Q179 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
The current in a primary coil of a step-down transformer having $N_p$ turns is given by $i = i_0 \cos\omega t$. The secondary coil of having $N_s$ turns is wound over the primary coil. The emf induced in the secondary is
- A$\dfrac{M i_0}{\omega} \sin\omega t$
- B$\omega^2 M i_0 \sin\omega t$
- C$-\omega M i_0 \sin\omega t$
- Correct answerD$\omega M i_0 \sin\omega t$
Solution
The current in the primary coil is given by $i = i_0 \cos\omega t$.
The magnetic flux linked with the secondary coil is $\phi_s = M i$, where $M$ is the mutual inductance between the coils.
According to Faraday's law of electromagnetic induction, the induced emf in the secondary coil is given by:
$e = -\dfrac{d\phi_s}{dt}$
$e = -M \dfrac{di}{dt}$
Differentiating the expression for current with respect to time $t$:
$\dfrac{di}{dt} = \dfrac{d}{dt}(i_0 \cos\omega t) = -i_0 \omega \sin\omega t$
Substituting this back into the equation for induced emf:
$e = -M (-i_0 \omega \sin\omega t)$
$e = \omega M i_0 \sin\omega t$
Answer: $\omega M i_0 \sin\omega t$
The magnetic flux linked with the secondary coil is $\phi_s = M i$, where $M$ is the mutual inductance between the coils.
According to Faraday's law of electromagnetic induction, the induced emf in the secondary coil is given by:
$e = -\dfrac{d\phi_s}{dt}$
$e = -M \dfrac{di}{dt}$
Differentiating the expression for current with respect to time $t$:
$\dfrac{di}{dt} = \dfrac{d}{dt}(i_0 \cos\omega t) = -i_0 \omega \sin\omega t$
Substituting this back into the equation for induced emf:
$e = -M (-i_0 \omega \sin\omega t)$
$e = \omega M i_0 \sin\omega t$
Answer: $\omega M i_0 \sin\omega t$
Hence, the correct answer is option: 'D'.