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
$B=\frac{{\mu }_{0}nq}{r^2}\underset{0}{\overset{r}{\int }}dx=\frac{{\mu }_{0}nq}{r^2}{\left(x\right)}_0^r=\frac{{\mu }_{0}nq}{r}$