Determine self inductance of coil:
A flux of 0.5 m Wb is generated by coil A of 600 turns wound on a ring with a current of 2 A in it. Determine (a) self inductance of coil A, (b) the e.m.f induced in coil A when a current of 6 A passing through it is switched off, supposing the current to fall to zero in 2 millisecond, and (c) the mutual inductance between the coils, if a second coil B of 400 turns is uniformly wound over the first coil A.
Solution
(a) Self inductance of coil A = N_{1} ( φ _{1}/ i_{1} )= (600 × 0.5 × 10^{- 3 })/2= 0.15 H
(b) e = L (di / dt) = 0.15 × (6 - 0) /2 × 10^{- 3} = 450 V
(c) M = N_{2} ( φ_{ 1} / i_{1} )= (400 × 0.5 × 10^{- 3}) /2= 0.1 H .