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Question: A disc of radius 0.1m is rotating with a frequency 10 rev/s in a normal magnetic field of strength 0...

A disc of radius 0.1m is rotating with a frequency 10 rev/s in a normal magnetic field of strength 0.1 T. Net induced emf is
A. 2π×10−22\pi \times {{10}^{-2}} V
B. 2π2×10−22{{\pi }^{2}}\times {{10}^{-2}} V
C. π×10−2\pi \times {{10}^{-2}} V
D. None of these

Explanation

Solution

First we will take a radial strip of the disc and find the emf between both the ends of the strip using the formula for Emf induced in a wire travelling in a magnetic field. Then as all the radial elements will produce the same emf between the centre of the disc and the edge we will take the emf we got between the edges of the strip and that will be the correct answer.

Formula used:
Emf induced in a wire travelling in a magnetic field
ε=Blv\varepsilon =Blv

Complete step by step answer:
The electro-motive force induced in a wire moving perpendicular to the magnetic field is given as the product of magnetic field, length of the wire and the velocity with which the wire moves.

In the disc there are no wire elements so we will take one radial element and find the emf on that element and the emf will be the same for all over the disc. The velocity of the different parts of that element will be different due to their radial distance from the centre of the disc. So, we will integrate over that region.

l=dr v=ωr dε=Bωrdr \begin{aligned} & l=dr \\\ & v=\omega r \\\ & d\varepsilon =B\omega rdr \\\ \end{aligned}
On integrating from r=0 to r=R we get
ε=∫0RBωrdr=BωR22\varepsilon =\int\limits_{0}^{R}{B\omega rdr}=\dfrac{B\omega {{R}^{2}}}{2}
We are given
B = 0.1 T
ω\omega = 10 rev/s = 20π20\pi rad/s
R = 0.1 meter
∴ε=(0.1)(20π)(0.1)22=0.01π\therefore \varepsilon =\dfrac{(0.1)(20\pi ){{(0.1)}^{2}}}{2}=0.01\pi
Hence, we find that the induced emf will be equal to 0.01π0.01\pi Volts for the radial strip. It will be the same for the whole disc and the emf will move the electrons from the centre of the disc towards the edge. Hence, the correct option is C, i.e. π×10−2\pi \times {{10}^{-2}} V.

Note:
Take care during integration over the radial strip must be done from radius r = 0 to radius r = R. The angular velocity is given in revolutions per second in the question and must be converted into radians per seconds and then used in the formula. Otherwise we will get the wrong answer.