Question:

A 250-Turn rectangular coil of length 2.1cm and width 1.25cm carries a current of 85µA and subjected to a magnetic field of strength 0.85T. Work done for rotating the coil by 180° against the torque is:

9.1µJ

1.15µJ

4.55µJ

23µJ

τ=BIA
Hence work done=∫τdθ
BIA∫₀^π sinθdθ=-BIA[cosθ]₀^π=2BIA
=2×0.85×85×10⁻⁶×250×(2.1×1.25×10⁻⁴)
=94828.125×10⁻¹⁰
=9.48×10⁻⁶