Question

The triangular coil of wire in the drawing is free to rotate about an axis that is attached along side AC. The current in the loop is 5.25 A, and the magnetic field (parallel to the plane of the loop and side AB) is B = 1.6 T. (a) What is the magnetic moment of the loop, and (b) what is the magnitude of the net torque exerted on the loop by the magnetic field?

Answer 1

Step-by-step explanation:

The magnetic moment of a current loop is simply the current times the area of the loop. In this case...

height of triangle = 2.00*tan(55) = 2.856 m

Area=0.5*base*height

=0.5*2*2.856

=2.856 m²

a)

magnetic moment = area * current = 2.856 * 5.26 = 14.994 A.m²

b)

then the torque is simply the B field times the magnetic moment

torque = B * mag moment = 1.6 * 15.022 = 24.03 N-m