Question

A copper disc of a radius $0.1m$ is rotated about its center with $20$ revolution per second in a uniform magnetic field of $0.1T$ with its plane perpendicular to the field. The emf induced across the radius of the disc is?

Answer 1

Hint: The equation for induced emf in a coil rotating in a magnetic field is induced emf $ = \dfrac{1}{2}B\omega {R^2}$. We also know that $\omega = 2\pi N$ ( $N = $ The revolutions per second). So the equation for induced emf becomes Induced emf $ = \dfrac{1}{2}B(2\pi \times N){R^2}$. Use this equation to reach the solution.

Complete step by step answer:
Magnetic flux density – The magnetic flux (denoted by $B$ ) through a surface is the surface integral of the magnetic field passing through the surface at a $90^\circ $ angle. Its unit is tesla.
Revolution per second – It is the angular speed that a body possesses divided by $2\pi $.
Induced emf – Whenever a coil (a conductor) is rotated in a constant magnetic field the magnetic field linking with it constantly changes and when this happens an emf is induced in the coil.
We know that the emf induced in a rotating circular coil is given by the following equation
Induced emf $ = \dfrac{1}{2}B\omega {R^2}$
Here, $B = $ Magnetic flux density
$\omega = $ Revolutions per second or the angular speed
$R = $ The radius of the circular coil
$\because \omega = 2\pi N$ ( $N = $ The revolutions per second)
$\therefore $ Induced emf $ = \dfrac{1}{2}B(2\pi \times N){R^2}$
Given in the problem
$B = 0.1T$
$N = 20rps$
$R = 0.1m$
So, the equation for the induced emf becomes
Induced emf $ = \dfrac{1}{2} \times 0.1(2\pi \times 20){\left( {0.1} \right)^2}$
Induced emf $ = 0.0628V$
So, the emf induced in the coil is $0.0628V$.

Note:
The concept of induced emf that we used to solve the problem given to us, is the basis of AC generator which uses the energy for a source (water, air, etc.) to rotate a conducting loop which is placed in a constant magnetic field, which will be used to produce an AC in an AC generator.