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Question: The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and ca...

The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is

A

1.6 N m

B

1.2 N m

C

1.4 N m

D

1.8 N m

Answer

1.6 N m

Explanation

Solution

τ=NIABsin⁡θ\tau = \mathrm { NIAB } \sin \theta

Here

∴τ=25×10×144×10−4×0.9×sin⁡30∘=1.6Nm\therefore \tau = 25 \times 10 \times 144 \times 10 ^ { - 4 } \times 0.9 \times \sin 30 ^ { \circ } = 1.6 \mathrm { Nm }