Question

A non-conducting disc of radius 2 $m$ is rotated about its axis of symmetry perpendicular to plane of disc with uniform angular speed $\omega$. The disc carries a charge whose density is given as $\sigma =A-Br,$ where $A$ and $B$ are positive constants and $r$ is distance from centre of the disc. The value of $\frac{A}{B}$ for which the magnetic field at the centre of the disc will be zero is

• A. 1
• B. 2
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A 1


$\begin{array}{cc}dI=& \frac{dq}{\frac{2\pi }{\omega }}=\frac{\omega }{2\pi }(\sigma \cdot 2\pi rdr)\\ & =\omega (A-Br)rdr\end{array}$
$\left(\begin{array}{cc}\because dq& =\sigma .2\pi rdr\\ & =(A-Br)\cdot 2\pi rdr\end{array}\right)$

For a circular current carrying loop,
magnetic field at center of loop will be
$dB=\frac{{\mu }_0}{2r}dI$
$\begin{array}{cc}& ={\mu }_0\omega (A-Br)rdr\\ \Rightarrow B& =\frac{{\mu }_0\omega }{2}{\int }_0^2(A-Br)dr\\ & =\frac{{\mu }_0\omega }{2}(2A-2B)\\ \therefore B& =0\text{for}\frac{A}{B}=1\end{array}$