Q139 / 180 · COMEDK 2024 (Evening Shift) · Physics
The current in a coil changes steadily from $3 \mathrm{~A}$ to $5 \mathrm{~A}$ in $0.2 \mathrm{~s}$ when an emf of $2 \mu \mathrm{V}$ is induced in it. The self-inductance of the coil is
- A2 $\mu H$
- B0.2 mH
- Correct answerC0.2 $\mu H$
- D20 $\mu H$
Solution
The induced emf $\varepsilon$ in a coil due to a change in current is given by the formula $\varepsilon = L \left| \dfrac{di}{dt} \right|$, where $L$ is the self-inductance of the coil.
Given values are:
Change in current $\Delta i = 5 \text{ A} - 3 \text{ A} = 2 \text{ A}$
Time interval $\Delta t = 0.2 \text{ s}$
Induced emf $\varepsilon = 2 \mu \text{V} = 2 \times 10^{-6} \text{ V}$
The rate of change of current is $\dfrac{di}{dt} = \dfrac{\Delta i}{\Delta t} = \dfrac{2 \text{ A}}{0.2 \text{ s}} = 10 \text{ A/s}$.
Substituting these values into the formula:
$2 \times 10^{-6} \text{ V} = L \times 10 \text{ A/s}$
$L = \dfrac{2 \times 10^{-6}}{10} \text{ H}$
$L = 0.2 \times 10^{-6} \text{ H} = 0.2 \mu \text{H}$.
Answer: 0.2 $\mu H$
Given values are:
Change in current $\Delta i = 5 \text{ A} - 3 \text{ A} = 2 \text{ A}$
Time interval $\Delta t = 0.2 \text{ s}$
Induced emf $\varepsilon = 2 \mu \text{V} = 2 \times 10^{-6} \text{ V}$
The rate of change of current is $\dfrac{di}{dt} = \dfrac{\Delta i}{\Delta t} = \dfrac{2 \text{ A}}{0.2 \text{ s}} = 10 \text{ A/s}$.
Substituting these values into the formula:
$2 \times 10^{-6} \text{ V} = L \times 10 \text{ A/s}$
$L = \dfrac{2 \times 10^{-6}}{10} \text{ H}$
$L = 0.2 \times 10^{-6} \text{ H} = 0.2 \mu \text{H}$.
Answer: 0.2 $\mu H$
Hence, the correct answer is option: 'C'.