Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:

• A. 0.4 mWb
• B. 0.004 mWb
• C. 4 mWb
• D. 0.04 mWb

Answer: (D)

Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA
No. of turns NA = NB = 1000
Flux linkage in a coil with B = Flux linkage in coil A × 80/100
= 0.8 × 5 × 10−5
= 4 × 10−5 wb =0.04 mwb