Question

A metal rod of resistance $20\Omega$ is fixed along a diameter of a conducting ring of radius $0.1m$ and lies on $x-y$ plane. There is magnetic field $\overline{B}=\left(50T\right)\hat{k}$. The ring rotates with an angular velocity $\omega =20rad{s}^{-1}$ about its axis. An external resistance of $10\Omega$ is connected across the centre of the ring and rim. The current through external resistance is

• A. $\frac{1}{2}A$
• B. $\frac{1}{3}A$
• C. $\frac{1}{4}A$
• D. Zero

Answer 1

Answer: (B)

$\frac{1}{3}A$

The potential difference is $V=\frac{1Bw{r}^2}{2}=\frac{1\ast 20\ast 50(0.1{)}^2}{2}=5V$
The equivalent circuit is
$i=\frac{V}{R_1+R_2}=\frac{5}{10+5}=\frac{1}{3}A$