Question

A metal rod of resistance $20\Omega$ is fixed along a diameter of a conducting ring of radius $10cm$ and lies on the $x-y$ plane. There is a magnetic field $\overrightarrow{B}=50\left(T\right)\hat{k}$. The ring spins with an angular velocity $20rad/s$ 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. $\frac{1}{3}A$
• B. $\frac{5}{3}A$
• C. $\frac{1}{4}A$
• D. $\frac{1}{2}A$

Answer 1

The correct option is A $\frac{1}{3}A$

Given that,

Rod of resistance $R=20\Omega$

Magnetic field $B=50T$

Angular velocity $\omega =20rad/s$

External resistance $r=5\Omega$

Now, the potential difference between center and rim of the ring is

$V=\frac{1}{2}B\omega r^2$

$V=\frac{1}{2}\times 50\times 20\times {\left(0.1\right)}^2$

$V=5V$

Now, the current through external resistance

$i=\frac{V}{R+r}$

$i=\frac{5}{10+5}$

$i=\frac{1}{3}A$

Hence, the current through external resistance is $\frac{1}{3}A$