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Question: A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a u...

A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines an angle of 30° with the normal of the coil that magnitude of the counter torque that must be applied to prevent the coil from turning is

A

33 N m

B

3.3 N m

C

3.3×10−2Nm3.3 \times 10 ^ { - 2 } \mathrm { Nm }

D

3.3×10−4Nm3.3 \times 10 ^ { - 4 } \mathrm { Nm }

Answer

3.3 N m

Explanation

Solution

N= 70

The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil

∴τ=NIABsin⁡θ=NIπr2 Bsin⁡30∘\therefore \tau = \mathrm { NIAB } \sin \theta = \mathrm { NI } \pi \mathrm { r } ^ { 2 } \mathrm {~B} \sin 30 ^ { \circ }

=70×8×3.14×(5×10−2)2×1.5×12=3.297Nm= 70 \times 8 \times 3.14 \times \left( 5 \times 10 ^ { - 2 } \right) ^ { 2 } \times 1.5 \times \frac { 1 } { 2 } = 3.297 \mathrm { Nm }

=3.3Nm= 3.3 \mathrm { Nm }