Question

A circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ , with its plane perpendicular to a uniform magnetic field $B$ . Then, the emf developed between the centre and the rim of the plate is
(A) $\pi \omega B{R^2}$
(B) $\omega B{R^2}$
(C) $\dfrac{{\pi \omega B{R^2}}}{2}$
(D) $\dfrac{{\omega B{R^2}}}{2}$

Answer 1

Hint : We need to analyse a differential element of the metal plate. The integral from zero to the radius will give us the emf between the centre and radius of the circular metal plate.

Formula used: In this solution we will be using the following formula;
$v = \omega r$ where $v$ is the linear speed of the metal plate, $\omega$ is the angular speed of rotation and $r$ is the radius of the plate.
$E = Blv$ where $E$ is the emf induced on a moving body in a magnetic field, $B$ is the magnetic field and $l$ is the length of the body.

Complete step by step answer
Generally, whenever a body moves in a magnetic field, emf is induced on the moving object. This emf is given by
$E = Blv$ where $E$ is the emf induced on a moving body in a magnetic field, $B$ is the magnetic field and $l$ is the length of the body.
For the rotating plate, picking a differential element of the plate we have
$dE = Bvdl$ , where $dE$ is an infinitesimal emf which is induced on an infinitesimal length $dl$ along the radius of the plate.
But from kinematics, $v = \omega r$ , for this plate we can rewrite and say
$v = \omega l$ . Hence, replacing this into above equation we have
$dE = Bwldl$
Now for the emf developed between the centre and the rim, we integrate from the centre to the radius i.e. from $l = 0$ to $l = R$ .
Hence
$E = \int_0^R {B\omega ldl} = B\omega \int_0^R {ldl}$ .
By performing the integration and substituting the limits, we have that
$E = \dfrac{{B\omega {R^2}}}{2}$
Hence, the correct option is D.

Note
For clarity, the integration $\int_0^R {ldl}$ is performed as follows:
By mathematical principle, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
Hence $\int {ldl} = \dfrac{{{l^{1 + 1}}}}{{1 + 1}} = \dfrac{{{l^2}}}{2}$
To integrate from zero to $R$ , is to subtract the value of the function when the length is replaced with $R$ form the function when the length is replaced with zero. i.e.
$\int_0^R {ldl} = \left[ {\dfrac{{{l^2}}}{2}} \right]_0^R = \dfrac{{{R^2}}}{2} - \dfrac{{{0^2}}}{2} = \dfrac{{{R^2}}}{2}$
Hence, $\int_0^R {ldl} = \dfrac{{{R^2}}}{2}$ .