The self inductance of a coil of 500 turns is 0.25 H. If 60% of the flux of the coil is linked with a second coil of 10000 turns. Calculate mutual inductance of the two coils and the emf induced in the second coil when current in the first coil changes at a rate of 100 A/sec.

 The self inductance of a coil of  500 turns is 0.25 H. If 60% of the flux of the coil is linked with a second coil of 10000 turns. Calculate 

i) mutual inductance of the two coils 

ii) the emf induced in the second coil when current in the first coil changes at a rate of 100 A/sec.

solution:

Let, current in first coil be    I1   

i) Mutual inductance between the two coils

    M= N2 φ2/I1

    Here, N1=500,    N2=10000

   Self inductance of coil one, 

    L = 0.25

=>N1 φ1/I1   = 0.25

=>φ1 = (0.25) I1/500 wb

So, electric flux in the second coil

    φ2= 60/100 × φ1 

          =0.6  ×  (0.25) I1/500 

          = 3 × 10^(-4) I1 wb

So, mutual inductance,

M= N2 φ2/I1

      = (1000 ×  3 ×  10^(-4) ×  I1) / I1

    = 3 H

ii) 

Given, rate of change of current in the first coil, d I1/dt = 100 A/s

e.m.f. induced in the second coil

E2=- M d I1/dt

|E2|= 3×  100

    =300 V