Q157 / 180 · COMEDK 2012 · Physics
The current in a coil changes from $1 \mathrm{~mA}$ to $5 \mathrm{~mA}$ in $4 \mathrm{milli}$ secon(d) If the coefficient of self-induction of the coil is $10 \mathrm{mH}$, then the magnitude of the self-induced emf is
- Correct answerA$10 \mathrm{mV}$
- B$5 \mathrm{mV}$
- C$2.5 \mathrm{mV}$
- D$1 \mathrm{mV}$
Solution
Given, $\quad I_{1}=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}$
$I_{2}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A}$
$t=4 \mathrm{~ms}=4 \times 10^{-3} \mathrm{~s}$
$L=10 \mathrm{~m} \mathrm{H}=10 \times 10^{-3} \mathrm{H}$
$\therefore$ Self-induced emf, $|e|=L \frac{d I}{d t}$
$=10 \times 10^{-3} \times \frac{(5-1) \times 10^{-3}}{4 \times 10^{-3}}=10 \mathrm{mV}$
Hence, the correct answer is option: 'A'.