Q131 / 150 · MHT CET 2012 · Physics

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

  • Correct answer
    A
    $4: 1$
  • B
    $1: 4$
  • C
    $1: 2$
  • D
    $2: 1$

Solution

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}$

Hence, the correct answer is option: 'A'.

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