Q. 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 $\overrightarrow{\mathbf{B}}$ = (50 T) $\widehat{k}$. The ring rotates with an angular velocity $\omega = 20\ rads^{- 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
Explanation
(b) Here, resistance of rod $= 2\Omega.r = 0.1\ m,\ B = 50\ T,\ along\ z - axis\ \omega = 20\ rads^{- 1}.$
Potential difference between centre of the ring and the rim is
$V = \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 figure
Current through external resistance,
$i = \ \frac{E}{R + r} = \frac{5}{10 + 5} = \frac{1}{3}$ A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.