Question

A counterclockwise current of 1.2 mA runs through a 100 turn coil. An external magnetic field of magnitude of 0.15 T is directed to the right as shown in the figure. If the radius of the coil is 4.0m what is the magnitude of the torque on the coil? a) zero N * m b) 0.90 N *m c) 0.23 N * m d) 0.45 N * m e) 3.6 N * m

Answer 1

Solution:

Using the formula for torque

=> torque = N * I * B * A sin(theta)

=> torque = 100 * 1.2 * 10^-3 , 0.15 * pi * 4^2 * sin(90) ( taking angle 90 as the figure is missing)

=> torque = 0.90 Nm

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