Question

Two coils placed near to each other have number of turns equal to 600 and 300 respectively. On passing a current of 3.0 A through coil A, the flux linked with each turn of coil A is 1.2*10⁻⁴ Wb and the total flux linked with coil B is 9.0*10⁻⁵ Wb. Find (1) self-inductance of A, (2) The mutual inductance of the system formed by A and B. [Ans : La= 24mH; Mb=30 μH]

Answer 1

The self inductance of coil A = 24mH and the mutual inductance of both the coil = 30μH

Explanation:

1) For self inductance of coil A, we have

NФ = LI

where , N = number of turns in A = 600

Ф = flux linked in each coil = 1.2 x 10⁻⁴ Wb

L = self inductance

I = current in coil A = 3A

Putting these value in the eqn we get

600 x 1.2 x 10⁻⁴ = L x 3

=> L = 2.4 x 10⁻² H = 24 mH

2) For coil B total flux linked by the current in coil A = 9 x 10⁻⁵ Wb = NФ

Hence ,

NФ = MIₐ

where M = mutual inductance between both the coils,

Iₐ = current in coil A = 3 A

=> 9 x 10⁻⁵ = M x 3

=> M = 3 x 10⁻⁵ H = 30μH