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]

The area of the semicircular region is $A = \dfrac{1}{2} \pi R^2$.

We can consider a closed loop formed by the semicircular wire and the straight segment joining its ends. Since this straight segment lies on the axis of rotation, its velocity is zero, and no motional emf is induced in it. Therefore, the emf induced in the semicircular wire is equal to the emf induced in the entire closed loop.

Assuming the magnetic field is perpendicular to the axis of rotation for maximum induced emf, the magnetic flux through the loop at any instant $t$ is given by $\Phi = B A \cos(\omega t)$, where $\omega = 2 \pi f$ is the angular frequency of rotation.

According to Faraday's law of induction, the induced emf is $\varepsilon = -\dfrac{d\Phi}{dt} = B A \omega \sin(\omega t)$.

The amplitude of the induced emf is $\varepsilon_{0} = B A \omega$.

Substituting $A = \dfrac{1}{2} \pi R^2$ and $\omega = 2 \pi f$, we get:

$\varepsilon_{0} = B \left(\dfrac{1}{2} \pi R^2\right) (2 \pi f) = \pi^2 R^2 f B$.