A coil having 100 turns, area of 5 10-3 m2, 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 .
Answer
100
Solution
M = NIA = 100 10-3 5 10-3 = 5 10-4 A m2
plane ⊥ B M B, so : 0 90
W = MB( 0 - 90 ) = MB
= 5 10-4 0.20 = 10-4 J = 100
Answer: 100