Two identical coils A and B each having 800 turns lie in parallel planes such that 60% of the flux produced by one coil links with the other. A current of 5 A in coil A produces a flux of 150 µ wb. The current in coil A changes from 20 A to -20A in 20 ms. Calculate i) self inductance of each coil ii) the mutual inductance iii) the voltage induced in coil B.

 Two identical coils A and B each having 800 turns lie in parallel planes such that 60% of the flux produced by one coil links with the other. A current of 5 A in coil A produces a flux of 150 µ wb. The current in coil A changes from 20 A to -20A in 20 ms. Calculate

i) self inductance of each coil

ii) the mutual inductance

iii) the voltage induced in coil B.

Solution:

Given,

N1=N2= 800 turns (no of turns)

Current in first coil, I1= 5A

Flux in first coil, φ₁= 150 µ wb

Now,

d I1/d t

=| (-20-20)/20×10^-3|

=2000 A/ sec

Flux in second coil, 

φ₂= 60/100 ×150 µ wb

    =90 µ wb

i) 

Self inductance of each coil

L= N₁φ1/I₁

    =800×150×10^-6/5

    =5 mH

ii)

mutual inductance, M

= N₂φ2/I₁

=(800×90×10^-6)/5

=14.4 mH

iii)

Voltage induced in coil B, E2

= |-M dI1/dt |

=14.4×10^-3×2000

=28.8 V