Question

A point charge of $0.1\text{C}$ is placed on the circumference of a non-conducting ring of radius $1m$ which is rotating with a constant angular acceleration of $1{\text{rad/sec}}^2$. If ring starts it's motion at $t=0$, the magnetic field at the centre of the ring at $t=10$ sec is ${10}^{-x}$. Then $x$ is

Answer 1

Angular velocity at $t=10s$
$\omega =0+1\times 10=10{\text{rad/sec}}^2$
Linear velocity of charge at $t=10s$
$\therefore v=r\omega =1\times 10=10\text{m/s}$
Magnetic field at the centre of the ring
$\overline{B}=\frac{{\mu }_0}{4\pi }\frac{q(\overline{v}\times \overline{r})}{r^3}$
$\Rightarrow \left|\overline{B}\right|=\frac{{\mu }_{0}qv}{4\pi r^2}$
$=\frac{{10}^{-7}\times 0.1\times 10}{(1{)}^2}={10}^{-7}\text{T}$