When a current of $4 \mathrm{~A}$ changes to $8 \mathrm{~A}$ in $0.6 \mathrm{~s}$ in a primary coil it…
When a current of $4 \mathrm{~A}$ changes to $8 \mathrm{~A}$ in $0.6 \mathrm{~s}$ in a primary coil it induces an e. m. f. of $50 \mathrm{mV}$ in secondary coil. The mutual inductance between two coils is
$1.2 \mathrm{mH}$
$3.33 \mathrm{mH}$
$7.5 \mathrm{mH}$
$10.5 \mathrm{mH}$
Solution
Change in current in primary coil,
$\mathrm{di}_1=(8-4) \mathrm{A}=4 \mathrm{~A}$
Time taken, $\mathrm{dt}=0.6 \mathrm{~s}$
Induced e. m. f in secondary coil,
$\mathrm{E}_2=50 \times 10^{-3} \mathrm{~V}$
So, mutual inductance is,
$M=\frac{E_2}{\mathrm{di}_1 / \mathrm{dt}}=\frac{\mathrm{E}_2 \mathrm{dt}}{\mathrm{di}_1}=\frac{50 \times 10^{-3} \mathrm{~V} \times 0.6 \mathrm{~s}}{4 \mathrm{~A}}=7.5 \times 10^{-3} \mathrm{H}=7.5 \mathrm{mH}$
.