Question

A thin disc having radius $r$ and charge $q$ distributed uniformly over the disc is rotated $n$ rotations per second about its axis. The magnetic field at the centre of the disc is

• A. $\frac{\mu_{0}qn}{2r}$
• B. $\frac{\mu_{0}qn}{r}$
• C. $\frac{3\mu_{0}qn}{r}$
• D. $\frac{3\mu_{0}qn}{4r}$

Answer 1

Answer: (B)

Consider a hypothetical ring of radius $x$ and thickness $dx$ on a disc as shown in figure.

Charge on the ring,
$dq=\frac{q}{\pi r^{2}}\times\left(2\pi xdx\right)$
Current due to rotation of charge on ring is
$dI=\frac{dq}{T}=\frac{dq}{1 /n}=ndq=\frac{nq2xdx}{r^{2}}$
Magnetic field at the centre $O$ due to current onring element is
$dB=\frac{\mu_{0} dI}{2x}=\frac{\mu_{0} nq2xdx}{r^{2}\left(2x\right)}=\frac{\mu_{0} nqdx}{r^{2}}$
Total magnetic field induction due to current on whole disc is
$B=\frac{\mu_{0}nq}{r^{2}} \int\limits_{0}^{r} dx=\frac{\mu_{0} nq}{r^{2}}\left(x\right)_{0}^{r}=\frac{\mu_{0} nq}{r}$