Question

An insulating thin rod of length $I$ has a linear charge density $\rho \left(x\right)={\rho }_0\left(\frac{x}{l}\right)$ on it. The rod is rotated about an axis passing through the origin $\left(x=0\right)$ and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time-averaged magnetic moment of the rod is :

• A. $\frac{\pi }{4}n\rho l^3$
• B. $n\rho l^3$
• C. $\pi n\rho l^3$
• D. $\frac{\pi }{3}n\rho l^3$

Answer 1

The correct option is A

$\frac{\pi }{4}n\rho l^3$

Step 1: Given data and assumption

Length of the rod$=l$

The linear charge density of the rod, $\rho \left(x\right)={\rho }_0\left(\frac{x}{l}\right)$

Number of rotations per second$=n$

Let when charge $q$ rotates with frequency $n$ then equivalent current $I=qn$

Step2: Find the time-averaged magnetic moment of the rod.

Formula used:

$M=IA$

Where $M$ is the magnetic moment and $A$ is the area of the coil.

From the figure, a small length $dx$ portion of magnetic moment is given as:

$dm=AdI$

Where $dm$ is the small magnetic moment and $dI$ is a small amount of current.

$dm=Andq$ $\left[\because I=nq\Rightarrow dI=ndq\right]$

$\Rightarrow$$dm=n\lambda dxA$ $\left[\because linearchargedensity,\lambda =\frac{dq}{dx}\right]$

$\Rightarrow$$dm=\frac{n{\rho }_{0}xdxA}{l}$ $\left[\because \lambda =\rho \left(x\right)={\rho }_0\left(\frac{x}{l}\right)\right]$

$\Rightarrow$$dm=\frac{n{\rho }_{0}xdx\pi x^2}{l}$ $\left[\because A=\pi x^2\right]$

$\Rightarrow$$dm=\frac{n\pi {\rho }_0{x}^{3}dx}{l}$ ……$\left(i\right)$

Now, integrating both side we get.

$\int dm={\int }_{x=0}^{x=l}\frac{n\pi {\rho }_0{x}^{3}dx}{l}$

$\Rightarrow$$M=\frac{n\pi {\rho }_0}{l}{\int }_{x=0}^{x=l}{x}^{3}dx$

$\Rightarrow$$M=\frac{n\pi {\rho }_0}{l}{\left[\frac{x^4}{4}\right]}_0^l$

$\Rightarrow$$M=n\pi {\rho }_0\frac{l^3}{4}$

At ${\rho }_0=\rho$

$\Rightarrow$$M=\frac{\pi }{4}n\rho l^3$

Hence, option A is correct.