Question

A rectangular coil of length 0.12 m and width 0.1 m having 50
turns of wire is suspended vertically in a uniform magnetic field
of strength 0.2 weber/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 : [2015 RS]
(a) 0.20 Nm (b) 0.24 Nm
(c) 0.12 Nm (d) 0.15 Nm

Answer 1

Answer:

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

Explanation:

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