A rod of length l is oscillating as a physica | vedclass

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(B) The rod oscillates as a physical pendulum. The maximum angular velocity $\omega$ occurs at the mean position.Using the conservation of energy: $mg

Vedclass · Accepted Answer

$B\alpha \sqrt {\frac{3}{8}g{l^3}} $

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(B) The rod oscillates as a physical pendulum. The maximum angular velocity $\omega$ occurs at the mean position.Using the conservation of energy: $mgh = \frac{1}{2} I \omega^2$,where $h = \frac{l}{2}(1 - \cos \alpha) \approx \frac{l \alpha^2}{4}$ for small $\alpha$.The moment of inertia about the end is $I = \frac{ml^2}{3}$.Thus,$mg \frac{l \alpha^2}{4} = \frac{1}{2} (\frac{ml^2}{3}) \omega^2$.Simplifying,$\frac{g \alpha^2}{4} = \frac{l}{6} \omega^2$,which gives $\omega^2 = \frac{6g \alpha^2}{4l} = \frac{3g \alpha^2}{2l}$,so $\omega = \alpha \sqrt{\frac{3g}{2l}}$.The induced $emf$ in a rotating rod is $\varepsilon = \frac{1}{2} B \omega l^2$.Substituting $\omega$: $\varepsilon = \frac{1}{2} B l^2 (\alpha \sqrt{\frac{3g}{2l}}) = \frac{1}{2} B \alpha \sqrt{\frac{3g l^4}{2l}} = B \alpha \sqrt{\frac{3}{8} g l^3}$.

$A$ rod of length $l$ is oscillating as a physical pendulum about one of its ends with a small angular amplitude $\alpha$ in a crossed magnetic field $B$. The maximum $emf$ induced in the rod will be:

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