Q143 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
A current of 2 A is passed through the primary coil. The total flux linked with the secondary coil, which is closely wound over the primary is $2000 \times 10^{-6}$ weber. What is the induced emf in the secondary if the current through the primary increases at a rate of $0.2 \mathrm{As}^{-1}$ ?
- Correct answerA$2 \times 10^{-4} V$
- B$4 \times 10^{-4} V$
- C$1 \times 10^{-4} V$
- D$8 \times 10^{-4} V$
Solution
The magnetic flux $\phi$ linked with the secondary coil is proportional to the current $I$ in the primary coil, given by $\phi = MI$, where $M$ is the mutual inductance between the coils.
Given the flux $\phi = 2000 \times 10^{-6} \text{ Wb}$ when the current $I = 2 \text{ A}$, we calculate the mutual inductance $M$ as:
$M = \dfrac{\phi}{I} = \dfrac{2000 \times 10^{-6}}{2} = 1000 \times 10^{-6} \text{ H} = 10^{-3} \text{ H}$
The induced electromotive force (emf) $e$ in the secondary coil is given by Faraday's law of induction, $e = -M \dfrac{dI}{dt}$.
Given the rate of change of current $\dfrac{dI}{dt} = 0.2 \text{ As}^{-1}$, the magnitude of the induced emf is:
$|e| = M \times \dfrac{dI}{dt} = 10^{-3} \times 0.2 = 0.2 \times 10^{-3} \text{ V} = 2 \times 10^{-4} \text{ V}$
Answer: $2 \times 10^{-4} V$
Given the flux $\phi = 2000 \times 10^{-6} \text{ Wb}$ when the current $I = 2 \text{ A}$, we calculate the mutual inductance $M$ as:
$M = \dfrac{\phi}{I} = \dfrac{2000 \times 10^{-6}}{2} = 1000 \times 10^{-6} \text{ H} = 10^{-3} \text{ H}$
The induced electromotive force (emf) $e$ in the secondary coil is given by Faraday's law of induction, $e = -M \dfrac{dI}{dt}$.
Given the rate of change of current $\dfrac{dI}{dt} = 0.2 \text{ As}^{-1}$, the magnitude of the induced emf is:
$|e| = M \times \dfrac{dI}{dt} = 10^{-3} \times 0.2 = 0.2 \times 10^{-3} \text{ V} = 2 \times 10^{-4} \text{ V}$
Answer: $2 \times 10^{-4} V$
Hence, the correct answer is option: 'A'.