Question

Figure shows a conducting circular loop of radius $a$ placed in a uniform, perpendicular magnetic field $B$. A thick metal rod $OA$ is pivoted at the centre $O$. The other end fo the rod touches the loop at $A$. The centre $O$ and a fixed point $C$ on the loop are connected by a wire $OC$ of resistance $R$. A force is applied at the middle point of the rod $OA$ perpendicularly so that the rod rotates clockwise at a uniform angular-velocity $\omega$. Suppose the wire connecting $O$ and $C$ has zero resistance but the circular loop has a resistance $R$ uniformly distributed along its length. The rod $OA$ is made to rotate with a uniform angular speed $\omega$ as shown in the figure. Find the current in the rod when $\angle AOC={90}^o$

Answer 1

The 2 resistances $r/4$ and $3r/4$ are in parallel
$R^{\prime }=\frac{r/4\times 3r/4}{r}=\frac{3r}{16}$
$ee=Bvl=B\times \frac{a}{2}\omega \times a=\frac{B^{}{a}^2\omega }{2}$
$i=\frac{e}{R^{\prime }}=\frac{B^{}{a}^2\omega }{2R^{\prime }}=\frac{B^{}{a}^2\omega }{2\frac{3r}{16}}=\frac{8}{3}\frac{B^{}{a}^2\omega }{r}$