Question

A rod of length $2a$ is free to rotate in a vertical plane, about a horizontal axis $O$ passing through its midpoint. A long straight, horizontal wire is on the same plane and is carrying a constant current $i$ as shown in figure. At initial moment of time, the rod is horizontal and starts to rotate with constant angular velocity $\omega$, calculate emf induced in the rod as a function of time.

Answer 1

$\theta =\omega t$
$de=B\left(\omega x\right)dx$
Here, $B=\frac{{\mu }_0}{2\pi }\frac{i}{d-x\sin \omega t}$
$\therefore$ $de=\frac{{\mu }_{0}i\omega }{2\pi }\frac{x}{d-x\sin \omega t}dx$
$V_{OA}=V_0-V_A={\int }_0^{a}de=\frac{{\mu }_{0}i\omega }{2\pi }{\int }_0^a\frac{x}{d-x\sin \omega t}dx$
$=-\frac{{\mu }_{0}i\omega }{2\pi \sin \omega t}\left[\frac{d}{\sin \omega t}\ln \left(\frac{d-a\sin \omega t}{d}\right)+a\right]$
Similarly,
$V_{OB}=V_O-V_B=\frac{{\mu }_{0}i\omega }{2\pi }{\int }_0^a\frac{x}{d+x\sin \omega t}dx$
$=\frac{{\mu }_{0}i\omega }{2\pi \sin \omega t}[a-\frac{d}{\sin \omega t}\ln \left(\frac{d-a\sin \omega t}{d}\right)]$

$\therefore$ $V_{AB}=V_{OB}-V_{OA}$
$=\frac{{\mu }_{0}i\omega }{2\pi \sin \omega t}[2a+\frac{d}{\sin \omega t}\ln \left(\frac{d-a\sin \omega t}{d+a\sin \omega t}\right)]$