Could you please derive the mutual induction and self induction for a conductor?

Let us deduce the same for conductor in the form of a solenoid:

**Self-Inductance of a Long Solenoid**

A long solenoid is one whose length is very large as compared to its radius of cross-section. The magnetic field *B* at any point inside such a solenoid is practically constant and is given by,

Where,

μ_{0} = Absolute magnetic permeability of free space

*N* = Total number of turns in the solenoid

∴ Magnetic flux through each turn of the solenoid coil = *B* × Area of each turn

Where,

*A* = Area of each turn of the solenoid

Total magnetic flux linked with the solenoid

If *L* is coefficient of self-inductance of the solenoid, then

∴Φ = *LI* … (iii)

Equating (ii) and (iii),

If core is of any other magnetic material, μ_{0} is replaced by

**Mutual Induction**

The phenomenon according to which an opposing *emf* is produced in a coil as a result of change in current, hence, the magnetic flux linked with a neighbouring coil is called mutual induction.

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)

Where, *M* is called coefficient of mutual induction

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

**Mutual Inductance of Two Long Solenoids**

_{}

Consider two long solenoids *S*_{1} and *S*_{2} of same length *l*, such that solenoid *S*_{2} surrounds solenoid *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}

Φ_{21} − 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 solenoids

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

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

*B*_{1} = μ_{0}*n*_{1}*I*_{1}

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

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

Therefore, total magnetic flux linked with the solenoid 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}*AI*_{1}

∴ *M*_{21} = μ_{0}*n*_{1}*n*_{2}*Al*

Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid *S*_{2} and induced *emf* is produced in solenoid *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 long solenoids

**
**