A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.28 T.The loop has an area of 0.22 m² and is mounted on an axis,perpendicular to the magnetic field,which allows the loop to rotate.If the plane of the loop is oriented parallel to the field,what torque is created by the interaction of the loop current and the field?

Question

Quizplus · Accepted Answer

The torque created is given by the equation τ = NIABsinθ, where N is the number of turns, I is the current, A is the area, B is the magnetic field strength, and θ is the angle between the normal to the loop and the magnetic field. In this case, since the loop is oriented parallel to the field, the angle θ is 0 and sinθ is also 0, which means the torque is 0. Therefore, we need to find the torque when the loop is oriented perpendicular to the field. Plugging in the given values, we get τ = (1.6 A)(1)(0.22 m^2)(0.28 T)sin90° = 0.10 N⋅m. Therefore, the best choice is C.

A Circular Loop Carrying a Current of 1