SKEDSOFT

Physics For Engineers - 2

If, on the other hand, the current $I_2$in the inner solenoid is varied, the field due to it $B_2 = \mu_0n_2I_2$which is non-zero only within the inner solenoid. The flux enclosed by the outer solenoid is, therefore,

\begin{displaymath}N_1\Phi_1 = (n_1L)\pi r_2^2 \mu_0n_2I_2 \end{displaymath}

If $I_2$is varied, the emf in the outer solenoid is  \begin{displaymath}{\cal E}_1 = -\mu_0n_1n_2L\pi r_2^2\frac{dI_2}{dt}\end{displaymath}  giving  \begin{displaymath}M_{12} = \mu_0n_1n_2L\pi r_2^2\end{displaymath}

One can see that  \begin{displaymath}M_{12}=M_{21}\end{displaymath} .

This equality can be proved quite generally from Biot-Savart's law. Consider two circuits shown in the figure.

The field at $\vec{r_2}$, due to current in the loop $C_1$(called the primary ) is \begin{displaymath}\vec B_1 = \frac{\mu_0}{4\pi}I_1\oint \frac{\vec{dl_1}\times\hat r}{\mid r\mid^2}\end{displaymath}

where $\vec r = \vec r_2 -\vec r_1$. We have seen that $\vec B_1$can be expressed in terms of a vector potential $\vec A_1$, where

, by Biot-Savart's law  \begin{displaymath}\vec A_1 = \frac{\mu_0I_1}{4\pi}\oint\frac{\vec{dl_1}}{\mid r\mid}\end{displaymath}

The flux enclosed by the second loop, (called the secondary ) is

\begin{eqnarray*} \Phi_2 &=& \int_{S_2}\vec B_1\cdot\vec{dS_2}\\ &=& \int_{S_2... ...I_1}{4\pi}\oint\oint\frac{\vec{dl_1}\cdot\vec{dl_2}}{\mid r\mid} \end{eqnarray*}

Clearly,  \begin{displaymath}M_{21} = \frac{\mu_0}{4\pi}\oint\oint\frac{\vec{dl_1}\cdot\vec{dl_2}}{\mid r \mid}\end{displaymath}

It can be seen that the expression is symmetric between two loops. Hence we would get an identical expression for $M_{12}$. This expression is, however, of no significant use in obtaining the mutual inductance because of rather difficult double integral.
Thus a knowledge of mutual inductance enables us to determine, how large should be the change in the current (or voltage) in a primary circuit to obtain a desired value of current (or voltage) in the secondary circuit. Since $M_{21}=M_{12}$, we represent mutual inductance by the symbol $M$. The emf ${\cal E}_s$in the secondary circuit is given by  \begin{displaymath}{\cal E}_s = -M\frac{dI_p}{dt}\end{displaymath} , where $I_p$is the variable current in the primary circuit.
Units of $M$is that of Volt-sec/Ampere which is known as Henry (h)