Question

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

• A. $\frac{{\mu }_{0}qn}{a}$
• B. $\frac{{\mu }_{0}qn}{2a}$
• C. $\frac{{\mu }_{0}qn}{4a}$
• D. $\frac{3{\mu }_{0}qn}{4a}$

Answer 1

Answer: (A)

$\frac{{\mu }_{0}qn}{a}$

Surface charge density $\sigma =\frac{q}{\pi a^2}$
Charge on the hypothetical ring $=\frac{q}{\pi a^2}2\pi xdx$
Frequency of rotation$=n,T=\frac{1}{n}$
$dI=\frac{q}{T}=\frac{q}{\frac{1}{n}}=nq$
Magnetic field due to the element
$dB=\frac{{\mu }_{0}dI}{2x}=\frac{{\mu }_{0}2xdxqn}{a^2\left(2x\right)}=\frac{{\mu }_{0}qndx}{a^2}$
$B=\int dB=\frac{{\mu }_{0}qn}{a^2}{\int }_0^{a}dx=\frac{{\mu }_{0}qn}{a^2}[x{]}_0^a=\frac{{\mu }_{0}qn}{a}$