A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
A) 0.15 N ∙ m
B) 0.088 N ∙ m
C) 0.29 N ∙ m
D) 0.40 N ∙ m
E) 0.076 N ∙ m

Question

Quizplus · Accepted Answer

The magnetic torque on a current loop is given by the equation:$$	au = NIAB\sin	heta$$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 problem, we have:* N = 200 turns* I = 4.0 A* A = πr² = π(0.01 m)² = 3.14 x 10^-4 m²* B = 0.70 T* θ = 30°Plugging these values into the equation, we get:$$	au = (200)(4.0 A)(3.14 x 10^-4 m²)(0.70 T)\sin 30° = 0.15 N ∙ m$$Therefore, the correct answer is A.

A Circular Coil of Wire of 200 Turns and Diameter