Module 3 : MAGNETIC FIELD
Lecture 19 : Mutual Inductance

\begin{eqnarray*} r^2 &=& d^2 + \rho^2\\ \tan\theta &=& \frac{\rho}{d} \end{eqnarray*}

so that,

\begin{eqnarray*} \cos\theta &=& \frac{d}{\sqrt{\rho^2+d^2}}\\ \sin\theta &=& \frac{\rho}{\sqrt{\rho^2+d^2}} \end{eqnarray*}

The angle between $\vec {dS_1}$and $\vec r$is $\theta$so that

\begin{eqnarray*} \vec B\cdot\vec{dS_1} &=& B_r\cos\theta \rho d\rho d\phi + B_... ...\frac{\mu_0}{4}I_2R_2^2\frac{\sin^2\theta}{r^3}\rho d\rho d\phi \end{eqnarray*}

where we have substituted $\mu = I_2R_2^2$for the magnetic moment of the current loop. The flux enclosed by $C_1$is obtaied by integrating over $\phi$from 0 to $2\pi$and over $\rho$from 0 to $R_1$.

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