Mutual Induction : Two coils C_{1} and C_{2} are placed very close to each other. A source of emf is connected in the coil C_{1}.

When current flows through the coil C_{1} magnetic field is produced . The coil C_{2} lies within the magnetic field of C_{1} and hence magnetic flux is associated with it. So long current changes in the coil C_{1} its magnetic field changes and hence flux associated with C_{2} also changes , which induces an emf in the coil . Induction of an emf due to variation of current in the neighbouring coil is known as Mutual Induction .

Flux associated with C_{2} depends on current in C_{1} and number of turns in C_{2}.

φ_{(C2)} = M I_{C1} , where M is the mutual inductance of the given pair of coils.

$ \displaystyle \xi = -\frac{d\phi}{dt} $

$ \displaystyle \xi_{C_2} = -M \frac{d I_{C_1}}{dt} $

SI unit of mutual inductance is henry.

Example : What is the self inductance of a system of co-axial cables carrying current in opposite directions as shown. Their radii are ‘ a ‘ and ‘ b ‘ respectively.

Solution : The ‘ B ‘ between the space of the cables is

$ \displaystyle B = \frac{\mu_0 I}{2 \pi r } $

The Ampere’s law tells that ‘ B ‘ outside the cables is zero, as the net current through the amperian loop would be zero. Taking an element of length l and thickness ‘ dr ‘ ; dφ through it is

$ \displaystyle d\phi = \frac{\mu_0 I}{2 \pi r} l dr $

$ \displaystyle \phi = \frac{\mu_0 I l}{2 \pi } \int_{a}^{b} \frac{1}{r} dr $

$ \displaystyle \phi = \frac{\mu_0 I l}{2 \pi } ln\frac{b}{a} $