Question

8. 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) EMF induced in the second coil, when current in the first coil changes at a rate of 100 A/Sec.​

Answer 1

Explanation:

The magnetic flux through an inductor is the self inductance of coil times the current through it.

Thus ϕ=LI

=50×10

−3

H×8×10

−3

A

=4×10

−4

Wb

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