Q163 / 180 · COMEDK 2023 (Evening Shift) · Physics
One volt induced emf is produced in the secondary coil when the current through the primary coil is changed from $3 \mathrm{~A}$ to $1 \mathrm{~A}$ in 100 milliseconds. the mutual inductance of the two coil is:
- A0.25 H
- B0.5 H
- C0.005 H
- Correct answerD0.05 H
Solution
The induced emf $\varepsilon$ in the secondary coil is given by the formula $\varepsilon = -M \dfrac{dI}{dt}$, where $M$ is the mutual inductance and $\dfrac{dI}{dt}$ is the rate of change of current in the primary coil.
Given values are:
$\varepsilon = 1 \text{ V}$
Change in current $\Delta I = 1 \text{ A} - 3 \text{ A} = -2 \text{ A}$
Time interval $\Delta t = 100 \text{ ms} = 0.1 \text{ s}$
The magnitude of the induced emf is $|\varepsilon| = M \left| \dfrac{\Delta I}{\Delta t} \right|$.
Substituting the values:
$1 = M \times \left| \dfrac{-2}{0.1} \right|$
$1 = M \times 20$
$M = \dfrac{1}{20} \text{ H}$
$M = 0.05 \text{ H}$
Answer: 0.05 H
Given values are:
$\varepsilon = 1 \text{ V}$
Change in current $\Delta I = 1 \text{ A} - 3 \text{ A} = -2 \text{ A}$
Time interval $\Delta t = 100 \text{ ms} = 0.1 \text{ s}$
The magnitude of the induced emf is $|\varepsilon| = M \left| \dfrac{\Delta I}{\Delta t} \right|$.
Substituting the values:
$1 = M \times \left| \dfrac{-2}{0.1} \right|$
$1 = M \times 20$
$M = \dfrac{1}{20} \text{ H}$
$M = 0.05 \text{ H}$
Answer: 0.05 H
Hence, the correct answer is option: 'D'.