Q190 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
A very small circular loop of radius 'a' is initially (at $t = 0$) coplanar and concentric with a much larger fixed circular loop of radius 'b'. A constant current I flows in the larger loop. The smaller loop is rotated with a constant angular speed $\omega$ about the common diameter. The emf induced in the smaller loop as a function of time t is
($\mu_0$ = permeability of free space)
($\mu_0$ = permeability of free space)
- A$\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega\cos\omega t$
- B$\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega\sin\omega^2 t^2$
- Correct answerC$\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega\sin\omega t$
- D$\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega(\sin)^2\omega t$
Solution
The magnetic field produced by the larger loop at its centre is given by
$B = \dfrac{\mu_0 I}{2b}$
Since the smaller loop is very small ($a \ll b$), the magnetic field over its area can be assumed to be uniform and equal to $B$.
The magnetic flux linked with the smaller loop at any time $t$ is
$\Phi = \vec{B} \cdot \vec{A} = B A \cos(\omega t)$
Substituting $A = \pi a^2$ and $B = \dfrac{\mu_0 I}{2b}$, we get
$\Phi = \left(\dfrac{\mu_0 I}{2b}\right) (\pi a^2) \cos(\omega t)$
According to Faraday's law of induction, the induced emf is
$e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt} \left[ \left(\dfrac{\mu_0 I \pi a^2}{2b}\right) \cos(\omega t) \right]$
$e = \left(\dfrac{\pi a^2 \mu_0 I}{2b}\right) \omega \sin(\omega t)$
Answer: $\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega\sin\omega t$
$B = \dfrac{\mu_0 I}{2b}$
Since the smaller loop is very small ($a \ll b$), the magnetic field over its area can be assumed to be uniform and equal to $B$.
The magnetic flux linked with the smaller loop at any time $t$ is
$\Phi = \vec{B} \cdot \vec{A} = B A \cos(\omega t)$
Substituting $A = \pi a^2$ and $B = \dfrac{\mu_0 I}{2b}$, we get
$\Phi = \left(\dfrac{\mu_0 I}{2b}\right) (\pi a^2) \cos(\omega t)$
According to Faraday's law of induction, the induced emf is
$e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt} \left[ \left(\dfrac{\mu_0 I \pi a^2}{2b}\right) \cos(\omega t) \right]$
$e = \left(\dfrac{\pi a^2 \mu_0 I}{2b}\right) \omega \sin(\omega t)$
Answer: $\left(\dfrac{\pi a^2\mu_0 I}{2b}\right)\omega\sin\omega t$
Hence, the correct answer is option: 'C'.