A circular loop carrying a current of 1.2 A is oriented in a magnetic field of 0.23 T.The loop has an area of 0.16 m² and is mounted on an axis,perpendicular to the magnetic field,which allows the loop to rotate.What is the torque on the loop when its plane is oriented at a 34° angle to the field?

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 the loop* A is the area of the loop* B is the magnetic field strength* θ is the angle between the normal to the loop and the magnetic fieldIn this problem, we have:* N = 1* I = 1.2 A* A = 0.16 m^2* B = 0.23 T* θ = 34°Plugging these values into the equation, we get:$$	au = (1)(1.2 A)(0.16 m^2)(0.23 T)\sin(34°) = 0.037 N⋅m$$Therefore, the correct answer is A.

A Circular Loop Carrying a Current of 1