Question

A disc of radius R is rotated about an axis perpendicular to its plane and passing through its centre. If the surface charge density σ at a distance r from centre is given by
\[\sigma \text{ }=\alpha \text{xr}\], find the magnetic dipole moment associated with the disk.
Assume angular speed to be ω.
\[\begin{align}
  & \text{A}.\alpha \pi \omega {{R}^{4}} \\
 & \text{B}.\dfrac{\alpha \pi \omega {{R}^{4}}}{2} \\
 & \text{C}.\dfrac{\alpha \pi \omega {{R}^{5}}}{5} \\
 & \text{D}\text{.}\dfrac{\alpha \pi \omega {{R}^{4}}}{4} \\
\end{align}\]

Answer 1

Hint: First, the surface charge density for a small area of element is calculated. Then, the area of the disc is calculated and later differentiated. After this, the total charge induced on the surface has to be calculated. Using all these factors along with the time and current, the magnetic moment can be obtained.

Complete answer:
Given data are
Radius of the disc R
Surface charge density σ
\[\sigma \text{ }=\alpha \text{xr}\]
Where,
r is the distance from the centre.
Let A be the area of the disc
For circle, the area is calculated as
\[A=~\pi {{r}^{2}}\]
The disc is usually circular in shape
Therefore, the area of the disc is expressed as
\[A=~\pi {{r}^{2}}\]---- (1)
Let this be equation (1)
Let us consider a small area element dA in the disc.
Differentiating equation (1) on both sides,
We get,
\[dA=2\pi rdr\]---- (2)
Let this be equation (2)

Let Q be the total charge induced on the surface
The total charge is calculated as the product of surface charge density and area.
That is,
\[Q=\text{ }surface\text{ }charge\text{ }density\text{ x}area\]
---(3)
Let this be equation (3)
 Let \[dq\] be the charge on an element of small area
Differentiating equation (3) on both sides,
We get,
\[dq~=~\sigma \text{x}dA\]---- (4)
Let this be equation (4)
Substitute the value of σ in equation (4),
We get,
\[~\sigma dA~=~\alpha \text{x}2\pi rdr~\]
\[\sigma dA=2\alpha \pi {{r}^{2}}dr\]--- (5)
Let this be equation (5)
Let T be the time period of rotation.
The time period of rotation is expressed as,
\[T=\dfrac{2\pi }{\omega }\] --- (6)
Let this be equation (6)
Let \[dI\] be the small current associated with an element of a small area.
\[dI\]can be written as,
\[dI=\text{ }\dfrac{dq}{T}\]--- (7)
Let this be equation (7)
Substitute the values of \[dq\] and T in equation (7)
We get,
\[dI=~\dfrac{2\pi \alpha {{r}^{2}}dr}{2\pi }\times \omega \]
\[dI=~\alpha {{r}^{2}}dr\omega \]--- (8)
Let this be equation (8)
The magnetic moment of the ring formed for a small area element dA is given by,
\[d\mu =~dI\times ~A~\]---(9)
Let this be equation (9)
Substitute the value of A in equation (9)
We get,
\[d\mu =~dI\left( \pi {{r}^{2}} \right)\] --- (10)
Let this be equation (10)
Substitute the value of \[dI\] in equation (10)
We get,
\[d\mu =\pi \alpha \omega {{r}^{4}}dr\]--- (11)
Let this be equation (11)
Let the magnetic moment of disc be µ
\[\mu =\pi \alpha \omega \int\limits_{0}^{R}{r4dr}~\] --- (12)
Let this be equation (12)
\[\mu =\dfrac{\pi \alpha \omega {{R}^{5}}}{5}~~\]

Therefore, the correct answer is option (c).

Note:
The time period of rotation is always to be expressed in terms of the function of angular velocity \[\omega \]. Most of the factors including current, surface charge, area, etc. have to be differentiated in order to get the right answer. Don’t directly substitute the values without differentiation or else the answer may be wrong.