A coil having 100 turns, area of 5×10−3m2, carrying current of 1 mA is placed in uniform magnetic field of 0.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90∘ is ______ μJ.
Answer: 100
Solution
The magnetic moment is μ=NIA=100×5×10−3×1×10−3=5×10−4 A m2. The work done in rotating the coil from θ=0 (plane ⊥B, i.e. μ∥B) to θ=90∘ equals the change in potential energy:
