e2)3/2
.a)
e2( 1 - u 2/2)
e
u2 F2 COS20
K KJ.L=
J.L
(l-u2/e2)
-
e2(I-u2/e2)
=
]
[
-m
2u.a
J
m( u
-
(u.a)
e (1 - u2/e2)2'
.a)
= ( 1 - u 2/2)3/2'SO
e
1 - (u2/e2) COS2O
(l-u2/e2)
F2.
qed
J
12.40
Problem
m
W= viI
u(u . a)
- U2/ e2 [ a+
.
Dotmu:(u.a)+
e2 - u2 J =q(E+uXB)=}a+(2
.
u2(u a)
e2( l-u
u(u . a)
q /
2)=-vl-u2/e2(E+uXB).
e - u
m
.
u a
e
2/2)=(
l-u
q
2 2
e
m
2/2)=-vll-u/e[u.E+.u.(uvXB~;
=0
:.
u(u . a)
(e2 - U2)
q
=-vll-u2/e2
m
u(u . E)
e2
.
q
Soa=-vll-u2/e2
m
[
E+uxB--u(u.E)
1
e2
].
qed
Problem 12.41
One way to see it is to look back at the general formula for E (Eq. 10.29). For a uniform infinite plane of
charge, moving at constant velocity in the plane, j = 0 and p = 0, whilep (or rather, a) is independent of t
(so retardation does nothing). Therefore the field is exactly the same as it would be for a plane at rest (except
that a itself is altered by Lorentz contraction).
A more elegant argument exploits the fact that E is a vector (whereas B is a pseudovector). This means that
any given component changes sign if the configuration is reflected in a plane perpendicular to that direction.
But in Fig. 12.35(b), if we reflect in the x y plane the configuration is unaltered, so the z component of E would
230
CHAPTER
12. ELECTRODYNAMICS
AND RELATIVITY
have to stay the same. Therefore it must in fact be zero. (By contrast, if you reflect in a plane perpendicular
to the y direction the charges trade places, so it is perfectly appropriate that the y component of E should
reverse its sign.)
Problem 12.42
(a) Field is ao/Eo, and it points perpendicular to the positive plate, so:
Eo = ao(cos45°x+sin45°y) =
EO
1
(b) From Eq. 12.108, Ex = Exo = -~;
;'0 (-x+y).
v 2 EO
So E
Ey = "IEyO= "I)i~o'
= tan-I
(c) From Frob. 12.10: tan 0 = "I,so 10
I
= ~(-x
"1.1
+ "Iy).1
y
(d) Let ii be a unit vector perpendicular to the plates in S-evidently
= - sinOx + cosOy; lEI = ;'°'0 ';1 + "12.
ii
x
So the angle 4>between ii and E is:
_E . ii
1.
lEI = cos4> =
= tan 0 = sinlJ
But'"
I
Evidently the field is
Problem 12.43
coslJ
~
2
I
1
E. da = q(1
cas
2
- v2/c2)
47rEO
47rEO
q(l-v2/c2)
=
2Eo
I
E.da=
So Icos 4>
=C
:"1"12
R2 sin2O dO d4>
~ sin2 0)3/2
"
sinO dO.
1
27r 0 (1 -
~ sin
(I
1-1 [1-
2
2
/'
Let u
0)3 2
du
c
(v) (C)
-1
2
=
2
(l-v2/c2)
1
C
V
(~c2 - 1) £.
v =
(-;;)
3
. 2
du
J-I (~
C2
-1 + u2 )3/2'
2
(1 - v2/ei) .
=. .(
q
-.2..~/Loq2(I-v2/c2)2vsinO
Rx(fi);
sin20)3 (~
(b) Using Eq. 12.111 and Eq. 12.92, S = /Lo(E X B) - /Lo47rEO47r R4(1 -
~
2
- sm0 dO,sm 0 = 1 - u .
3 {I
(V)
2EO
c
1
= cos0, so du =
q(l-v2/c2)
~ + ~u2J3/2 =
(~c2 - 1) .ju~c 2 - 1 + U2
C 3
q(1 - v2/C2)
~ 3
2EO
.
=>
R2(1 -
The integral is:
I
1
yI+'Y2
~
+1
So
I
perpendicular to the plates in S.
q(1 - V2/C2)
=
= ~cosO
1 + "12
-1 => ...L20 = ",2 + 1 => cosO =
V cos21J
q(1 - v /c)
R E 1292
(a ) E -- ~
47rEO ( 1 - ~v2. sm 20 )3/2 R2 ( q. . )
I
2"1
= V~(tanO+"I)
1 + "12
~(smO+"IcosO)
1 + "12
= v'I-cos21J=
coslJ
cosO
-(}
).
231
S
= 1-
Problem
q2
(l-v2je2)2vsinOA
~2 sin 2 0)3 (J.
2
1611" £0 R4(1-
12.44
(a) Fields of A at B: E = 4;fO ~Yj B
= O. So
=
force on qB is IF
4:£0
q~;B Y.
~
*-x
-L
y
2 2
.
) 1 - 'Y qA
E q. 12.92 , WIth 0 -- 90°.. E- -- - 1 qA(1 -2 v 2je3 2
.n y
.ny - _411"£0u4 11"£01
( - v j e ) / u-
..
(11) Fr am
X
V qA
~Ty.1 (Note: here the particle is at rest in S.)
=
(b) (i) From Eq. 12.68: IF
W d
A
(this also follows from Eq. 12.108).
B i- 0, but since VB = 0 in S, there is no magnetic force anyway,and IF = ~TY
Problem
Here 0
'Y
B = -~
A
= (~)2
Note that (E2 - B2e2) = (~)2'Y2(1-~)
VA
= v, so E = - q 1
411"£0 r2
F
VB=
= q [E + (-vx)
v+v
l+v2je2
~
'YB
=
vI
-
-
y,
F
-
B2C2
SystemC:
ve=O.
=
2-
2
!Ll
[y - ~(x
~~~
~
(1 + v2jc2)
c
C4
V1-2~+V4
2
,:y41 + 2 v
= (-L41rfor ) I (
~
2
+
1
v2jc2
2-
X
2
z)] = !Ll
~~~
(1+ ~)y.
~
411"£0r2
=
(1 + v2jc2)
(l-v2je2)
= -2
'Y (1+
2),Vb'YB=2v12.
C
V2.
A.Bq --z.
2v12A
(I+v2e2 )y,
- -- 4 11"£0 e2 r2
4
4 2
v - v
C4
~
v2
+ "2)Y.
(+q at rest
c
E=-~~Yj
Fe = -y2(t+~2/c2)FB'
./]
is invariant, because it doesn't depend on v. We can
2v
(t+V2/C2)2
[The relative velocity of Band
=>
where 'Y = VI - v2je2.
- (1+V2jc2)
14v2/c2
q2 12
= qE = --411"£0
"2(1
r
z,
.
:.E=-~~12
411"£0
r2
E2
1
A
B --411"£0
c22r1 Z, where 1 = Vlq ~
A
B] =
X
v 'Y
411"£0 C2 r2
use this as a check.
[Check:
(as before).
12.45
411"£0r2 y,
System B:
I
= 90°, ~ = y, lb = Z,Iz-= r, so (using e2 = IjJ-lo£o):
E=-~
System A:
A
=>
B=O;
2
2
I
- = (-L- ) . ./ ]
) = (-L41rfor ) 14 ¥
41TfOr
no magnetic force). [Check: Eq. 12.68 => FA = J-FB.
./]
-y
F=qE=_L~y.
411"£0r2
C is 2vj(1 + v2jc2), and the corresponding 'Y is 12(1 + v2je2). So Eq. 12.68
232
CHAPTER
12. ELECTRODYNAMICS
AND RELATIVITY
Summary:
(-),y
( - 47rfQor2
)
+
(-
)2V
2A
2"/
Z
Z
"/
(- 47r:r2),(1 +
Problem
(-),y2(1
()-
)Y
47rfOr
47rfOr
"/
)y
(- 47r;Or2)Y
0
2(1 + 2v2) Y
A
(- 47r;or2)Y
}~
";'-~'I"
~
12.46
(a) From Eq. 12.108:
--
--
--
--
E. B = ExBx + EyBy + EzBz
2
= ExBx +
V
v
(Ey - vBz)(By + czEJ + ,,/(Ez + vBy)(Bz - ZEy)
c
v ~
v2
V~
V2
= ExBx + ,,/2{EyBy + ci/'yEz - vP(Bz - C2EzBz + EzBz - ci/'yEz + vP(Bz - c2EyBy}
(
= ExBx + ,,/2 [EyBy 1 (b) P;2 - C2jp
=
[E; + ,,/2(Ey
-
"/
(1 -
~:) + EzBz
+
VBz)2
+
,,/2(Ez
~:)] = ExBx + EyBy + EzBz = E
-
VBy)2]
c2
[B; + ,,/2(By +
= E2 + ",2 (E2 - 2E £B + v2 B2 + E2 + 2E £B + v2 B2
x
/
Y
"')fv, z
z
z
7v, y
y
~
V2
-c2-E2-C2B2+C22
c4Z
=E
2
x
-cB2
2
x
z
V2
+"/ 2 E 2 1-[ Y
= (E2x + E2Y + E2z ) -
(
c2 (B2x
c2
)
-c2-E2
) -C2B2
c4Y
+E
2
z
1--
(
V2
C2
)
+ B2Y + B2z ) = E2 -
(c) No.1 For if B = 0 in one system, then (E2
any system. Therefore E =I- 0 in all systems.
I
Problem
V2
E
zy
-
-cB2
B2C2
;
. B.
Ez)2 + ,,/2(Bz
- C2B2 - c22 v
y
k
;li Uy
E
qed
-
;
Ey)2]
z
x
2
Y
.
1--
(
V2
c2
)
-cB2
2
z
I--
(
v2
C2
)]
q ed
C2B2) is positive. Since it is invariant, it must be positive in
12.47
(a) Making the appropriate
modifications
E(x, y, z, t) = Eo cos(kx - wt) Y,
in Eq. 9.48 (and picking 8
= 0 for
convenience),
B(x, y, z, t) = Eo
c cos(kx - wt) z,
where k ==~.
c
(b) Using Eq. 12.108 to transform the fields:
Ex
= Ez = 0, Ey = ,,/(Ey- vBz) = ,,/Eo[cos(kx- wt) - ~ cos(kx- Wi)] = o:Eo cos(kx -
v
1
v
Bx = By = 0, Bz = ,,/(Bz - ZEy)
= ,,/Eo [ -c cos(kx - wt) - zc cos(kx - wt)]
c
where
0:=="/(1-~)=
1 - vlc
l+vlc.
wi),
Eo
= 0:-c
cos(kx - wi),
233
Now the inverse Lorentz transformations
(Eq. 12.19) =* x = 'Y(x + vf)
;x)] ='Y[(k-~~)x-(w-kv)t]
kx-wt='Y[k(X+Vf)-w(f+
where (recalling that k = wie): k =='Y (k
- ~ ) = 'Yk(1 -
E(x, y, z, f) = Eo cos(kx - wf)y,
wave in S is
vlc
w-
I
= ak
and
w=='Yw(1 - vi c) = aw..
w=
- aw, and a -= 1 /11 +v
- vlcc .
I
-\
A
I
- -211"-= 211"
k - ak
.
~
a
The velocity of the
w
v = -A
= \"A = @]
211"
same in any inertial
vie)
This is the Doppler shift for light.
l+vlc'
=kx-wf,
B(x, y, z, f) = Eo
c cos(kx - wf)z,
whereEo
- = aEo, k- = ak,
1-
so
WV
Conclusion:
(c) Iw = w
; x),
and t = 'Y(f +
I
Yup, this is exactly what I expected (the velocity of a light wave is the
I
system).
(d) Since intensity goes like E2, the ratio is
i=
~~
= a2 = ~ ~ ~~~.
I
Dear AI,
The amplitude, frequency, and intensity of the light wave will all decrease to zero as you
run faster and faster. It'll get so faint you won't be able to see it, and so red-shifted even your
night-vision goggles won't help. But it'll still be going 3 x 108mls relative to you. Sorry about
that.
I
I
Sincerely,
David
Problem
[02
12.48
= A~A;tA<7= A8A~tO2+ A~A~t12 = 'YtO2+ (-'Y{3)t12 = 'Y(tO2- {3t12).
~3 = A~A~tA<7 = A8A~tO3 + A~A~t13 = 'YtO3+ (-'Y{3)t13 = 'Y(tO3 - {3t13) = 'Y(tO3 + {3t31).
~f23 = A~A~tA<7 = A~A~t23 = t23.
fH = A1A~tA<7 = A~AAt30 + A~Att31 = (-'Y{3)t30 + 'Yt31 = 'Y(t31 + {3tO3).
f12
Problem
= AiA;tA<7 = AAA~tO2+ AtA~t12 = (-'Y{3)tO2+ 'Yt12= 'Y(t12- {3tO2).
12.49
Suppose t"l-' = :HI-'" (+ for symmetric, - for antisymmetric).
fl 8z (Bz) + 8x (B:zJ+ 8y (By) = 0 =>V . B = O.
235
One temporal:
say, J1.=
0, v = 1, A = 2 (other
permutations of these indices yield the same result, or minus
it).
~FOl + 8oFl2 + 8lF20
( ) + 8(~) (B%)+ :x (~y ) = 0,
= 0 ~ :y - ~:r:
= VxE.
or -!!..fft+ (§j: - 8ff:r:.)= 0, which is the z component of -~~
component; for v
= 2, A = 3 we get
(If J1.= 0, v
= 1,A = 2, we get
the y
the x component.)
Conclusion: 8>.Fp.v+ 8p.Fv>.+ 8vF>.p.= 0 is equivalent to V.B = 0 and ~~ =
-
V X E, and hence to
8vGp.v = O. qed
Problem 12.54
.
Ko = q'f/vFov= q('f/lFo1+ 'f/2Fo2+ 'f/3FO3)= q(TJ E)je
=
= ~ dd'; ,. where
= ~dt, we
Ko
W is the energy of the particle.
-1'1edtdW
Since dr
= -1'(u.
e
q
E) ~
I
E.! Now from Eq. 12.71 we know that
~1'u,
- = q(u.
dt
IdW
have:
E).
This says the power delivered to the particle is force (qE) times velocity (u)
Problem 12.55
8°1jJ
8t
-
8lIjJ
-
which is as it should be.
= ~1jJ
8_1jJ= _!(81jJ8~ + 81jJ8~+ 81jJ8~+ 81jJ8:).
~ = _!em
emm
&m ~m
&m
From Eq. 12.19, we have:
So 8°1jJ
I
8x
8y
8z
8t = 1', 8t = 1'V, 8t = 8t = O.
1 81jJ+ v-81jJ.
81jJ- -v81jJ
= --1'(_
8t
8x ) or (smce et = XO = -xo): 8°1jJ= 1'( 8 Xo e 8x l)
e
_
= ~1>
8x = 81jJ8t
8t 8Xl
+ 81>8x + 81jJ8y + 81>8z
8x 8x
8y 8x
8z 8x
= l'
[(8oljJ)
-
~(811jJ)].
= 1'3!..
81jJ - ~ 8cp) = [(8lljJ) e2 88tcp + l' 81>
8x = 1'( 8Xl
e 8xo
l'
~1jJ=~=~m+~&+~~+~&=~=~~
8y
8t 8y 8x 8y 8y 8y
8z 8y
8y
~1jJ=~=~m+~&+~~+~&=~=~~
8z
8t 8z 8x 8z
8y 8z
8z 8z
8z
1-(:onclusion: 8p.1>transforms in the same way as ap. (Eq. 12.27)-and
Problem 12.56
~(8°1jJ)].
hence is a contravariant 4-vector. qed
According to Prob. 12.53, 8ff;"~= 0 is equivalent to Eq. 12.129. Using Eq. 12.132, we find (in the notation
of Prob. 12.55):
8Fv>. +
-8Fp.v
\ + -
8x"
8 xp.
_
8F>.p. 8
8
8
8XV = >.Fp.v+ p.Fv>.+ vF>.p.
= 8>.(8p.Av - 8vAp.) + 8p.(8vA>. - 8>.Av) + 8v(8>.Ap. - 8p.A>.)
=
(8>.8p.Av - 8p.8>.Av) + (8p.8vA>. - 8v8p.A>.) + (8v8>.Ap. - 8>.8vAp.)
[Note that 8>.8p.Av= 88~v
x xp. = 88~v
x x >. = 8p.8>.Av,by equality of cross-derivatives.]
=
O.
qed
CHAPTER 12. ELECTRODYNAMICS
2:~6
Problem
AND RELATIVITY
12.57
y
Step 1: rotate from xy to XY, using Eq. 1.29:
~
y
x = cos ifJx + sin ifJy
Y = - sin ifJx + cosifJy
Step 2: Lorentz-transform
<{
from XY to
XY,
x
using
Eq. 12.18:
(p
X = 1'(X - vt) = 1'[cosifJx + sinifJy - .Bet]
Y = Y = - sin ifJ x + cos ifJY
x
Z=Z=z
et = 1'(et - .BX) = 1'[et- .B(cosifJx+ sin ifJy)]
Step 3: Rotate from
XY
to xii, using Eq. 1.29 with negative ifJ:
= cosifJX - sin ifJY = 1'cosifJ[cosifJx+ sinifJy - .Bet]- sinifJ[- sin4>x + cos4>Y]
= C'YCOS2ifJ+ sin2 ifJ)x + C'Y 1) sin ifJcos ifJy - ,.B cos (et)
ii = sin ifJX + cos ifJY = l' sin ifJ(cos ifJx + sin ifJy - .Bet) + cos ifJ(- sin ifJe + cos y)
= C'Y - 1) sinifJcosifJx + C'Ysin2ifJ+ COS24»y - ,.Bsin 4>(ct)
x
ifJ
..:...
ifJ
l'
-1'.B cos ifJ
-1'.B c~s ifJ
-1'.BsmifJ
C'YCOS2ifJ+ sin2 ifJ)
C'Y
C'Y-1)sin4>cos4>
C'Ysin24>+cos2ifJ)
0
C
In matrix
Problem
form:
~
~
zOO
()
=
(
-1'.B sin ifJ
- 1) sin 4>cos 4>
12.58
1f
In center-of-momentum system, threshold occurs when incident energy is just sufficient to cover the rest energy of the resulting particles,
with none "wasted" as kinetic energy. Thus, in lab system, we want
the outgoing K and E to have the same velocity, at threshold:
(X)---+0-0
1f
P
KE
momentum:
0--
--0
00
KE
P
before (CM)
after (CM)
After
Before
Initial
~) m.
P7T;initial energy of 1f: E2
-
p2e2
= m2e4
:::}
E; = m~c4+ p~c2.
Total initial energy: mpe2 + vm;c4 + p;e2. These are also the final energy and momentum: E2
(mK + m~:Ye4.
(mpe2
+ vm;c4 + p;c2) 2 - p;C2 = (mK + m!Yc4
2m e2
m;!, + -;f-vm;e2
2:p vm;e2
+ p; e + m;!' + rJ!c2 -
+ p;
rJ!c2
= (mK + mE)2!,
= (mK + mE)2- m; -m;
-
p2e2
=
237
2
4
2 2
2 4m;
4
2
2
(m".c +p". ) ~c = (mK +mI; ) - 2( mp+m". )( mK +mI; ) +mp+m".
4m2
-TP;
c
= (mK + mI;)4 - 2(m; + m;)(mK + mI;)2 + (m;
-
4
2
2
+ 2 mpm".
m;)2
P". = 2~ p v(mK + mI;)4 - 2(m~ + m~)(mK + mI;)2 + (m~ - m~)2
= (2m~c2)c
v(mKc2 + mI;c2)4 - 2[(mpc2)2+ (m".c2)2](mKc2 + mI;c2)2 + [(mpc2)2 - (m".c2)2]2
= 2C(~OO)
V(1700)4
= ~V(8.35
Problem
-
2[(900)2 + (150)2](1700)2+ [(900)2 - (150)2]2
X 106) = 11133MeVIc.!
X 1012)- (4.81 X 1012)+ (0.62 x 1012)= ~(2.04
12.59
rP
P
P
0--
In CM:
(p = magnitude
Lx
--0
in CM,
u;;;~u
Before
4>
of 3-momentum
angle)
= CM scattering
sp)
After
Outgoing4-momenta:
rP
= (~,pcos4>,psin4>,O)jsP = (~,-pcos4>,-psin4>,O)o
TP
0---+-
In Lab:
0
Problem: calculate 0, in terms of p,
~-"
Before
Lorentz transformation:
Tx = -y(rx
S
- {3rO)j Ty = rvj
Bx= -y(sx - {3s0)j By= By.
Now E = -ymc2jP = --ymv (v here is to the left); E2 - p2c2 = m2c4, so {3= -1jfo
:. Tx= -y(pcos 4> + 1jf~)
cosO
= -yp(1+ COg4»j
= -r . S =
TB
-
-
Tv
= psin 4>jBx= -yp(1-
-y2p2(1
Vh2p2(1
- p2 sin2
+ COg4»2 + p2 sin2 4>][-y2p2(1
(-y2
-
I) sin2
-
4»; By
= -psin
4>.
4>
cos 4»2 + p2 sin2 4>]
4>
Vh2(1 + cos4»2 + sin2 4>]h2(1(-y2- I)
-
COS2 4»
COg
cos4»2 + sin2 4>]
-
(-y2- I)
[-y2e;i~o:.
238
CHAPTER 12, ELECTRODYNAMICS
cos ()
AND RELATIVITY
w
=
(where w ==,.,? - 1)
V(1 + cot2 ~+ wcot2 ~)(1 + tan2 ~ + wtan2~)
w
w sin 1!.cos 1!.
2
2
- V(CSc2~ + wcot2 ~)(sec2 ~ + wtan2~) - V(1+ WCOS2 ~)(1+ wsin2~)
-
~wsin
-
sin
- V[1 + ~w(1 + cos][(~ + 1) - cos]
-sin
-sin
-1
h
, were
r 2 -- -4 + -,4
V(~+1)2-cos2
,
sm
() -
r2
-
4
- sin"" r - i? (1
+ w) --
(
* =K
'J!..
' 'dt
(
12.60
u
)
~
dx dt
,x/
. 2
tan () =
I
"(V22c2
sin .
dt -
u
(VI - U2j C2)
= Kmw~;
= 2Kx+
m
dx
wdw
dx
1
'=
mu
P
0/
." .
\ <0'.
r j sin
fL
1
,
Vl-u2/c2
u
(VI - U2j c2) = Km -/1- u'
u2jc2 Let w
= ~!£W2
= !:.-;
2dx
m
w
d(W2)
dx
=~
VI - U2j C2.
= 2K
m
=>
d(w2)
= 2K
m (dx),
constant. But at t = 0, x = 0 and u = 0 (so w = 0), and hence the constant is O.
2K
W2
2
U
2,
Let mc2
2K =
- a ,
= !£
W2
/$
-2r..
- (-y2-1)sin""
=>'!lfdT = K, But dr - Vl-u2/c2'
= Kyl-u2jc2,
Multiply by ~; =~:
m
dw
dx
At t
() -
, SO t an
d dt
Vl-u2/c2
ct
2
VI + (rj sin
=
m
-x
U2
=
2Kxjm
= 1+
ct=
= J -/y2y+
-
c2
dx .
a2 2ydy
,
,
mc2
- 1 + ( 2Kx
)
2Kx
mc2
J~ ..;x
2
2Kx
u =---u
m
1 -u 2j c 2 j
dx
+ a2 dy
= 2Kx,
; u2(1 + 2Kx
mc2)
m
c
;
= 2ydyj
..;x
+ 1)
Let: z ==yja = ..;xv -?!b= V~'!j, Then ¥!d = z~
I
ct
=J
mc2
1 + (2Kx) dx.
= y.
o~
= [YVy2 + a2 + a2ln(y +
= 0, x = 0 => y = 0, so 0 = a2Ina+ constant =>constant =
:. ct = YVy2 + a2 + a2ln(yja + v(yja)2
2
mc2
dt = VI + (~;~)
Let x ==y2j dx
= 2 J Vy2
2Kx
Vy2 + a2)] + constant,
-a2 Ina.
= a2 [ (~) V (~)
2
+ 1 + In
+ In(z+ ~),I
(~+/ (~) + 1) ],
2
239
Problem
12.61
......--........
(a) x(t) = f; [VI + (at)2 -1], where a = ::c' The force of +q on
-q will be the mirror image of the force of -q on +q (in the x axis),
so the net force is in the x direction (the net magnetic force is zero).
All we need is the x component of E.
-d/2""".""'.:'q
The field at +q due to -q is: (Eq. 10.65)
q
,z
47f/Oo~.
U)3 [u(c2_v2)+u(~'a)-a(~'u)].
E=--(
U
= Cot-
v ~ Ux
= ci - v = t(cl-
............-..-........
v,z)j~. U = 0'1- ~. v = (0'1-Iv); ~. a = la. So:
Ex = -~
,z
~(cl47f/Oo(0'1-- vl)3 [ ,z
+ ~(cl - )4)la -" a(O'1-- jAJ)]
m)(c2 - V2)
"z
,
~
tca(12 - ~2)
=--
x
= -carP /~
47f/Oo
q (0'1-- 1 V 1)3 [(cl-m)(c2-v2)-cacP.]
The force on +q is qEx, and there is an equal force on -q, so the net force on the dipole is:
2q2
F
v(t)
a(tr)
1
= - 47f/Oo(c,z -lv)3
c 1
= -= -dx
2
d
t
dv
a
ca
It remains to determine ,z, 1,
v, and a, and plug these in.
[(cl - V,z)(C2 - V2) - cad2]x.
1
2
V1 + (at)2 2a t
1
( )
cat
=
2a2tr
= dtr = T + catr -2 ~
catr
V1 - (at)2 ; v
ca
/
' where T ==vI + (atr)2.
= v(tr) = -T
ca
2
2
= T3[1+ (atr) - (atr)] = f3'
Now calculate tr: c2(t - tr)2 =,z2 = 12+ rP; 1= x(t) - x(tr) = f; [VI + (at)2 - VI + (atr)2] , so
y - 2ttr +Yr = ~ [1 + (~2 + 1 + (a}{)2- 2Vl + (at)2vl + (atr)2] + (d/C)2
(*) VI + (at)2vl + (atr)2 = 1 + a2ttr + ~ (':,d)2. Square both sides:
1 ad
.4.h.4.h
,X
+ (at)2+ (atr)2+ a/ t; =,X+ a/ t; +:1 (~
2
ad
2
t + tr - 2ttr - ttr
2
d
4
)
2
(~ ) - (~)
ad
+ 2a2ttr+ (~
a2
d
2
)
ad
+ a2ttr (~
2
)
4
- 4 (~) = O.
At this point we could solve for tr in terms of t, but since v and a are already expressed in terms of tr it is
simpler to solve for t (in terms of tr), and express everything in terms of tr:
t2-ttr[2+(:d)2]+[t;_(~)2
_~2(~)4]
=O~
t = ~{tr [2+ (:dr] ~ t~r +4(acdr + (:df] - ~ + 4(~r + a2(~f}
= tr [1+ ~ (:dr]
~
[1+ (atr)2](~r [1+ (~:r]
CHAPTER 12. ELECTRODYNAMICS
240
AND RELATIVITY
Which sign? For small a we want t ~ tr + die, so we need the + sign:
2
1 ad
= tr [ 1 +"2 (7
t
)
I
d
]
ad
+ ~TD, where D ==VI +
2
( 2e ) .
So1-= e(t - tr) =?1-= ~(Q:)2 + dTD. Now go back to Eq. (*) and solve for 0+(at)2:
VI + (at)2
= ~{1+
(
~ aed
f + a2tr [tr (1 + ~ (:d f)
= ~{ II + ~atr)2J[1 +
+ ~TD] }
~ (aedf] + a2~rdTD } = [1 + ~ (aedf] T + a2~rdD.
T2
e
e
l=
2
1 ad
rv
d.
a2trd
] a { r +"2( 7 ) ] T + --;;-D -;t } = ad ( 2eT + trD ) .
a[VI + (at2)- VI + (atr)2.=
Putting all this in, the numerator in square brackets in F becomes:
[ ]= {cad(:eT+trD)
d
= cad [2eT + vb e2a~
= 'f3
.
..
F
=
q2
c2ad2
1
["2T2~
- e~r [e~r(:df +dTD]}(e2 - e2;:t~) -e~~d2
d(atr)2
e2
2
2
e2a~
2eT
1
-
vb ] T2 [1 + (¥'r)
e2ad2
]
-
(¥'r)
] - 'f3
[
] =-
"2(atr)2- 1 = 2T3 1+ (atr)2- (atr)2 - 2
c2a~
2T3 .
It remains to compute the denominator:
3x.
47r€o[(e1--lv)T]
ctr
(c1--lv)T=
{ e""2
[
ad
2
(7 )
=[~¥d2
d
+dTD ] -ad
eatr
.
(2eT+trD ) T
}T
+cdTD- ~¥d2 - Cd(~r)2D]T = edD[
~
1+(%)2 _(%)2
q2
:. F
e2d2a
= 47r€o c3d3D3
~
X
q2
a
X
(a
= 47"€ocd[I + (adI2e)2]3/2
I
=
:J.
Energy must come from the "reservoir" of energy stored in the electromagnetic fields.
ad 2 3/2
1 q2
a
q2
Jloq2
(b) F = mea = - =? 1+ =
= -.
[ ( 2e ) ]
2 47r€ocd[I + (adI2e)2]3/2
t
81f€omc2d 87rmd
(force on one end only)
. ..a -
2e
d
Jloq2
2/3
(87rmd )
-
1, so F
-
2mc2
d
JlOq2
(87rmd)
2/3
-
1.
] = dcD.
241
Problem 12.62
(a) AI' = (Vie, Ax,Ay,Az) is a 4-vector (like xl' = (et, x, y, z)), so (using Eq. 12.19): V = ,(if + vAx). But
if = 0, and
A - JLo(m X f)x
x -
= myz - mzij = myz
Now (m X f)x
411"
.
1'3
- mzY. So
= ,vJLo
(myz 1'3
- mzy) .
411"
= ,Rx, fj = y = Ry,Z = z = Rz, where R is the vector (in S)
V
Now
x = ,(x
- vt)
from the (instantaneous)
location of the dipole to the point of observation. Thus
V2
1'2= ,2 R; + R~ + R~ = ,2(R; + R~ + R~) + (1 - ,2)(R~ + R~) = ,2 (R2 - c2R2sin2B)
(where B is the angle between R and the x axis, so that R~ + R~
:. V = J.Lo v,(myRz
411",3R3(1-
. but v. (m
- mzRy)
~sin2B)3/2'
X
= v(m
R)
V = J.Lov . (m X R)(1
411"
or, using J.Lo=
~
.
and v (m X R)
= R. (v
= R2 sin2 B).
-
X R)x = v(myRz - mzRy),
so
~)
2 n 3/2 '
R 3 (1 - 112.
~ sm u )
X m):
V
=
1
Ii. (v X m) (1 -
411"fOC2R2 (1 -
~)
~ sin2 B) 3/2 .
(b) In the nonrelativistic limit (V2« e2):
V= ~Ii.(vx
411"fo
e2R2
m) = ~Ii.p,
411"fOR2
with p
=v
Xm
e2
'
which is the potential of an electric dipole.
Problem 12.63
(a) B
IN
(b)
= -EfKy
(Eq. 5.56); N
=m
X B (Eq. 6.1), so
N = -EfmK(z
X y).
= TmKx = ~(Avl2)(UV)X = EfAUV2l2X.
I
Charge density on the front side: AO(A = ,AO);
v/T7
LJv
-
,= VI -
Charge density on the back side:
1
(1+v2
C2)2 =
4v2/c2
X= 7AO,whereii =
(l+v2/c2)
l+v2/c2
1+;~/C2 =>
(l+v2/c2)
(
V2
)
~ + 114
C4 - 4112
~ = VI - 2v2+
~
C4
VI + 2112
114= (1- v2/e2) = ,2 1+ e2 .
Length of front and back sides in this frame: lIT. So the net charge on the back side is:
-l
q+ = A- =,
,
2
V2
Al
(1+ -e2)--"
V2
= (1+ - )Al.
e2
242
CHAPTER
12. ELECTRODYNAMICS
AND RELATIVITY
Net charge on front side is:
l
q-
>..l
1
= >"0-"( = -= ->"l.
"( "(
"(2
So the dipole moment (note: charges on sides are equal):
l
= (q+)-2 Y-
P
E
u
h
= .£.JL
2 Z, were
<0
l
(q-)-2 y =
l
V2
[(
1+
a = "(ao,so N = p
X
1
l
>"l2
V2
~
V2
l2V2
-C2 ) >"l-->"l- y = -2 (1+ -C2 -1 + -C2 )Y = -y.
2 "(2 2 ]
C2
E
.
_2
>"l2V2 a
= - C2
~
€o"(
~
(Y X z )
= -"(l{Lo\
_2 Aa l 22~v x.
So apart from the relativistic factor of "( the torque is the same in both systems-but
in S it is the torque
exerted by a magnetic field on a magnetic dipole, whereas in S it is the torque exerte<;lby an electric field on
an electric dipole.
Problem
12.64
Choose axes so that E points in the z direction and B in the yz plane: E = (0,0, E); B = (0, B cos
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