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

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.

  1. 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.

  1. 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.

Solution

a) The forces on wire M due to the other wires are repelling forces, away from the other wires. Use equation 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.

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.

;

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.

  1. 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.

  1. 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.