Homework 7
Ch20: P 33, 41, 43, 45, 47, 53, 59; 87
33. (II) Two long thin parallel wires 13.0 cm apart carry 25-A currents in the same
direction. Determine the magnetic field at point P, 12.0 cm from one wire and 5.0
cm from the other (Fig. 20–55).
B1
B2
1
5.0 cm
90o  1
13.0 cm
12.0 cm
1
Solution
The field from each wire makes concentric circles about the source wire. The individual fields at
point P are perpendicular to the lines from the wires to point P. We define B 1 as the field from
the closer point, and B 2 as the field from the further point. Use Eq. 20-6 to calculate each
individual field.
 0 I 4 10 7 Tm / A25 A
B1 

 1.000 10 4 T
2r1
2 0.050m 
 0 I 4 10 7 Tm / A25 A
B2 

 4.167 10 5 T
2r2
2 0.120m 

 
Bnet  B1  B2


Take into account that 12 2  5 2  132 . This means that B1  B2 . Using Pythagorean theorem we
have
0 I 1 1 0 I
r12 4 10 7 Tm / A25 A
5
 B B 
 2 
1 2 
1     1.110 4 T
2
2
2 r1 r2
2r1
r2
2  5 10 m
 12 
2
Bnet
2
1
2
2

To find direction of Bnet we can calculate x and y components:
1  arctan( 5 / 12)  22.6










Bnet x  B1 cos 90o  1  B2 cos 1  1.0  10 4 T cos 67.4o  4.167  10 5 T cos 22.6o
 4.1  10 8 T  0 T


Bnet y  B1 sin 90o  1  B2 sin 1  1.0  104 T sin 67.4o  4.167  10 5 T sin 22.6o
 1.1  104 T
Bnet  1.1 104 T @ 90o
41. (II) A rectangular loop of wire lies in the same plane as a straight wire, as shown in
Fig. 20–57. There is a current of 2.5 A in both wires. Determine the magnitude and
direction of the net force on the loop.
Solution
The magnetic field at the loop due to the long wire is into the page, and can be calculated
by Eq. 20-6. The force on the segment of the loop closest to the wire is towards the wire,
since the currents are in the same direction. The force on the segment of the loop farthest
from the wire is away from the wire, since the currents are in the opposite direction.
Because the magnetic field varies with distance, it is difficult to calculate the total force
on either the left or right segments of the loop. Using the right hand rule, the force on
each small piece of the left segment of wire is to the left, and the force on each small
piece of the right segment of wire is to the right. If left and right small pieces are chosen
that are equidistant from the long wire, the net force on those two small pieces is zero.
Thus the total force on the left and right segments of wire is zero, and so only the parallel
segments need to be considered in the calculation. Use Eq. 20-7.
Fnet  Fnear  Ffar 

 1
 0 I1 I 2
 II

1 
lnear  0 1 2 lfar  0 I1 I 2l 


2 d near
2 d far
2
 d near d far 
4  107 T m A
2


 2.6  10 6 N, towards wire

 0.030 m 0.080 m 
 2.5 A   0.100 m  
2
1

1
43. (II) Two long wires are oriented so that they are perpendicular to each other. At their
closest, they are 20.0 cm apart (Fig. 20–59). What is the magnitude of the magnetic
field at a point midway between them if the top one carries a current of 20.0 A and
the bottom one carries 5.0 A?
Solution
The magnetic fields created by the individual currents will be at right angles to each
other. The field due to the top wire will be to the right, and the field due to the bottom
wire will be out of the page. Since they are at right angles, the net field is the hypotenuse
of the two individual fields.
2
 0 I top   0 I bottom 
0
4  107 T m A
2
2
Bnet  


I

I

 

top
bottom
2  0.100 m 
 2 rtop   2 rbottom  2 r
 4.12  105 T
2
 20.0 A 2   5.0 A 2
45. (II) Three long parallel wires are 3.8 cm from one another. (Looking along them, they
are at three corners of an equilateral triangle.) The current in each wire is 8.00 A, but its
direction in wire M is opposite to that in wires N and P (Fig. 20–60). Determine the
magnetic force per unit length on each wire due to the other two.
Solution
a) The forces on wire M due to the other wires are repelling forces, away from
the other wires. Use Eq. 20-7 to calculate the force per unit length on wire M
due to each of the other wires, and then add the force vectors together. The
horizontal parts of the forces cancel, and the sum is vertical.
FM net y
lM


FMN
lM
cos 30o 
FMP
lM
FMP
FMN
30 30o
o
cos 30o
0 I M I N
 I I
1m  cos 30o  0 M P 1m  cos 30o
2 d MN
2 d MP
2
 4 10
7
TmA
2
 8.00 A 
2
 0.038 m 
cos 30o  5.8  10 4 N m , 90 o
b) The forces on wire N due to the other wires are an attractive force towards wire P
and a repelling force away from wire M. Use Eq. 20-7 to calculate the force per
unit length on wire N due to each of the other wires, and then add the force vectors
together. From the symmetry, we expect the net force to lie exactly between the
two individual force vectors, which is 60o below the horizontal.
FNM
We can see that
 0 I M I P 4  10 7 Tm / A8.00 A2
Fnet  FMN  FMP 

 3.4  10 4 N / m ;
2d MP
2 0.038m
c) The forces on wire P due to the other wires are an attractive force
towards wire N and a repelling force away from wire M. Use Eq. 20-7 to
calculate the force per unit length on wire P due to each of the other wires,
and then add the force vectors together. From symmetry, this is just a
mirror image of the previous solution, and so the net force is as follows.
FP net  3.4  104 N m
  240o
FNP
90o
30o
  300 
FPN
90o
FPM
30o
47. (II) Let two long parallel wires, a distance d apart, carry equal currents I in the same

direction. One wire is at x  0, the other is at x  d , Fig. 20–61. Determine B along
the x axis between the wires as a function of x.
50
25
0
0
0.2
0.4
0.6
-25
-50
Solution
The left wire will cause a field on the x-axis that points in the y-direction, and the right
wire will cause a field on the x-axis that points in the negative y-direction. The distance
from the left wire to a point on the x-axis is x, and the distance from the right wire is
d  x.
I
0 I
 I 1
1 
Bnet  0 
 0  

2 x 2  d  x  2  x d  x 
A graph of the above function is shown above.
0.8
1
53.
(III) (a) Use Ampère’s law to show that the magnetic field between the
conductors of a coaxial cable (Fig. 20–63) is B   0 I 2r if r is greater than the radius of
the inner wire and less than the radius of the outer cylindrical braid. (b) Show that B  0
outside the coaxial cable.
Solution
(a) Ampere’s Law says that along a closed path,  B l  0 I encl . For the path choose a
circle of radius r, centered on the center of the wire, greater than the radius of the
inner wire and less than the radius of the outer cylindrical braid. Because the wire is
long and straight, the magnetic field is tangent to the chosen path, and so B  B .
The current enclosed by the path is I.
I
0 I encl   B l   Bl  B  l  B  2 r   B  0
2 r
(b) We make a similar argument, but now choose the path to be a circle of radius r,
greater than the radius of the outer cylindrical braid. Because the wire and the braid
are long and straight, the magnetic field is tangent to the chosen path, and so B  B .
The current enclosed by the path is zero, since there are two equal but oppositely
directed currents.
I
0 I encl   B l   Bl  B  l  B  2 r   B  0 encl  0
2 r
59. (II) A circular coil 16.0 cm in diameter and containing nine loops lies flat on the
ground. The Earth’s magnetic field at this location has magnitude 5.50 10 5 T and
points into the Earth at an angle of 56.0° below a line pointing due north. If a 7.20-A
clockwise current passes through the coil, (a) determine the torque on the coil, and
(b) which edge of the coil rises up: north, east, south, or west?
Solution
(a) The torque is given by Eq. 20-10. The angle is the angle between the B-field and the
perpendicular to the coil face.
2
  NIAB sin   9  7.20 A    0.080 m    5.50  105 T  sin 34o  4.01  10 5 m N
(b) If the coil is free to turn, it will rotate toward the orientation so that the angle is 0
(torque is 0). In this case, that means the north edge of the coil will rise, so that a
perpendicular to its face will be parallel with the Earth’s magnetic field.

87. An electron enters a uniform magnetic field B  0.23 T at a 45° angle to B.
Determine the radius r and pitch p (distance between loops) of the electron’s helical
path assuming its speed is 3.0 106 m s . See Fig. 20–72.
Solution
The centripetal force is caused by the magnetic field, and is given by Eq. 20-3.
F  qvB sin   qv B  m
r
mv
v2
r

 9.1110 kg  3.0 10 m s  sin 45

1.60 10 C   0.23T 
31
6
19
qB
o
 5.251 105 m  5.3  105 m
The component of the velocity that is parallel to the magnetic field is unchanged, and so
the pitch is that velocity component times the period of the circular motion.
T
2 r
v
2

mv
qB
v

2 m
qB
2  9.11  1031 kg 
 2 m 
6
o
p  v T  v cos 45 
  3.0  10 m s  cos 45
 3.3  10 4 m

19
1.60 10 C   0.23T 
 qB 
o
Note!
p  v||T  2r
v||
v
 2r tan 45  2r