Question

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)

• A. \(\left(\frac{\pi a^{2}\mu _{0}I}{2b}\right)\omega \cos \omega t\)
• B. \(\left(\frac{\pi a^{2}\mu _{0}I}{2b}\right)\omega \sin \omega ^{2}t^{2}\)
• C. \(\left(\frac{\pi a^{2}\mu _{0}I}{2b}\right)\omega \sin \omega t\)
• D. \(\left(\frac{\pi a^{2}\mu _{0}I}{2b}\right)\omega (\sin )^{2}\omega t\)

Hint:

The large loop gives a uniform field at the centre; flux through the rotating small loop varies as the cosine of the angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The field at the centre of the large loop is \(B=\dfrac{\mu_0I}{2b}\). Since \(a\ll b\), this field is uniform over the small loop.

Step 2: Flux
The small loop starts coplanar and rotates at \(\omega\), so \(\phi=B\pi a^{2}\cos\omega t\).

Step 3: Induced emf
\(\varepsilon=-\dfrac{d\phi}{dt}=B\pi a^{2}\omega\sin\omega t\)

Step 4: Result
\(\varepsilon=\left(\dfrac{\pi a^{2}\mu_0I}{2b}\right)\omega\sin\omega t\), option (C).

Final Answer:
The induced emf is \(\left(\dfrac{\pi a^{2}\mu_0I}{2b}\right)\omega\sin\omega t\). \[ \boxed{\text{(C) }\left(\dfrac{\pi a^{2}\mu_0I}{2b}\right)\omega\sin\omega t} \]