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 $\bar{B}$ = (50 T) $\hat{k}$ . The ring rotates with an angular velocity $\omega=20$ rad / s about its axis. An external resistance of $10\,\Omega$ is connected across the centre of the ring and rim. The current through external through external resistance is :
Electromagnetic Induction
Solution:
emf induced between centre of the ring and the rim is
$e = \frac{1}{2} B \omega R^2 = \frac{1}{2} (50 )(20)(20)^2$ = 5V
Now the circuit can be drawn as follows :
$\therefore \, i = \frac{5}{10 +5} = \frac{1}{3}A$
