A circular loop carrying a current of 1.25 A is oriented in a magnetic field of 0.22 T.The loop has an area of 0.10 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? A)0.018 N⋅m B)2.8 N⋅m C)0.028 N⋅m D)0.014 N⋅m E)57 N⋅m

Question

Quizplus · Accepted Answer

The torque, τ, created by the interaction of the loop current and the magnetic field is given by the formula:

τ = NIA × B

where N is the number of turns in the loop (which is 1 in this case), I is the current in the loop, A is the area of the loop, B is the magnetic field, and × denotes the vector cross product.

When the loop is oriented parallel to the magnetic field, the angle between the normal to the loop (i.e. the axis of rotation) and the magnetic field is 0 degrees, so the torque is given by:

τ = NIA × B × sin(0°) = 0

Therefore, the correct answer is:

Answer: C
 When the loop is oriented parallel to the magnetic field, the torque created by the interaction of the loop current and the field is zero.

A Circular Loop Carrying a Current of 1