Question

A metallic rod of length R is rotated with an angular frequency $\omega$ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R, about an axis passing through the centre and perpendicular to the plane of the ring.There is a magnetic field B, perpendicular to the plane of the ring. The emf induced between the centre and the metallic ring is

• A. $Bsin\omega t$
• B. $\frac{B{R}^2\omega }{2}$
• C. $2B{R}^2\omega$
• D. $B{R}^2\omega$

Answer 1

The correct option is B $\frac{B{R}^2\omega }{2}$

From fig

Emfis induced is given by

$E=\frac{dQR}{dt}=\frac{d}{dt}(BA=B\frac{dA}{dt})$

Where,$\frac{dA}{dt}$ is the rule of change of area of the loop formed by the sector OPQ

At any instant of time t, let be the angle between the rod and the radius of the circle at P

Area of the sector OPQ$=R^2\frac{\theta }{2\pi }=\frac{1}{2}{R}^2\theta$

R be the radius of the circle

therefore, emf induced is,

$E=B\times \frac{d}{dt}\left(\frac{1}{2}{R}^2\theta \right)E=\frac{1}{2}B{R}^2\frac{d\theta }{dt}=\frac{B\omega R^2}{2}$