Question

The surface charge density of a ring of radius $a$ and thickness $d$ is $\sigma$ as shown in figure. If rotates with a frequency $f$ about an axis passing through its centre and perpendicular to its plane. Assume that charge is present only on the outer surface. The magnetic field induction at a point $\text{P}$ on axis at a very large distance $L$ from its centre will be:
$[$Assume that $(d<<a)]$

• A. $\frac{\pi {\mu }_0\sigma df}{L^2}$
• B. $\frac{\pi {\mu }_0{\sigma }^{2}adf}{L^3}$
• C. $\frac{\pi {\mu }_0\sigma dfa}{L^3}$
• D. $\frac{\pi {\mu }_0{a}^{3}d\sigma f}{L^3}$

Answer 1

The correct option is D $\frac{\pi {\mu }_0{a}^{3}d\sigma f}{L^3}$

Surface charge density of ring is $\sigma$.

The surface area of ring can be given as,

$A=$ Perimeter $\times$ width

$A=2\pi a\times d=2\pi ad$

Thus, the total charge on surface of ring is,

$q=\sigma A=2\pi ad\sigma$

The equivalent current in the circular loop due to rotation of ring is,

$i=\frac{q}{T}=qf(\because d<<a)$

$\Rightarrow i=\left(2\pi ad\sigma \right)f$

Now, the point $\text{P}$ lies at a very large distance $L(L>>a)$ on the axis of rotating ring.

Thus magnetic field at $\text{P}$ due to rotating ring will be,

$B_P=\frac{{\mu }_{0}i{a}^2}{2L^3}$

$(\because$ radius of ring is $a)$

$\Rightarrow B_P=\frac{{\mu }_0\left(2\pi ad\sigma f\right)a^2}{2L^3}$

$B_p=\frac{2\pi {\mu }_0{a}^{3}d\sigma f}{2L^3}$

$\therefore B_P=\frac{\pi {\mu }_0{a}^{3}d\sigma f}{L^3}$

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