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Physics Question on Moving charges and magnetism

A rectangular coil, of sides 2 cm2 \,cm and 3 cm3\, cm respectively, has 1010 turns in it. It carries a current of 1 A1 \,A, and is placed in a uniform magnetic field of 0.2 T0.2 \,T in such a manner that its plane makes an angle 60∘60^{\circ} with the field direction. The torque on the loop is

A

6.0×10−4Nm6.0 \times 10^{-4} Nm

B

6.0×10−5Nm6.0 \times 10^{-5} Nm

C

1.2×10−3Nm1.2 \times 10^{-3} Nm

D

6.0 Nm6.0\, Nm

Answer

6.0 Nm6.0\, Nm

Explanation

Solution

T=NIABsin⁡θ,θ=60T = NI A B \sin \theta, \theta=60
T=10×1×(2×10−2×3×10−2)×0.2×32T =10 \times 1 \times\left(2 \times 10^{-2} \times 3 \times 10^{-2}\right) \times 0.2 \times \frac{\sqrt{3}}{2}
10×6×10−4×3×10−110 \times 6 \times 10^{-4} \times \sqrt{3} \times 10^{-1}
63×10−4N−M.6 \sqrt{3} \times 10^{-4} N - M .