A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of $0.60 \mathrm {~T}$ with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?
A) 2.1 N ∙ m
B) 1.2 N ∙ m
C) 3.9 N ∙ m
D) 5.4 N ∙ m
E) 1.0 N ∙ m

Question

Quizplus · Accepted Answer

The magnetic torque on a current 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 problem, we have:* N = 200 turns* I = 7.0 A* A = πr² = π(0.03 m)² = 2.83 × 10⁻³ m²* B = 0.60 T* θ = 30°Plugging these values into the equation, we get:$$	au = (200)(7.0 A)(2.83 × 10⁻³ m²)(0.60 T)sin 30° = 2.1 N ∙ m$$Therefore, the magnitude of the magnetic torque on the coil is 2.1 N ∙ m.

A Flat Circular Coil of Wire Having 200 Turns and Diameter