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} $ .