Question
Download Solution PDFTwo 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:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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
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
Last updated on Jun 16, 2025
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