Question

A disc of radius R rotates at an angular velocity $\omega$ about the axis perpendicular to its surface and passing through its centre. If the disc has a uniform surface charge density $\sigma$ find the magnetic induction on the axis of rotation at a distance x from the centre.

Answer 1

In the above question it is asked to find the magnetic field of the disc on the axis of rotation at a distance x, so, we will first discuss what is magnetic induction then we will further proceed towards the solution where we will use different formulas like of current and magnetic field.

Complete step-by-step solution:
In order to solve the above question where we have to find the magnetic induction on the axis of rotation at a distance x from the centre, if it is given that radius of the disc is R, the angular velocity is $\omega$ and the uniform charge density of the disc is $\sigma$.
Let us first discuss what magnetic induction is?
Magnetic induction was given by Michael Faraday in 1831. Later on, Maxwell described magnetic induction mathematically and it was known as Faraday’s law of induction. Faraday had performed three experiments to describe electromagnetic induction. Then Faraday’s law became crucial to understand induction which now has so many practical applications like in generators, transformers, etc.
So, let us now start solving the question, we can think that a disc is made up of elementary rings and when the disc rotates about its axis which is passing through the centre and also id perpendicular to the plane of the disc then each elementary ring will constitute current in it. The magnetic field which is along the axis of rotation that is due to each elementary ring is considered to be
The distance of disc from the centre is x, radius of the ring is r and width is dr
Charge on the ring is,
$dq=(2\pi rdr)\sigma$
Also, the velocity,
$V=\dfrac{2\pi R}{T}=\omega R$
Also, the current is defined as rate of change of charge so,
$dI=\dfrac{dq}{dt}$
$=\dfrac{\omega dq}{2\pi }=\sigma \omega rdr$
The magnetic field due to the ring at point P on the axis of disc is,

& dB=\dfrac{{{\mu }_{0}}dI{{r}^{2}}}{2{{({{r}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}or \\\ & B=\int{dB=\dfrac{{{\mu }_{0}}\sigma \omega }{2}}\int\limits_{0}^{R}{\dfrac{{{r}^{3}}dr}{{{({{r}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}}.........(i) \\\ \end{aligned}$$ When we put $$({{r}^{2}}+{{x}^{2}})$$ $$={{t}^{2}}$$and $$2rdr=2tdt$$ also, integrating equation (i) we will get, $$B=\dfrac{{{\mu }_{0}}\sigma \omega }{2}[\dfrac{{{R}^{2}}+2{{x}^{2}}}{\sqrt{{{R}^{2}}+{{x}^{2}}}}-2x]$$ **Note:** Magnetic induction which is also known as electromagnetic induction is defined for the production of voltage (or EMF) across an electrical conductor placed inside any varying magnetic field. According to Faraday’s law, for any closed circuit, the induced electromotive force is always equal to the rate of change of the magnetic flux enclosed within the circuit.