Homework Solutions (Week of 2.23)
25.2: a) Current is given by I =
Q
t
=
420 C
80 ( 60 s )
= 8.75 10 2 A.
b) I = nqvd A
I
8.75 102 A
=
nqA (5.8 10 28 )(1.6 1019 C)( (1.3 103 m) 2 )
= 1.78 106 m s .
vd =
EA (0.49 V m)( 4 (0.84 103 m) 2 )
=
= 11.1 A.
25.23
Physics 151 Homework Set 3
23.7: a)
qq
qq
U = k 1 2 + 1 2
r
r13
12
(4.00 nC)(3.00 nC) (4.00 nC)(2.00 nC)
+
(0.200 m)
(0.100m)
q1q2
=k
+
r23
(3.00 nC)(2.00 nC)
+
(0.100 m)
= 3.60 10 7 J.
qq
q q
q q
b) If U = 0, 0 = k 1 2 + 1 3 + 2 3 . So solving for
Physics 151, Homework Solutions to Chapter 29
29.24: a) = vBL = (7.50 m s)(0.800 T)(0.500 m) = 3.00 V.
b) The current flows counterclockwise since its magnetic field must oppose the
increasing flux through the loop.
LB (3.00 V)(0.500 m)(0.800 T)
c) F = IL
Homework Set 6: Chapter 27, Magnetic Forces
27.4:
a=
27.7:
qv B
, and since = 0
j j
m
C)(3.0 10 4 m s )(1.63 T)( i )
j
2
= (0.330 m s )k.
3
1.81 10 kg
F = ma = q v B a =
(1.22 10 8
F = q v B sin v =
F
4.60 10 15 N
=
q B sin (1.6 10 -19 C)(3.5 10 3 T) si
Physics 151 Homework Solutions
24.6: (a) 12.0 V since the plates remain charged.
(b) (i) V = Q
C
Q does not change since the plates are disconnected from the battery.
A
C=
d
If d is doubled, C 1 C , so V 2V = 24.0 V
2
(ii) A = 7 2 , so if r 2r , then A 4
Chapter 22 Homework solutions
22.20: a) For points outside a uniform spherical charge distribution, all the charge can
be considered to be concentrated at the center of the sphere. The field outside the sphere
is thus inversely proportional to the square
Physics 151, Homework Solutions Chapter 28
28.11: a) At the point exactly midway between the wires, the two magnetic fields are in
opposite directions and cancel.
b) At a distance a above the top wire, the magnetic fields are in the same
I I I
0 I 2 0
Physics 151, Chapter 32 Homework Solutions
32.6: a) x direction.
2 2 f
kc (1.38 10 4 rad m) (3.0 10 8 m s)
=
f =
=
= 6.59 1011 Hz.
b) k =
c
2
2
c) Since the magnetic field is in the + y -direction, and the wave is propagating in the
x -direction, then t
Here are the solutions to the Amperes Law problems in Chapter 28:
28.32: Consider a coaxial cable where the currents run in OPPOSITE directions.
r
I
a) For a < r < b, I encl = I B dl = 0 I B 2r = 0 I B = 0 .
2r
b) For r > c, the enclosed current is zero,
Physics 151, Homework Set 1 Solutions
21.8: a) The total number of electrons on each sphere equals the number of protons.
ne = np = 13 N A
0.0250 kg
= 7.25 1024.
0.026982 kg mol
b) For a force of 1.00 10 4 N to act between the spheres,
1 q2
F = 10 N =
q