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 a magnetic field $\vec{B}=(50T)\hat{k}$. 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)

Here, resistance of $\mathrm{rod}=20\Omega ,r=0.1m$,
$B=50T$ along $z$ -axis; $\omega =20rad{s}^{-1}$.
Potential difference between centre of the ring and the rim is
$\epsilon =\frac{1}{2}B\omega r^2=\frac{1}{2}\times 50\times 20\times (0.1{)}^2=5V$
The equivalent circuit of the arrangement is shown in figures.


$\frac{1}{R_P}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}\text{ or }{R}_P=5\Omega$
Current through external resistance,
$I=\frac{\epsilon }{R+r}=\frac{5}{10+5}=\frac{1}{3}A$