Question

A metal rod of resistance 20$\Omega$ is fixed along a diameter of conducting ring of radius $0.1m$ and lies in x-y plane. There is a magnetic field $\overrightarrow{B}=\left(50T\right)\hat{k}$ . The ring rotates with an angular velocity $\omega =20rad/s$ 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. $\frac{1}{4}A$
• B. $\frac{1}{2}A$
• C. $\frac{1}{3}A$
• D. zero

Answer 1

Answer: (C)

$\frac{1}{3}A$

Given : A metal rod of resistance 20ΩΩ is fixed along a diameter of conducting ring of radius 0.1m0.1m and lies in x-y plane. There is a magnetic field B⃗ =(50T)k^B→=(50T)k^ . The ring rotates with an angular velocity ω=20rad/sω=20rad/s about its axis. An external resistance of 10ΩΩ is connected across the centre of the ring and rim. The current through external resistance is

Emf induced between center and rim of circle E=$\frac{B\omega r^2}{2}$

E=$\frac{1}{2}\left(50\right)(0.1{)}^2$=5V

As we can see in above equivalent figure we can write $I=\frac{5}{10+5}$=$\frac{1}{3}$A

The Correct Opt:C