Question

A metal disc of radius R rotates with an angular velocity, $\omega =1rad/s$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction B acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is:

• A. $B{R}^2$
• B. $2B^2{R}^2$
• C. $B{R}^3$
• D. $B{R}^2/2$

Answer 1

The correct option is D $B{R}^2/2$

As shown in figure 1.AB is the axis of rotation, $\overrightarrow{B}$ is directed into the plane of disc consider a ring between the disc of thickness $dx$ at a distance $x$ from centre 0. The velocity of a ring $V=wx$ where w = angular velocity of a disc.

Now emf induced due to this ring is

$de=VBdx=Bwxd{\int }_0^{e}de={\int }_0^{R}Bxdxe=B{\left[\frac{x^2}{2}\right]}_0^{R}e=\frac{B{R}^2}{2}$as w= 1