Module 3 : MAGNETIC FIELD
Lecture 19 : Mutual Inductance

If the current in the outer solenoid varies with time, the emf in the inner solenoid is

\begin{displaymath}{\cal E}_2 = -N_2\frac{d\Phi_2}{dt}= -\mu_0n_1n_2L\pi r_2^2\frac{dI_1}{dt}\end{displaymath}

so that  \begin{displaymath}M_{21} = \mu_0n_1n_2L\pi r_2^2\end{displaymath}

\includegraphics{fig3.48.eps}

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} .

1
4
6
8