A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A.If it is lying flat on the ground, and the Earth's magnetic field points due north, has a magnitude of 5.8 x 10^-5 T, and makes a downward angle of 25° with the vertical, what is the torque on the loop? A)1.7 x 10^-2 N·m west B)3.6 x 10^-2 N·m west C)1.7 x 10^-2 N·m east D)3.6 x 10^-2 N·m east E)3.6 x 10^-2 N·m south

Question

Quizplus · Accepted Answer

The torque on the loop is given by:$$	au = NIAB\sin	heta$$where:* N is the number of turns in the coil* I is the current in the coil* A is the area of the coil* B is the magnetic field strength* θ is the angle between the magnetic field and the normal to the coilIn this case, we have:* N = 1000 turns* I = 15 A* A = πr^2 = π(0.12 m)^2 = 0.0452 m^2* B = 5.8 x 10^-5 T* θ = 25°Plugging these values into the equation, we get:$$	au = (1000)(15 A)(0.0452 m^2)(5.8 x 10^-5 T)\sin 25° = 1.7 x 10^-2 N·m$$The torque is in the west direction, so the correct answer is A.

A Coil of 1000 Turns of Wire Has a Radius