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Two different coils of self inductance $L_{1}$ and $L_{2}$ are placed close to each other so that the effective flux in one coil is completely linked with other. If $M$ is the mutual inductance between them, then
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The correct answer is:
$M=\sqrt{L_{1} L_{2}}$
We know that, $M=-\frac{e_{2}}{d i_{1} / d t}=-\frac{e_{1}}{d i_{2} / d t}$
Also, $e_{1}=-L_{1} \frac{d i_{1}}{d t}$ and $e_{2}=-L_{2} \frac{d i_{2}}{d t}$
$\begin{array}{ll}
\Rightarrow & M^{2}=\frac{e_{1} e_{2}}{\left(\frac{d i_{1}}{d t}\right)\left(\frac{d i_{2}}{d t}\right)}=L_{1} L_{2} \\
\Rightarrow & M=\sqrt{L_{1} L_{2}}
\end{array}$
Also, $e_{1}=-L_{1} \frac{d i_{1}}{d t}$ and $e_{2}=-L_{2} \frac{d i_{2}}{d t}$
$\begin{array}{ll}
\Rightarrow & M^{2}=\frac{e_{1} e_{2}}{\left(\frac{d i_{1}}{d t}\right)\left(\frac{d i_{2}}{d t}\right)}=L_{1} L_{2} \\
\Rightarrow & M=\sqrt{L_{1} L_{2}}
\end{array}$
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