6.9 Inductance

An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. These two situations are described separately in the next two sub-sections. However, in both the cases, the flux through a coil is proportional to the current. That is, ${\mathrm{\Phi }}_B\propto I$.

Further, if the geometry of the coil does not vary with time then,

$$\frac{\mathrm{d}{\mathrm{\Phi }}_B}{\mathrm{d}t}\propto \frac{\mathrm{d}I}{\mathrm{d}t}$$

For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. When the flux ${\mathrm{\Phi }}_{\mathrm{B}}$ through the coil changes, each turn contributes to the induced emf. Therefore, a term called flux linkage is used which is equal to $N{\mathrm{\Phi }}_{\mathrm{B}}$ for a closely wound coil and in such a case

$$N{\mathrm{\Phi }}_{\mathrm{B}}\propto I$$

he constant of proportionality, in this relation, is called inductance. We shall see that inductance depends only on the geometry of the coil and intrinsic material properties. This aspect is akin to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the dielectric constant K of the intervening medium (intrinsic material property).

Inductance is a scalar quantity. It has the dimensions of $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}{\mathrm{A}}^{-2}\right]$ given by the dimensions of flux divided by the dimensions of current. The SI unit of inductance is henry and is denoted by H. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA, independently of Faraday in England.

6.9.1 Mutual inductance

Consider Fig. 6.15 which shows two long co-axial solenoids each of length l. We denote the radius of the inner solenoid $S_1$ by $r_1$ and the number of turns per unit length by $n_1$ . The corresponding quantities for the outer solenoid $S_2$ are $r_2$ and $n_2$ , respectively. Let $N_1$ and $N_2$ be the total number of turns of coils $S_1$ and $S_2$ , respectively.

When a current $I_2$is set up through $S_2$ , it in turn sets up a magnetic flux through $S_1$ . Let us denote it by${\mathrm{\Phi }}_1$ . The corresponding flux linkage with solenoid $S_1$ is

$$N_1{\mathrm{\Phi }}_1=M_{12}{I}_2$$

$M_{12}$ is called the mutual inductance of solenoid $S_1$ with respect to solenoid $S_2$ . It is also referred to as the coefficient of mutual induction.

For these simple co-axial solenoids it is possible to calculate $M_{12}$ . The magnetic field due to the current $I_2$ in ${\mathrm{S}}_2$ is ${\mu }_0{n}_2{I}_2$ . The resulting flux linkage with coil $S_1$ is,

$$N_1{\mathrm{\Phi }}_1=\left(n_{1}l\right)\left(\pi r_1^2\right)\left({\mu }_0{n}_2{I}_2\right)$$ $$={\mu }_0{n}_1{n}_2\pi r_1^{2}l{I}_2$$

where $n_{1}l$ is the total number of turns in solenoid $S_1$ . Thus, from Eq. (6.9) and Eq. (6.10),

$$M_{12}={\mu }_0{n}_1{n}_2\pi r_1^{2}l$$

Note that we neglected the edge effects and considered the magnetic field ${\mu }_0{n}_2{I}_2$ to be uniform throughout the length and width of the solenoid $S_2$ . This is a good approximation keeping in mind that the solenoid is long, implying l >> $r_2$

We now consider the reverse case. A current $I_1$ is passed through the solenoid $S_1$ and the flux linkage with coil $S_2$ is,

$$N_2{\mathrm{\Phi }}_2=M_{21}{I}_1$$

$M_{21}$ is called the mutual inductance of solenoid $S_2$ with respect to solenoid $S_1$ .

The flux due to the current $I_1$ in $S_1$ can be assumed to be confined solely inside $S_1$ since the solenoids are very long. Thus, flux linkage with solenoid $S_2$ is

$$N_2{\mathrm{\Phi }}_2=\left(n_{2}l\right)\left(\pi r_1^2\right)\left({\mu }_0{n}_1{I}_1\right)$$

where $n_{2}l$ is the total number of turns of $S_2$ . From Eq. (6.12),

$$M_{21}={\mu }_0{n}_1{n}_2\pi r_1^{2}l$$ Using Eq. (6.11) and Eq. (6.12), we get $$M_{12}=M_{21}=M\text{ (say) }$$

We have demonstrated this equality for long co-axial solenoids. However, the relation is far more general. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage $N_1{\mathrm{\Phi }}_1$ because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. In this case, the calculation of $M_{12}$ would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. Therefore, the calculation of $M_{21}$ would also be extremely difficult in this case. The equality $M_{12}$ =$M_{21}$ is very useful in such situations.

We explained the above example with air as the medium within the solenoids. Instead, if a medium of relative permeability µ r had been present, the mutual inductance would be

$$M={\mu }_r{\mu }_0{n}_1{n}_2\pi r_1^{2}L$$

It is also important to know that the mutual inductance of a pair of coils, solenoids, etc., depends on their separation as well as their relative orientation.

Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil $C_1$ wherever there was any change in current through coil $C_2$ . Let ${\mathrm{\Phi }}_1$ be the flux through coil $C_1$ (say of $N_1$ turns) when current in coil $C_2$ is $I_2$. Then, from Eq. (6.9), we have

$$N_1{\mathrm{\Phi }}_1=M{I}_2$$ For currents varrying with time, $$\frac{\mathrm{d}\left(N_1{\mathrm{\Phi }}_1\right)}{\mathrm{d}t}=\frac{\mathrm{d}\left(M{I}_2\right)}{\mathrm{d}t}$$ Since induced emf in coil C 1 is given by $${\epsilon }_1=-\frac{\mathrm{d}\left(N_1{\mathrm{\Phi }}_1\right)}{\mathrm{d}t}$$ We get, $${\epsilon }_1=-M\frac{\mathrm{d}{I}_2}{\mathrm{d}t}$$

It shows that varying current in a coil can induce emf in a neighbouring coil. The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils.

6.9.2 Self-inductance

In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as

$$\begin{array}{l}N{\mathrm{\Phi }}_B\propto I\\ N{\mathrm{\Phi }}_B=LI\end{array}$$

where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.15), the induced emf is given by

$$\begin{array}{l}\epsilon =-\frac{\mathrm{d}\left(N{\mathrm{\Phi }}_{\mathrm{B}}\right)}{\mathrm{d}t}\\ \epsilon =-L\frac{\mathrm{d}I}{\mathrm{d}t}\end{array}$$

Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil.

It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross- sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is $B={\mu }_{0}nI$ (neglecting edge effects, as before). The total flux linked with the solenoid is

$$N{\mathrm{\Phi }}_{\mathrm{B}}=(nl)\left({\mu }_0\pi I\right)(\mathrm{A})$$ $$={\mu }_0{n}^{2}AlI$$

where nl is the total number of turns. Thus, the self-inductance is,

$$\text{[}\begin{array}{rl}L& =\frac{N{\mathrm{\Phi }}_B}{I}\\ & ={\mu }_0{n}^{2}Al\end{array}$$ \]

If we fill the inside of the solenoid with a material of relative permeability ${\mu }_r$ (for example soft iron, which has a high value of relative permeability), then,

$$L={\mu }_r{\mu }_0{n}^{2}Al$$

The self-inductance of the coil depends on its geometry and on the permeability of the medium

The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf ( $\epsilon$ ) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is

$$\frac{\mathrm{d}W}{\mathrm{d}t}=|\epsilon |I$$

If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.16),

$$\frac{\mathrm{d}W}{\mathrm{d}t}=LI\frac{\mathrm{d}I}{\mathrm{d}t}$$

Total amount of work done in establishing the current I is

$$W=\int \mathrm{d}W={\int }_0^{I}LI\mathrm{d}I$$

Thus, the energy required to build up the current I is,

$$W=\frac{1}{2}L{I}^2$$

This expression reminds us of $m{v}^2/2$ for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit).

Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. Equation (6.9) would be modified into

$$N_1{\mathrm{\Phi }}_1=M_{11}{I}_1+M_{12}{I}_2$$

where $M_{11}$ represents inductance due to the same coil.

Therefore, using Faraday’s law,

$${\epsilon }_1=-M_{11}\frac{\mathrm{d}{I}_1}{\mathrm{d}t}-M_{12}\frac{\mathrm{d}{I}_2}{\mathrm{d}t}$$

$M_{11}$ is the self-inductance and is written as $L_1$ . Therefore,

$${\epsilon }_1=-L_1\frac{\mathrm{d}{I}_1}{\mathrm{d}t}-M_{12}\frac{\mathrm{d}{I}_2}{\mathrm{d}t}$$