Question

A metal rod $\frac{1}{\sqrt{\pi }}m$ long rotates about one of its ends perpendicular to a plane whose magnetic induction is $4\times {10}^{-3}T$. Calculate the number of revolutions made by the rod per second if the e.m.f induced between the ends of the rod is $16mV$.

Answer 1

$E=\frac{1}{2}B\omega l^2$
$\Rightarrow 16\times {10}^{-3}=\frac{1}{2}\times 4\times {10}^{-3}\times \omega \times {\left(\frac{1}{\sqrt{\pi }}\right)}^2$
$\Rightarrow 16\times {10}^{-3}=\frac{1}{2}\times 4\times {10}^{-3}\times \omega \times \left(\frac{1}{\pi }\right)$
$\Rightarrow n=\frac{16\times {10}^{-3}\times 2}{4\times {10}^{-3}\times 2}$
$n=4$