Question

A square current loop 5.5 cm on each side carries a 550 mA current. The loop is in a 1.1 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction. Part A What is the magnitude of the torque on the current loop?

Answer 1

Answer:

$\tau =0.000915\ Nm$

Explanation:

given,                        

side of loop = 5.5 cm = 0.055 m

current on each side of loop =  550 m A = 0.55 A

magnetic field on the loop = 1.1 T

axis of the loop make an angle of = 30°

torque = ?

$\tau = i \times A\times B \times sin \theta$

$\tau = 0.55 \times 0.055^2\times 1.1 \times sin 30^0$

$\tau =0.000915\ Nm$

hence, the torque on the current loop is $\tau =0.000915\ Nm$