Question

A square current loop 4.9 cm on each side carries a 480 mA current. The loop is in a 0.60 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30° away from the field direction. What is the magnitude of the torque on the current loop? Express your answer in newton-meters.

Answer 1

Answer: 0.000346 Nm

Step-by-step explanation:

T = u X B

u = i x A = magnetic moment

T = i x A x B x sin(30)

T = 0..48 x 0.049^2 x 0.6 x 0.5 = 0.000346 Nm