#### Question

Consider two concentric circular coils, one of radius r_{1} and the other of radius r_{2} (r_{1} < r_{2}) placed coaxially with centers coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.

#### Solution

Coefficient of mutual induction − consider two coils *P *and *S*. Suppose that a current *I* is flowing through the coil* P *at any instant i.e.,

Φ ∝ *I*

Φ = *MI*… (i)

If ‘*e*’ is the induced *emf* produced in the *S*-coil, then

`e=(dphi)/dt=-d/dt(MI)=-M(dl)/dt`

Mutual Inductance of two concentric coils, one of radius *r*_{1} and the other of radius *r*_{2} (*r*_{1} < *r*_{2}) placed coaxially with centers coinciding with each other:

Consider two circular coil *S*_{1} and *S*_{2} of same length *l*, such that coil *S*_{2} surrounds coil *S*_{1} completely

Let

*n*_{1} − Number of turns per unit length of *S*_{1}

*n*2 − Number of turns per unit length of *S*2

*I*_{1} − Current passed through solenoid *S*_{1}

Φ21 − Flux linked with *S*2 due to current flowing through *S*1

Φ_{21} ∝ *I*_{1}

Φ_{21} = *M*_{21}*I*_{1}

Where *M*_{21} is the coefficient of mutual induction of the two coils

When current is passed through *S*_{1}, an *emf* is induced in *S*_{2}.

Magnetic field produced inside *S*_{1} on passing current through it,

*B*_{1} = μ_{0}*n*_{1}*I*_{1}

Magnetic flux linked with each turn of *S*_{2 }will be equal to *B*_{1} times the area of the cross-section of *S*_{1}.

Magnetic flux linked with each turn of the *S*_{2} = *B*_{1}*A*

*Therefore, total magnetic flux linked with the S _{2},*

*Φ _{21} = B_{1}A × n_{2}l = μ_{0}n_{1}I_{1} × A× n_{2}l*

*Φ _{21} = μ_{0}n_{1}n_{2}AlI_{1}*

*∴ M _{21} = μ_{0}n_{1}n_{2}Al*

*Similarly, the mutual inductance between the two coils, when current is passed through coil S _{2} and induced emf is produced in coil S_{1}, is given by*

*M _{12} = μ_{0}n_{1}n_{2}Al*

*∴M _{12} = M_{21} = M (say)*

Hence, coefficient of mutual induction between the two coil will be

`M=mu_0n_1n_2Al` |