Question

$\large\green{QUESTION:-}$

A rectangular coil of length 0.12m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2Weber/m2. The coil carries a current of 2A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be:​

Answer 1

Correct option is C)

The torque on a loop in uniform magnetic field is given by

τ= M × B

τ=MBsinθ [where θ is angle between normal and the plane of loop]

As the coil is inclined at an angle of 30

o

with magnetic field so area vector will be inclined at an angle of 60

o

with the field.

M=NI×πr 2 =50×2×.12×.1=1.2

B=.2T

τ =1.2 ×.2×sin60°

τ=0.2Nm