A conducting loop is made in the form of two squares of sides s1 = 2cm and s2 =

Question

A conducting loop is made in the form of two squares of sides s1 = 2cm and s2 = 5.5 cm as shown. At time t = 0, the loop enters a region of length L = 14.5 cm that contains a uniform magnetic field B = 1.4 T, directed in the positive z-direction. The loop continues through the region with constant speed v = 48 cm/s. The resistance of the loop is R = 1.6 .

1) At time t = t1 = 0.014 s, what is I1, the induced current in the loop? I1 is defined to be positive if it is in the counterclockwise direction.

2)At time t = t2 = 0.382 s, what is I2, the induced current in the loop? I2 is defined to be positive if it is in the counterclockwise direction.

3) What is Fx(t2), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 48 cm/s at t = t2 = 0.382 s?

4) At time t = t3 = 0.316 s, what is I3, the induced current in the loop? I3 is defined to be positive if it is in the counterclockwise direction.

5) Consider the two cases shown above. How does II, the magnitude of the induced current in Case I, compare to III, the magnitude of the induced current in Case II? Assume s2 = 3s1.

A) II < III

B) II = III

C) II > III

Explanation / Answer

Here,

s1 = 2 cm = 0.02 cm

s2 = 5.5 cm = 0.055 m

L = 14.5 cm

B = 1.4 T

v = 48 cm/s

R = 1.6 Ohm

1)

at t = 0.014 s

distance moved by the loop = 0.014 * 48 = 0.672 cm

hence , the induced current = B * s1 * v/R

induced current = 1.4 * 0.48 * 0.02/1.6

induced current = 8.4 *10^-3 A

2)

at t = 0.382 s

distance moved by the loop = 0.382 * 48 = 18.336 cm

as the s2 part is moving out of the magnetic field area

hence , the induced current = B * s2 * v/R

induced current = 1.4 * 0.48 * 0.055/1.6

induced current = 23.1 *10^-3 A

3)

at the time t = 0.382 s

force applied to keep constant motion = B * I * L

force applied to keep constant motion = 1.4 * 23.1 *10^-3 * 0.055

force applied to keep constant motion = 1.779 * 10^-3 N

4)

at t = 0.316 s

distance moved by the loop = 0.316 * 48 = 15.2 cm

as the s2 part is moving out of the magnetic field area

hence , the induced current = B * (s2 - s1) * v/R

induced current = 1.4 * 0.48 * (0.055 - 0.02)/1.6

induced current = 0.0147 A

the induced current in the loop is 0.0147 A

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