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A rectangular coil of length 0.12 m and...

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

A

0.20 Nm

B

0.24 Nm

C

0.12 Nm

D

0.15 Nm

Text Solution

Verified by Experts

The correct Answer is:
A
`tau = MB sin theta [as theta = 90^(@) - 30^(@) = 60^(@)]`
= NIAB sin `60^(@) = 50 xx 2 xx 0.12 xx 0.1 xx 0.2 xx (sqrt3)/(2)`
`= 0.20` Nm
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