Question

A metal rod of resistance $20\, \Omega$ is fixed along a diameter of a conducting ring of radius $0.1\, m$ and lies on $x-y$ plane. There is a magnetic field $\vec{B}=(50 T ) \hat{k}$. The ring rotates with an angular velocity $\omega=20\, rad\, s ^{-1}$ 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 $\operatorname{rod}=20\, \Omega,\, r=0.1\, m$,
$B=50\, T$ along $z$ -axis; $\omega=20\, rad\, s ^{-1}$.
Potential difference between centre of the ring and the rim is
$\varepsilon=\frac{1}{2} B\, \omega r^{2}=\frac{1}{2} \times 50 \times 20 \times(0.1)^{2}=5\, V$
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,
$I=\frac{\varepsilon}{R+r}=\frac{5}{10+5}=\frac{1}{3}A$