A rod of length ' $l$ ' is rotated with angular velocity ' $\omega$ ' about its one end, perpendicular to a…
- $\mathrm{B} l^2 \omega$
- $0.5 \mathrm{~B} l^2 \omega$
- $\mathrm{B} / \omega$
- $0.5 \mathrm{~B} / \omega$
Solution

E.M.F induced, $\mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n}=\frac{\mathrm{B} \pi l^2}{\mathrm{~T}}$ $\therefore \quad \mathrm{e}=\frac{1}{2} \mathrm{~B} l^2 \omega=0.5 \mathrm{~B} l^2 \omega \quad \ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)$
Asked in: MHT CET 2024 (03 May Shift 2)
Practice more Electromagnetic Induction questions on Aicharya