# Self and Mutual Inductance

## Self induction

Self-induction occurs when a changing current in a coil produces an opposing EMF in the same conductor.

When current flows through a coil, it creates a magnetic field around it. If the current in the coil changes, the magnetic field also changes. This change in the magnetic field induces an opposing EMF in the coil itself, which resists the change in current. This opposing EMF is called the self-induced EMF or the back EMF.

The magnetic flux through a coil is proportional to the current through it. The constant of proportionality is called self inductance $L$ i.e., \begin{align} \phi=LI. \end{align}

If current through the coil varies with time then an emf is induced in it. Faraday's law gives induced emf as \begin{align} \phi=-\frac{\mathrm{d}\phi}{\mathrm{d}t}=-L\frac{\mathrm{d}I}{\mathrm{d}t}. \end{align}

The self inductance of a solenoid of radius $r$, length $l$ and number of turns per unit length $n$ is given by \begin{align} L=\mu_0 n^2(\pi r^2 l). \end{align}

The energy stored in an inductor of inductance $L$ when a current $I$ flows through it is given by \begin{align} U=\frac{1}{2}LI^2. \end{align}

## Mutual induction

Let two coils, 1 and 2, carrying currents $I_1$ and $I_2$ are placed close to each other. The magnetic flux through the coil 2 is proportional to the current flowing through the coil 1 i.e., \begin{align} \phi_2=M_{21}I_1. \end{align}

Similarly, the magnetic flux through the coil 1 is proportional to the current flowing through the coil 2 i.e., \begin{align} \phi_1=M_{12}I_2. \end{align} It can be proved that $M_{12}=M_{21}$. Thus, flux through a coil is proportional to the current flowing through the other coil \begin{align} \phi =MI. \end{align} The constant of proportionality $M$ is called mutual inductance of the coils 1 and 2.

If current through one coil varies with time then an emf is induced in the other coil. The induced emf is given by Faraday's law, \begin{align} e=-\frac{\mathrm{d}\phi}{\mathrm{d}t}=-M\frac{\mathrm{d}I}{\mathrm{d}t}. \end{align}

## Problems from IIT JEE

Problem (JEE Mains 2019): A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of~3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil

1. decrease by a factor of 9
2. increase by a factor of 27
3. increase by a factor of 3
4. decrease by a factor of$9\sqrt{3}$

Solution: The wooden frame is made by joining three wooden logs that form three sides of an equilateral triangle. We assume these logs to be cylindrical with length $l$ and radius $a$. The copper wire is wound over these logs.

The wooden frame with copper winding is equivalent to three solenoids connected in series. The self-inductance of each solenoid is (see problem \OPBhyperlink{lre} on page~\pageref{lre}), \begin{align} L=\mu_0\pi n^2 a^2 l.\nonumber \end{align} The flux through each solenoid is $Li$ when a current $i$ flow through it. The total flux through the wooden frame (three solenoids in series) is $\phi=L_\text{eq} i=3Li$. Thus, self-inductance of the wooden frame is, \begin{align} L_\text{eq} &=3L \\ &=3\mu_0\pi n^2 a^2 l.\nonumber \end{align} Increasing the linear dimension of each side by a factor of three means each side is increased from $l$ to $3l$. The logs are three times longer but are of the same radius $a$. The number of turns per unit length $n$ remains the same. Thus, the self-inductance of the wooden frame $L_\text{eq}$ increases by a factor of three. The self-inductance will increase by a factor of 27 if logs are three times longer and three times thick.

Problem (IIT JEE 2001): Two circular coils can be arranged in any of the three situations shown in figure. Their mutual inductance will be

1. maximum is situation (i).
2. maximum is situation (ii).
3. maximum is situation (iii).
4. the same in all situations.

Solution: The mutual inductance $M$ is given by, $\phi=Mi$, where $\phi$ is the flux through a coil due to the current $i$ in another coil. The flux through an area $\vec{A}$ due to a magnetic field $\vec{B}$ is given by \begin{align} \phi=\vec{B}\cdot\vec{A}. \end{align} The $\vec{B}$ and $\vec{A}$ are parallel in configuration (i) but perpendicular in configuration (ii) and configuration (iii). Hence, the flux and mutual inductance are maximum in configuration (i).

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