Question

Semicircular wire of radius R is rotated in a uniform magnetic field about an axis passing through its ends. If the frequency of rotation is f, the amplitude of induced e.m.f. in the wire is [B\(=\)magnetic field induction]

• A. \(\pi ^{2}RfB\)
• B. \(\pi ^{2}R^{2}fB\)
• C. \(2\pi ^{2}RfB\)
• D. \(2\pi ^{2}Rf^{2}B\)

Hint:

Close the semicircle with its diameter on the axis. The diameter adds no emf, so use the flux through the half disc.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The diameter lies on the rotation axis and has no velocity, so it adds no emf. The wire's emf equals that of the closed half-disc.

Step 2: Flux
Area \(A=\dfrac{\pi R^{2}}{2}\), \(\phi=BA\cos\omega t\) with \(\omega=2\pi f\).

Step 3: Amplitude
\(e=-\dfrac{d\phi}{dt}=BA\omega\sin\omega t\)
\(e_{0}=B\cdot\dfrac{\pi R^{2}}{2}\cdot2\pi f=\pi^{2}R^{2}fB\)

Final Answer:
The amplitude of the induced emf is \(\pi^{2}R^{2}fB\). \[ \boxed{\text{(B) }\pi^{2}R^{2}fB} \]