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Physics for Engineers - 2

Mutual Inductance

Mutual Inductance: According to Faraday's law, a changing magnetic flux in a loop causes an emf to be generated in that loop. Consider two stationary coils carrying current. The first coil has $N_1$turn and carries a current $I_1$. The second coil contains $N_2$turns. The current in the first coil is the source of a magnetic field $\vec B_1$in the region around the coil. The second loop encloses a flux $N_2\Phi_2 = N_2\int_S\vec B_1\cdot\vec{dS_2}$, where $S$is the surface of one turn of the loop. If the current $I_1$in the first coil is varied, $\vec B_1$, and consequently $\Phi_2$will vary with time.

The variation of $\Phi_2$causes an emf to be developed in the second coil. Since $B_1$is proportional to $I_1$, so is $\Phi_2$. The emf, which is the rate of change of flux is, therefore, proportional to $dI_1/dt$,  \begin{displaymath}{\cal E}_2 = -M_{21}\frac{dI_1}{dt}\end{displaymath}

where $M_{21}$is a constant, called the mutual inductance of the two coils, which depends on geometrical factors of the two loops, their relative orientation and the number of turns in each coil.

Analogously, we can argue that if the second loop carries a current $I_2$which is varied with time, it generates an induced emf in the first coil given by \begin{displaymath}{\cal E}_1 = -M_{12}\frac{dI_2}{dt}\end{displaymath}

For instance, consider two concentric solenoids, the outer one having $n_1$turns per unit length and inner one with $n_2$turns per unit length. The solenoids are wound over coaxial cylinders of length $L$each. If the current in the outer solenoid is $I_1$, the field due to it is $B_1=\mu_0n_1I_1$, which is confined within the solenoid. The flux enclosed by the inner cylinder is

\begin{eqnarray*} N_2 \Phi_2 &=& N_2\pi r_2^2B_1 \\ &=& n_2L\pi r_2^2\cdot(\mu_0n_1I_1)\\ &=& \mu_0n_1n_2L\pi r_2^2I_1 \end{eqnarray*}

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}