Given:

• Rod length $L$
• Angular speed $\omega$
• Uniform magnetic field $B$
• Rod rotates about one end, with plane of rotation perpendicular to $B$

Take a small element of rod at distance $r$ from the axis, of thickness $dr$.

Its linear speed is $v = \omega r$

Small induced emf across this element: $d\mathcal{E} = Bv\,dr = B(\omega r)dr$

Integrate from $r=0$ to $r=L$: $\mathcal{E} = \int_0^L B\omega r\,dr$

$\mathcal{E} = B\omega \int_0^L r\,dr$

$\mathcal{E} = B\omega \left[\frac{r^2}{2}\right]_0^L$

$\mathcal{E} = \frac{1}{2}B\omega L^2$

Match with options: $\boxed{\mathcal{E} = \frac{1}{2}BL^2\omega}$ This is Option B.

Verification with stored answer: Stored correct answer is B, which matches the derived result.