Question

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

• A. B $\alpha \sqrt{\frac{1}{2}g{\ell }^3}$
• B. B $\alpha \sqrt{\frac{3}{8}g{\ell }^3}$
• C. B $\alpha \sqrt{\frac{1}{3}g{\ell }^3}$
• D. B $\alpha \sqrt{g{\ell }^3}$

Answer 1

The correct option is B B $\alpha \sqrt{\frac{3}{8}g{\ell }^3}$

$Conservationofenergy$
$\frac{1}{2}I{\omega }^2=\frac{mgl(1-cos\alpha )}{2}\Rightarrow \omega =\sqrt{\frac{mgl(1-cos\alpha )}{I}}$
$1-cos\alpha =2{sin}^2\left(\frac{\alpha }{2}\right)\approx 2{\left(\frac{\alpha }{2}\right)}^2=\frac{{\alpha }^2}{2}$
$I=\frac{m{l}^2}{3}$
$\omega =\sqrt{\frac{mgl\left(\frac{{\alpha }^2}{2}\right)}{\frac{m{l}^2}{3}}}=\alpha \sqrt{\frac{3g}{2l}}$
$EMF=\frac{1}{2}B\omega l^2=\frac{1}{2}B\alpha \sqrt{\frac{3g}{2l}}{l}^2=B\alpha \sqrt{\frac{3g{l}^3}{8}}$