Question

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

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

Answer 1

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


EMF developed across rod $PQ$ is,

$E_{PQ}=\frac{B\omega l^2}{2}=\frac{B\omega r^2}{2}[\because l=r]$

$=\frac{50\times 20\times {0.1}^2}{2}=5\text{V}$

Similarly, EMF developed across the rod $QR$ is,

$E_{QR}=5\text{V}$

Since, the ends of the rod are joined together by a conducting ring, therefore, the equivalent circuit of the given setup is as shown below.


The equivalent internal resistance of the above circuit is,

$r_{eq}=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$

The equivalent emf of the circuit is,

$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{\frac{5}{10}+\frac{5}{10}}{\frac{1}{10}+\frac{1}{10}}=5\text{V}$

Now, current flowing in the $10\Omega$ external resistor is,

$i=\frac{E_{eq}}{R+r_{eq}}=\frac{5}{10+5}=\frac{1}{3}\text{A}$

Hence, option $\left(C\right)$ is the correct answer.