Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must
be directed due south. Since B = 0i 2 r ,
b
g
c
h
6
2 rB 2 0.080 m 39 10 T
i=
=
= 16 A.
0
4 10 7 T m A
(b) The current must be from west to east to produce

Chapter 24
1. If the electric potential is zero at infinity then at the surface of a uniformly charged
sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius.
Thus q = 40RV and the number of electrons is
q 4 0 R V
( 1.0 106

Chapter 23
1. The vector area A and the electric field E are shown on the diagram below. The angle
between them is 180 35 = 145, so the electric flux through the area is
2
= E A = EA cos = ( 1800 N C ) ( 3.2 10 3 m ) cos145 = 1.5 10 2 N m 2 C.
z
2. We u

Chapter 22
1. (a) We note that the electric field points leftward at both points. Using F = q0 E , and
orienting our x axis rightward (so points right in the figure), we find
i
N
F = ( +1.6 1019 C ) 40 = (6.4 10 18 N)
i
i
C
which means the magnitude of t

Chapter 21
1. Eq. 21-1 gives Coulombs Law, F = k
k | q1 | q2 |
r=
=
F
( 8.99 10 N m
9
2
q1 q2
r2
, which we solve for the distance:
C 2 ) ( 26.0 10 6 C ) ( 47.0 10 6 C )
5.70N
= 1.39m.
2. (a) With a understood to mean the magnitude of acceleration, Newton

Chapter 30
1. (a) The magnitude of the emf is
=
c
bg
h
d B
d
=
6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV.
dt
dt
(b) Appealing to Lenzs law (especially Fig. 30-5(a) we see that the current flow in the
loop is clockwise. Thus, the current is to left

Chapter 28
1. (a) The force on the electron is
(
)(
)
r
r
rr
FB = qv B = q vx + v y Bx + By j = q ( v x B y v y Bx ) k
i
j
i
(
(
)(
)
(
)
= 1.6 1019 C 2.0 106 m s ( 0.15 T ) 3.0 106 m s ( 0.030 T )
14
= 6.2 10 N k.
)
r
r
Thus, the magnitude of FB is 6.2

Chapter 27
1. (a) The energy transferred is
2t
(2.0 V) 2 (2.0 min) (60 s / min)
U = Pt =
=
= 80 J .
r+R
1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s / min) = 67 J.
U = i Rt = G J Rt = G
H+ R K H + 5.0 J
r

Chapter 25
1. (a) The capacitance of the system is
C=
q
70 pC
=
= 3.5 pF .
V
20 V
(b) The capacitance is independent of q; it is still 3.5 pF.
(c) The potential difference becomes
V =
q 200 pC
=
= 57 V .
C 3.5 pF
2. Charge flows until the potential differ

Chapter 32
1. We use
6
n =1
Bn = 0 to obtain
5
B 6 = Bn = ( 1Wb + 2 Wb 3 Wb + 4 Wb 5 Wb ) = +3 Wb .
n =1
2. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the
bottom is +0.70 mWb as given in the problem statement. Since

Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and
time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s,
q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The number of electrons N is given by q = Ne, where e is t