A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T.The loop has an area of 0.14 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 21° angle to the field? A)2.4E-2 N⋅m B)6.3E-2 N⋅m C)3.7E-2 N⋅m D)5.6E-2 N⋅m E)9.4E-1 N⋅m

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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 = 1.6 A* A = 0.14 m^2* B = 0.30 T* θ = 21°Plugging these values into the equation, we get:$$	au = (1)(1.6 A)(0.14 m^2)(0.30 T)\sin(21°) = 2.4	imes10^{-2} N\cdot m$$Therefore, the correct answer is A.

A Circular Loop Carrying a Current of 1