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

Concept:

Consider two coils having self-inductance L1 and L2 placed very close to each other. Let the number of turns of the two coils be N1 and N2 respectively. Let coil A carries current I1 and coil B carries current I2.

Due to current I1, the flux produced is ϕ1 which links with both the coils. Then the mutual inductance between two coils can be written as

\(M = \frac{{{N_1}{ϕ _{12}}}}{{{I_1}}}\)

Here, ϕ12 is the part of the flux ϕ1 linking with the coil 2

Calculation:

Flux produced in coil X (ϕ1) = 0.05 mWb

As we are just required to find the flux linked with the second coil, and we are given that 80% of the flux produced by one coil links with the other. 

∴ Flux linked with Y (ϕ12) = 80% of flux produced in coil 1 

= 0.05 × 0.8 mWb

0.04 mWb