Question

A non-conducting thin disc of radius R charged uniformly over one side with surface charge density $\sigma $ rotates about its axis with an angular velocity $\omega $.
Find:
The magnetic induction at the centre of disc;
The magnetic moment of the disc.

Answer 1

Hint: In this problem, the thin disc that is rotating about its axis produces a current. This current gives rise to a magnetic field, whose magnitude is given by the equation of Biot-Savart’s law –
The magnetic field due to a line element at a point distant r from the line element is given by –
Magnetic field, $dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I.dl \times r}}{{{r^3}}}$
where ${\mu _0}$= absolute permeability of free space, $dl$= line element and r = distance of the point from the line element.
The magnetic moment is a quantity that represents the strength and orientation of the magnet to produce a magnetic field.

Complete step-by-step answer:
i) Magnetic induction:
Let us consider a ring element of radius R and thickness $dr$with uniform surface charge density $\sigma $.
The surface charge density represents the distribution of the net charge over the given area. Hence, the net charge on the thin disc is given by –
$q = \sigma \times a$
where a = area of the ring element, which is the product of circumference of the circle and the ring element.
Area, $a = 2\pi Rdr$
Thus, the net charge, $q = \sigma \left( {2\pi R} \right)dr$
Since the disc is spinning, there is a flow of charges within the disc resulting in formation of electric current. The current is defined as the charge per unit time. In this problem, the angular frequency of rotation, $\omega $ is given which is –
$\omega = \dfrac{{2\pi }}{t}$
Hence, the current in the ring element is equal to,
$di = \dfrac{q}{t} = q \times \dfrac{\omega }{{2\pi }}$
Substituting the net charge,
$di = 2\pi \sigma Rdr \times \dfrac{\omega }{{2\pi }} = \omega \sigma Rdr$
This current set up in the ring element induces a magnetic field, whose value is given by the Biot-Savart’s law –
The magnetic field due to a line element at a point distant r from the line element is given by –
Magnetic field, $dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I.dl \times r}}{{{r^3}}}$
where ${\mu _0}$= absolute permeability of free space, $dl$= line element and r = distance of the point from the line element.
Applying the above Biot-Savart law to this ring element of radius R at the centre of the ring is,
Magnetic field, $dB = \dfrac{{{\mu _0}di}}{{2R}}$
Substituting the value of $di$, we get –
$dB = \dfrac{{{\mu _0}}}{{2R}}\left( {\omega \sigma R} \right)dr$
To find the magnetic field due to the entire disc, we have to integrate the above expression from the centre of the disc to the extreme end i.e. the radius of the disc. Thus, we get –
Induced magnetic field, $B = \int\limits_0^R {dB} $
Substituting,
$B = \int\limits_0^R {\dfrac{{{\mu _0}}}{{2R}}\left( {\omega \sigma R} \right)dr} $
$ \Rightarrow B = \int\limits_0^R {\dfrac{{{\mu _0}}}{2}\left( {\sigma \omega } \right)dr} $
$ \Rightarrow B = \dfrac{{{\mu _0}}}{2}\left( {\sigma \omega } \right)\int\limits_0^R {dr} $
Integrating,
$ \Rightarrow B = \dfrac{{{\mu _0}}}{2}\left( {\sigma \omega } \right)\left[ r \right]_0^R$
$ \Rightarrow B = \dfrac{{{\mu _0}}}{2}\left( {\sigma \omega } \right)\left[ {R - 0} \right]$
$ \Rightarrow B = \dfrac{{{\mu _0}\sigma \omega R}}{2}$
Therefore, the induced magnetic field, $B = \dfrac{{{\mu _0}\sigma \omega R}}{2}$
ii) Magnetic moment
The strength of a magnetic pole is given by the quantity called magnetic moment. The mathematical definition of magnetic moment is given by relating the torque applied to align a magnet in an external magnetic field.
If $\tau $ is the torque applied to align a magnet of magnetic moment m in an external field B, we have –
$\tau = m \times B$
The unit of magnetic moment is $\dfrac{{N - m}}{T}$ where T = tesla.
By solving the unit $\dfrac{{N - m}}{T}$, we get it as $A - {m^2}$ where A is the unit of current amperes and area unit metre-square.
Thus, we can say, $m = I \times A$
where I = current and A = area
The magnetic moment of the ring element in this problem, is –
$dm = di \times A$
Area of the ring element, $A = \pi {R^2}$
As solved above, the current, $di = \omega \sigma Rdr$
Substituting,
$dm = \omega \sigma Rdr \times \pi {R^2}$
$ \Rightarrow dm = \pi \omega \sigma {R^3}dr$
The net magnetic moment of the disc can be obtained by integrating the magnetic moment of the ring element from the centre of the disc to the extreme end i.e. radius R.
Thus,
$m = \int\limits_0^R {dm} $
Substituting,
$m = \int\limits_0^R {\pi \omega \sigma {R^3}dr} $
$ \Rightarrow m = \pi \omega \sigma \int\limits_0^R {{R^3}dr} $
Integrating,
$ \Rightarrow m = \pi \omega \sigma \left[ {\dfrac{{{R^4}}}{4}} \right]_0^R$
$ \Rightarrow m = \pi \omega \sigma \left[ {\dfrac{{{R^4}}}{4} - 0} \right]$

Hence, the magnetic moment of the disc, $m = \dfrac{{\pi \omega \sigma {R^4}}}{4}$

Note: The students might have confused the conversion of the unit of magnetic moment from $\dfrac{{N - m}}{T}$ to $A - {m^2}$. Let us understand the method of deducing the same.
As per the Lorentz’s force law, the magnetic force on a charge moving at a velocity in a magnetic field of strength, B is given by –
$F = q\left( {v \times B} \right)$
The SI units of force is newton, charge is coulomb, velocity is metre-per-second and magnetic field is tesla (T). Substituting the units.
$N = C \times \dfrac{m}{s} \times T$
Multiplying metre on both sides,
$Nm = \dfrac{C}{s} \times {m^2} \times T$
$ \Rightarrow \dfrac{{Nm}}{T} = \dfrac{C}{s} \times {m^2}$
The unit coulomb per second is written as 1 ampere, the unit of current.
Thus, we have proved that –
$\dfrac{{Nm}}{T} = A{m^2}$