Above, a long straight wire carries a current of 12 A to the left Below is a rec

Question

Above, a long straight wire carries a current of 12 A to the left Below is a rectangular loop abcd which carries a current of 20 A clockwise. a) Find the magnetic force by the top wire on segment ab. Give magnitude and direction b) Find the magnetic force by the top wire on segment cd. Give magnitude and direction. c) Find the net magnetic force by the top wire on the segments ac and bd. Give Magnitude and direction. d) Do segments ab and cd exert attractive or repulsive forces on each other? What is the magnitude of this force e) What is the total force on the loop as a whole? Give the magnitude and direction.

Explanation / Answer

a)The magnetic force between segment ab and wire will be given by

F = u0 I1 I2 L/2 pi R

F = 4 pi x 10^-7 x 20 x 12 x 0.6 / (2pi x 0.2) = 1.44 x 10^-4 N

Since the wires are carrying current in the apposite direction, the force will be repulsive and downwards.

Hence, F = 1.44 x 10^-4 N ; downwards

b)The magnetic force between segment bc and wire will be given by

F = u0 I1 I2 L/2 pi R

R = 0.2 + 0.15 = 0.35 m

F = 4 pi x 10^-7 x 20 x 12 x 0.6 / (2pi x 0.35) = 0.822 x 10^-4 N

Since the wires are carrying current in the same direction, the force will be attractive and upwards

Hence, F = 1.37 x 10^-4 N ; upwards

c)The net magnetic force will be the diffrence of two as they are in the apposite direction

Fnet = (1.44 - 0.822) x 10^-4 = 6.18 x 10^-5 N

Since the force between ab and wire is stronger so the direction would be upwards.

Hence, Fnet = 6.18 x 10^-5 N ; upwards

d)The segments ab and cd repels each other, since the current is in apposite direction.

F = u0 I1 I2 L/2 pi R

R = 0.15 m

F = 4 pi x 10^-7 x 20 x 12 x 0.6 / (2pi x 0.15) = 1.92 x 10^-4

Hence, F = 1.92 x 10^-4 N

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