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Question

Find the mutual inductance between the circular coil and the loop shown in figure.

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Solution



The magnetic field due to coil 1 at the centre of coil 2 is given by
B=μ0 Nia22 (a2+x2)3/2

The flux linked with coil 2 is given by
ϕ=B.A'=μ0 Nia22 (a2+x2)3/2πa'2

Now, let y be the distance of the sliding contact from its left end.
Given:
v=dydt
Total resistance of the rheostat = R
When the distance of the sliding contact from the left end is y, the resistance of the rheostat is given by
r'=RLy

The current in the coil is the function of distance y travelled by the sliding contact of the rheostat. It is given by
i=ERLy+r

The magnitude of the emf induced can be calculated as:
e=dϕdt=μ0 Na2 a'2π 2 (a2+x2)3/2didt

e=μ0 N πa2 a'22 (a2+x2)3/2ddtERLy+re=μ0 N πa2 a'22 (a2+x2)3/2E-RLvRLy+r2

emf induced, e=μ0 N πa2 a'22 (a2+x2)3/2E-RLvRLy+r2
The emf induced in the coil can also be given as:
didt=E-RLvRLy+r2
e=Mdidt , didt=E-RLvRLy+r2M=edidt=Nμ0πa2a'22(a2+x2)3/2

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