90
CHAPTER
5.
MAGNETOSTATICS
E
E
E
E.
E
E
y(t)
=

wB cos(wt)
+
Bt
+
wB; z(t)
=
wB
sm(wt).
y(t)
=
wB
[1+
wt

cos(wt)]
j
z(t)
=
wB sin(wt).
Let /3 ==
EjwB;
then
[y
 /3(1 +
wi)]
=
(3cos(wt),
z
=
{3sin(wt)j
[y
 {3(1+
wtW
+
z2
=
{32. This
is a circle
of
radius
{3 whose
center
is at
Yo
=
(3(1 +
r.vt),
Zo = O.
z
4~A
y
/3
(c)
Problem
5.3
(a) From Eq. 5.2, F =
q[E
+ (v x B)] = 0 =>
E
=
vB
=>
I
v
= ~
.1
q
v
m
(b) FromEq. 5.3,
mv
=
qBR
=>;;, =
FiR
=
~
Problem
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
=
I
Ika2
z.1
Problem
5.5
(a)
I
K
=
1
,
I
because the lengthperpendiculartoflow
is the circumference.
21ra
(b)
J
= ~ =>
1
=
/
J da
=
a
/
!s
ds de/>
=
21ra
/
ds
=
21raa
=>
a
=
2
1
;
J
=
1
2
1
.1
s
s
1ra
1ras
Problem
5.6
(a)
v
=
r.vr,
so
I
K
=
f7r.vr.1
(b) v
=
r.vrsin8~
=>
I
J
=
pr.vrsin8~,
I
where
p
==
Qj(4j3)1rR3.
Problem
5.7
:
= ~
Iv
pr
dr
=
/
(a;:)
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 divergencetheorem), and since J is entirely
insideV,
it is zeroon the surface
S.
Therefore
Iv(V'
J)xdr
=

Iv
Jx dr,
or, combiningthis with the
y
and
z
components,
Iv(V'
J)rdr
=

Iv
J
dr.
Or, referring back to the first line,
ii
=
/
J
dr.
Qed
Problem
5.8
.
.
I~oll
(a) Use Eq. 5.35, wIth
z
=
R,82
= 81 = 45°, and four sIdes:
B
=
~.
(b)
z
=
R,
82= 81 = ;, and
n
sides:
B
=
~~~ sin(1r/n).