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Homework Solution
Chapter 28
Young and Freedman 13 th Edition
28.2.IDENTIFY: A moving charge creates a magnetic field as well as an electric field.
 qv sin 
SET UP: The magnetic field caused by a moving charge is B  0
, and its electric field is
4 r 2
1 e
since q  e.
4 e0 r 2
EXECUTE: Substitute the appropriate numbers into the above equations.
E
B
0 qv sin  4  107 T  m/A (160  1019 C)(22  106 m/s)sin 90

 13 T, out of the page.
4 r 2
4
(53  1011 m) 2
1 e (900  109 N  m 2 /C2 )(160  1019 C)

 51 1011 N/C, toward the electron.
4 e0 r 2
(53  1011 m) 2
EVALUATE: There are enormous fields within the atom!
E
28.3.
IDENTIFY: A moving charge creates a magnetic field.
SET UP: The magnetic field due to a moving charge is B 
0 qv sin 
4
EXECUTE: Substituting numbers into the above equation gives
(a) B 
r2
.
0 qv sin  4  107 T  m/A (16  1019 C)(30  107 m/s)sin 30

.
4 r 2
4
(200  106 m) 2
B  6.00 10–8 T, out of the paper, and it is the same at point B.
(b) B  (1.00 10–7 T  m/A)(1.60 10–19 C)(3.00 107 m/s)/(2.00 10–6 m)2
B  1.20  10 –7 T, out of the page.
(c) B  0 T since sin(180)  0.
EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the earth’s
magnetic field.
28.14.IDENTIFY: A current segment creates a magnetic field.
0 Idl sin 
.
4 r 2
Both fields are into the page, so their magnitudes add.
EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives
SET UP: The law of Biot and Savart gives dB 
4  107 T  m/A
dB 
4
 250 cm 
(120 A)(000150 m) 

 800 cm   8.79  10 –8 T
(00800 m)2
The field from the 24.0-A segment is twice this value, so the total field is 2.64  10 –7 T, into the page.
EVALUATE: The rest of each wire also produces field at P. We have calculated just the field from the
two segments that are indicated in the problem.
28.26.
IDENTIFY: Each segment of the rectangular loop creates a magnetic field at the center of the loop, and
all these fields are in the same direction.
 I
2a
SET UP: The field due to each segment is B  0
. B is into paper so I is clockwise around
4 x x 2  a 2
the loop.
EXECUTE: Long sides: a  475 cm. x  210 cm. For the two long sides,
B  2(100  107 T  m/A) I
2(475  102 m)
(210  102 m) (00210 m) 2  (00475 m) 2
 (1742  10 5 T/A) I 
Short sides: a  210 cm. x  475 cm. For the two short sides,
B  2(100  107 T  m/A) I
2(210  102 m)
(475  10
2
m) (00475 m)  (00210 m)
2
2
 (3405  10 6 T/A) I 
Using the known field, we have B  (2082 105 T/A) I  550 105 T, which gives I  264 A.
EVALUATE: This is a typical household current, yet it produces a magnetic field which is about the
same as the earth’s magnetic field.
28.27.IDENTIFY: The net magnetic field at the center of the square is the vector sum of the fields due to each
wire.
 I
SET UP: For each wire, B  0 and the direction of B is given by the right-hand rule that is
2 r
illustrated in Figure 28.6 in the textbook.
EXECUTE: (a) and (b) B  0 since the magnetic fields due to currents at opposite corners of the
square cancel.
(c) The fields due to each wire are sketched in Figure 28.27.
 I
B  Ba cos45  Bb cos45  Bc cos45  Bd cos 45  4Ba cos45  4 0  cos45 .
 2 r 
r  (10 cm) 2  (10 cm) 2  10 2 cm  010 2 m, so
B4
(4  107 T  m/A)(100 A)
cos 45  40  104 T, to the left.
2 (010 2 m)
EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the
square would be to the right.
Figure 28.27
28.30.IDENTIFY: Apply Eq. (28.11) for the force from each wire.
SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel
conductors carrying currents in opposite directions repel each other.
EXECUTE: On the top wire
F 0 I 2  1 1  0 I 2

, upward. On the middle wire, the magnetic
 

L
2  d 2d  4 d
F 0 I 2  1 1  0 I 2

, downward.
 

L
2  d 2d  4 d
EVALUATE: The net force on the middle wire is zero because at the location of the middle wire the
net magnetic field due to the other two wires is zero.
forces cancel so the net force is zero. On the bottom wire
28.36.IDENTIFY: The magnetic field at the center of a circular loop is B 
0 I
. By symmetry each segment of
2R
the loop that has length l contributes equally to the field, so the field at the center of a semicircle is
1 that of a full loop.
2
SET UP: Since the straight sections produce no field at P, the field at P is B 
EXECUTE: B 
0 I
0 I
4R
.
. The direction of B is given by the right-hand rule: B is directed into the page.
4R
EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is
 I
B 0 .
8R
28.39.IDENTIFY: Apply Eq. (28.16).
SET UP: At the center of the coil, x  0. a is the radius of the coil, 0.020 m.
 NI  (600) (0500 A)
 942  103 T.
EXECUTE: (a) Bcenter  0  0
2a
2(0020 m)
(b) B( x) 
0 NIa 2
2( x  a )
2
2 3/2
. B(008 m) 
0 (600)(0500 A)(0020 m) 2
2((0080 m)  (0020 m) )
2
2 3/2
 134  104 T.
EVALUATE: As shown in Figure 28.14 in the textbook, the field has its largest magnitude at the center
of the coil and decreases with distance along the axis from the center.
0 Idl  rˆ
.
4 r 2
SET UP: The two straight segments produce zero field at P. The field at the center of a circular loop
 I
 I
of radius R is B  0 , so the field at the center of curvature of a semicircular loop is B  0 .
2R
4R
EXECUTE: The semicircular loop of radius a produces field out of the page at P and the semicircular
1   I  1 1   I  a 
loop of radius b produces field into the page. Therefore, B  Ba  Bb   0     0 1   ,
2  2  a b  4a  b 
out of page.
EVALUATE: If a  b, B  0.
28.74.IDENTIFY: Apply dB 