Physics EM current

Question

A square, 45-turn coil 10.0 cm on a side with a resistance of 0.720  is placed between the poles of a large electromagnet. The electromagnet produces a constant, uniform magnetic field of 0.600 T directed into the page. As suggested by the figure below the field drops sharply to zero at the edges of the magnet. The coil moves to the right at a constant velocity of 2.00 cm/s.

1)What is the current through the wire coil before the coil reaches the edge of the field?

2)What is the current through the wire coil while the coil is leaving the field?

3)What is the current through the wire coil after the coil leaves the field?

4)What is the total charge that flows past a given point in the coil as it leaves the field?

Answer 1

The induced voltage is equal in magnitude to the rate of change in flux, Φ, times the number of loops of wire.The flux is equal to BA.

ξ = -Nd(BA)/dt = NBdA/dt =NBvL where L is the length across and v is the velocity of the loop into the field or out of the field and N is the number of loops. There is not induced current when the loop is out of the B field or completely in the B field. Answers to a and c are 0.

ξ = 45* .6 Tesla* 0.1 m* 0.02 m/s = ,054 V. With the field into the plane and the loop is leaving to the right. Lenz's Law states that the induced current will run so as to support the B field that is reducing because the area of the loop in the field is going down. As a result, the current will run clockwise.

I = V/R which I leave to you to calculate.

d) To solve for total charge we need to multiply the current times the time for the loop to leave the field.

The time will be 10 cm/ 2 cm/s = 5 seconds.

Good luck!