Calculate the mutual inductance between two coils if a current $10 A$ in the primary coil changes the flux by $500 W b$ per turn in the secondary coil of $200$ turns. Also determine the induced $e . m . f$ across the ends of the secondary coil in $0.5$ second:

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Solution

given $I = 10 A$
$t = 0.5 A$
$\begin{matrix} ϕ = 500 w b / t u r n N = 200 \end{matrix}$
total flux $= 500 \times 200$
$= 10^{5} w b$
mutual inductance $= r^{2}$
induced emf $e = ?$
$\begin{matrix} Q = M I m = \frac{Q}{I} = \frac{10^{5}}{10} m = 10^{4} H e = \frac{m d I}{d t} = \frac{10^{5} \times 10}{0.5} e = 2 \times 10^{6} V \end{matrix}$

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