Module 3 : MAGNETIC FIELD
Lecture 19 : Mutual Inductance

The integration over $\phi$gives $2\pi$. Using the expressions for $\sin\theta, \cos\theta$and $r$in terms of the variable $\rho$, we get, for the flux $\Phi = \int\vec B\cdot\vec{dS_1} $

\begin{eqnarray*} \Phi &=& \mu_0I_2R_2^2\pi\int_0^{R_1}\left[\frac{d^2\rho}{(\rh... ...}\int_0^{R_1} \frac{2d^2+\rho^2}{(\rho^2+d^2)^{5/2}} \rho d\rho \end{eqnarray*}

The integration can be easily performed by substituting $x=d^2+\rho^2$. After a bit of an algebra one gets

\begin{displaymath}\Phi = -\frac{\mu_0I_2R_2^2\pi}{2}\left[ (\rho^2+d^2)^{-1/2}+\frac{d^2}{3} (\rho^2+d^2)^{-3/2}\right]_0^{R_1}\end{displaymath}

On substituting the limits and using a binomial expansion to retain leading order term when $d>>R_1$, we get

\begin{displaymath}\Phi = \frac{\mu_0R_1^2R_2^2\pi}{2d^3}I_2\end{displaymath}

which gives the same expression for mutual inductance as in the previous example.

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