Module 3 : MAGNETIC FIELD
Lecture 19 : Mutual Inductance
  Example 22 :
  Consider two parallel rings $C_1$and $C_2$with radii $R_1$and $R_2$respectively with a separation $d$between their centres. Radius of $C_2$is much smaller than that of $C_1$, so that the field experienced by $C_2$due to a current in $C_1$may be taken to be uniform over its area. Find the mutual inductance of the rings.
  Solution :
The field experienced by the smaller ring may be taken to be given by the expression for the magnetic field of a ring along its axis. We had earlier shown that at a distance $d$from the centre of the ring, the field along the axis is given by
 
\begin{displaymath}B_1 = \frac{\mu_0}{2}\frac{R_1^2}{(R_1^2+d^2)^{3/2}}I_1\end{displaymath}
  The flux enclosed by $C_2$is  \begin{displaymath}\Phi_2 = \frac{\mu_0}{2}\frac{R_1^2}{(R_1^2+d^2)^{3/2}}I_1\cdot\pi R_2^2\end{displaymath}
  By Faraday's law the emf in $C_2$is  \begin{displaymath}{\cal E}_2 = -\frac{\mu_0}{2}\frac{R_1^2}{(R_1^2+d^2)^{3/2}}\pi R_2^2 \frac{dI_1}{dt}\end{displaymath}

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