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 = N₁  ( φ ₁/ i₁ )= (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 = N₂  ( φ ₁ / i₁  )= (400 × 0.5 × 10^{- 3}) /2= 0.1 H .