A circular loop of wire of cross-sectional area 0.12 m² consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic field of 0.050 T oriented at 30° to the plane of the loop. What torque acts on the loop?

Question

Quizplus · Accepted Answer

The torque on a current loop in a magnetic field is given by:$$	au = NIAB\sin	heta$$where:* N is the number of turns in the loop* I is the current in each turn* A is the area of the loop* B is the magnetic field strength* θ is the angle between the magnetic field and the normal to the loopIn this problem, we have:* N = 200 turns* I = 0.50 A* A = 0.12 m^2* B = 0.050 T* θ = 30°Plugging these values into the equation, we get:$$	au = (200)(0.50)(0.12)(0.050)\sin(30°) = 0.52 	ext{ m}\cdot	ext{N}$$Therefore, the correct answer is B) 0.52 m∙N.

A Circular Loop of Wire of Cross-Sectional Area 0