Question

Consider a long cylindrical shell of non-conducting material which carries a surface charge fixed in outer periphery with surface charge density $\sigma \text{C}/{\text{m}}^2$ as shown in figure.


The cylinder is suspended in a manner such that it is free to revolve about its axis, without friction. The cylinder is rotating with tangential speed $V_0$, then magnetic field inside the cylinder will be:

• A. $2\sigma {\mu }_0{V}_0$
• B. $\sigma {\mu }_0{V}_0/2$
• C. zero
• D. $\sigma {\mu }_0{V}_0$

Answer 1

The correct option is D $\sigma {\mu }_0{V}_0$

For the element shown in the figure, the charge is:

$dq=\sigma \times \left(2\pi Rdx\right)=2\pi \sigma Rdx$


The equivalent current due to rotation of this element will be

$i=\frac{dq}{T}$

$i=\frac{2\pi \sigma Rdx}{\left(\frac{2\pi R}{V_0}\right)}=\sigma V_{0}dx$

The current per unit width of cylinder along axis on the circumference,

$K=\frac{i}{dx}$

$K=\frac{\sigma V_{0}dx}{dx}=\sigma V_0$

Thus, the given situation is equivalent to a long solenoid with current per unit length $K=ni$.

Using the relation of magnetic field inside the core of solenoid,

$B={\mu }_{0}ni={\mu }_{0}K$

$\therefore B={\mu }_0\left(\sigma V_0\right)=\sigma {\mu }_0{V}_0$

Hence, option $\left(d\right)$ is correct.

Why this question? When current is along the circumference of a cylindrical shape, then it can be related as analogy of a solenoid with current per unit width $\left(ni\right)$ as $\frac{i}{dx}$, where $dx$ is a small strip on the circumference along the axis of cylinder.