A rectangular 30-turn coil of wire,with width of 12.0 cm along the y-axis and height of 25.0 cm along the z-axis,is placed in a uniform external magnetic field with an x-component of 0.0951 T and a y-component of 0.182 T.The coil is hinged along the y-axis and is located in the yz-plane.A current of 0.350 A flows through the wire.What is the magnitude of the torque that B exerts on the coil?

Question

Quizplus · Accepted Answer

The torque on a current-carrying loop in a magnetic field can be calculated as τ = IABsinθ, where I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the normal to the loop and the direction of the magnetic field. In this case, the loop is rectangular with dimensions 12.0 cm and 25.0 cm, so the area is A = (12.0 cm)(25.0 cm) = 300 cm^2 = 0.03 m^2. The magnetic field has x- and y-components given, so the total magnetic field B is B = (0.0951^2 + 0.182^2)^0.5 = 0.206 T. The angle between the normal to the loop and the direction of the magnetic field is 90 degrees, so sinθ = 1. Finally, the current is given as 0.350 A. Plugging in these values, we get τ = (0.350 A)(0.03 m^2)(0.206 T)(1) = 0.0573 Nm, which corresponds to choice C.

A Rectangular 30-Turn Coil of Wire,with Width of 12