A loop with a length of 25 cm (the vertical dimension in the figure) and a width

Question

A loop with a length of 25 cm (the vertical dimension in the figure) and a width of 8 cm (the horizontal dimension in the figure) is moving (with a speed of v = 15 m/s) out of the magnetic field (with B = 5.78 T out of the page) as shown below. The total resistance of the loop is 0.04 ohms.

1. What is induced current in the loop?

a) 291 A, clockwise

b) 173 A, clockwise

c) 0 A

d) 173 A, counterclockwise

e) 291 A, counterclockwise

2. What is the magnetic force on the loop?

a) 80 N upwards

b) 50 N upwards

c) 0 N

d) 50 N downwards

e) 80 N downwards

3.  The same loop, oriented in the same manner as shown in the figure, is now pulled out of the field to the left. Compare the magnitude and direction of the induced current compared to above:

a) the current has a greater magnitude and is clockwise

b) the current has a greater magnitude and is counterclockwise

c) the current has a smaller magnitude and is clockwise

d) the current has a smaller magnitude and is counterclockwise

e) the current is the same (magnitude and direction) as above

Explanation / Answer

Here,

L = 25 cm = 0.25 m

B = 5.78 T

v = 15 m/s

w = 8 cm

R = 0.04 Ohm

1. induced current in the loop = L * B * v/R

induced current in the loop = 5.78 * 0.08 * 15/.04

induced current in the loop = 173 A

Using Lenz's law , the direction of current is clockwise

the induced current in the loop is 173 A clockwise.

2.

magnetic force on the loop = B *I*w

magnetic force on the loop = 5.78 * 173 * 0.08

magnetic force on the loop = 80 N

as magnetic force = i L X B

the magnetic force on the loop is 80 N upwards

3.
here,

as the magnetic is decresing in this case also ,

e) the current is the same (magnitude and direction) as above

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