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Two different coils have self inductance $8 \mathrm{mH}$ and $2 \mathrm{mH}$. The current in both coils are increased at same constant rate. The ratio of the induced emf's in the coil is
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The correct answer is:
$4: 1$
The induced emf, $e=\frac{L d i}{d t}$
Here, $L_{1}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$
and $L_{2}=2 \mathrm{mH}=2 \times 10^{-3} \mathrm{H}$
So, ratio of induced emfs, $\frac{e_{1}}{e_{2}}=\frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}}=\frac{L_{1}}{L_{2}}$
(since, current is increased at same rate)
$\Rightarrow$$\frac{\mathrm{q}}{e_{2}}=\frac{8 \times 10^{-3} \mathrm{H}}{2 \times 10^{-3} \mathrm{H}}=\frac{4}{1}$
Here, $L_{1}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$
and $L_{2}=2 \mathrm{mH}=2 \times 10^{-3} \mathrm{H}$
So, ratio of induced emfs, $\frac{e_{1}}{e_{2}}=\frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}}=\frac{L_{1}}{L_{2}}$
(since, current is increased at same rate)
$\Rightarrow$$\frac{\mathrm{q}}{e_{2}}=\frac{8 \times 10^{-3} \mathrm{H}}{2 \times 10^{-3} \mathrm{H}}=\frac{4}{1}$
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