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Physics Question on torque

A closely wound solenoid of 20002000 turns and area of cross-section 1.5×10−4m21.5 \times 10^{-4} m ^{2} carries a current of 2.0A2.0 A. It is suspended through its center and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5×10−2T5 \times 10^{-2} T, making an angle of 30∘30^{\circ} with the axis of the solenoid. The torque on the solenoid will be

A

3×10−3N−m3 \times 10^{-3} N - m

B

1.5×10−3N−m1.5 \times 10^{-3} N - m

C

1.5×10−2N−m1.5 \times 10^{-2} N - m

D

3×10−2N−m3 \times 10^{-2} N - m

Answer

1.5×10−2N−m1.5 \times 10^{-2} N - m

Explanation

Solution

τ=NiBAsin⁡θ\tau =N i B A \sin \theta =2000×2×5×10−2×1.5×10−4×sin⁡30∘=2000 \times 2 \times 5 \times 10^{-2} \times 1.5 \times 10^{-4} \times \sin 30^{\circ} =2000×50×10−6×12=2000 \times 50 \times 10^{-6} \times \frac{1}{2} =1.5×10−2N−m=1.5 \times 10^{-2} N - m