Question

A non-conducting disc of radius 1m has charge distributed over its surface. The surface density of charge at a point is given by $\sigma =A–Br;0\le r\le 1$ where r is the distance of point from the centre of the disc. The disc is rotated with an angular velocity $\omega$ about an axis passing through its centre and perpendicular to disc. Then

• A. Magnetic field at the centre of the disc is zero if $\frac{A}{B}=\frac{4}{5}$.
• B. Magnetic field at the centre of the disc is zero if $\frac{A}{B}=\frac{1}{2}$.
  
• C. Magnetic moment due to disc is zero if $\frac{A}{B}=\frac{1}{2}$.
• D. Magnetic moment due to disc is zero if $\frac{A}{B}=\frac{4}{5}$.

Answer 1

Answer: (B), (D)

The correct options are
B

Magnetic field at the centre of the disc is zero if $\frac{A}{B}=\frac{1}{2}$.

D

Magnetic moment due to disc is zero if $\frac{A}{B}=\frac{4}{5}$.

$B=\int \frac{{\mu }_0}{2}\frac{di}{r}di=\frac{\omega }{2\pi }dq=\frac{\omega }{2\pi }dq=\frac{\omega }{2\pi }(A-Br)\left(2\pi r\right)drB={\int }_0^1\left(\frac{{\mu }_0}{2}\right)\left(\frac{\omega }{2\pi }\right)\frac{(A-Br)\left(2\pi r\right)dr}{r}=\left(\frac{{\mu }_0}{2}\right)\left(\frac{\omega }{2\pi }\right)\left(2\pi \right)[\int Adr-\int Brdr]=\left(\frac{{\mu }_0}{2\pi }\right)\left(\frac{\omega }{2\pi }\right)\left(2\pi \right)[A-\frac{B}{2}]A-\frac{B}{2}=0,ifmagneticfield=0\frac{A}{B}=\frac{1}{2}M=\int \left(di\right)A=\int \left(\pi r^2\right)\frac{\omega }{2\pi }dq=\int \frac{\left(\pi r^2\right)\omega (A-Br)\left(2\pi r\right)dr}{2\pi }=\pi \omega {\int }_0^1(A{r}^3-B{r}^4)dr=\pi \omega (\frac{A}{4}-\frac{B}{5})$
If $\frac{A}{B}=\frac{4}{5},M=0$