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Physics 112
Homework 7 (Ch20)
1. A rectangular loop of wire lies in the same plane as a straight wire, as shown in figure
below. 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 equation
F
 0 I1 I 2
l
2 d
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.
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
Physics 112
Homework 7 (Ch20)
2. Two long wires are oriented so that they are perpendicular to each other. At their
closest, they are 20.0 cm apart (see figure below). 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
2
 20.0 A    5.0 A 
2
2
 4.12  105 T
3. 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 (see figure below). Determine
the magnetic force per unit length on each wire due to the other two.
Physics 112
Homework 7 (Ch20)
Solution
a) The forces on wire M due to the other wires are repelling forces, away from FMP
FMN
 0 I1 I 2 l
30o 30o
the other wires. Use equation F 
to calculate the force per unit
2r
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
cos 30o
0 I M I N
 I I
1m  cos 30o  0 M P 1m  cos 30o
2 d MN
2 d MP
 4 10
2
7
TmA
2
 8.00 A 
2
 0.038 m 
cos 30o  5.8  104 N m , 90o
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. 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.

FNP
90o
FNM
30o

2
Fnet FMN FMP  0 I M I P
4  10 7 Tm / A 8.00 A




 3.4  10  4 N / m ;
l
l
l
2d MP
2 0.038m
  300 
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. 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
FPN
90o
FPM
30o
4. The north pole of the magnet in figure below is being inserted into the coil. In which
direction is the induced current flowing through the resistor R?
Solution
As the magnet is pushed into the coil the magnetic flux increases to the right. To oppose
this increase, flux produced by the induced current must be to the left, so the induced
current in the resistor will be from right to left.
Physics 112
Homework 7 (Ch20)
5. What is the direction of the induced current in the circular loop due to the current
shown in each part of figure below?
Solution
(a) The increasing current in the wire will cause an increasing field out of the page
through the loop. To oppose this increase, the induced current in the loop will
produce a flux into the page, so the direction of the induced current will be
clockwise.
(b) The decreasing current in the wire will cause a decreasing field out of the page
through the loop. To oppose this decrease, the induced current in the loop will
produce a flux out of the page, so the direction of the induced current will be
counterclockwise.
(c) The decreasing current in the wire will cause a decreasing field into the page through
the loop. To oppose this decrease, the induced current in the loop will produce a flux
into the page, so the direction of the induced current will be clockwise.
(d) Because the current is constant, there will be no change in flux, so the induced
current will be zero.