Question

A charge of $3.14\times {10}^{-6C}$ is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60.0 rad $s^{-1}.$ Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the centre.

Answer 1

Given $q=3.14\times {10}^{-6}C,$

$r=20cm=20\times {10}^{-2}m$

$\omega =60rad/sec.$

So, time to take 1 revolution = $\frac{2\pi }{60}$

$\therefore$ Current $i=\frac{q}{t}=\frac{3.14\times {10}^{-6}\times 60}{2\pi }$

$=30\times {10}^{-6}A.$

Then electric field

$E=\frac{q}{4\pi {\in }_0{r}^2}$

$=\frac{9\times {10}^9\times 3.14\times {10}^{-8}}{25\times {10}^{-4}}$

Magnetic field

$B=\frac{{\mu }_0}{2}.\frac{i{r}^2}{(x^2+r^2{)}^{3/2}}$

$=\frac{4\pi \times {10}^{-7}\times 30\times {10}^{-6}\times 20\times 20\times {10}^{-4}}{2[5\times 5\times {10}^{-4}+20\times 20\times {10}^{-4}{]}^{3/2}}$

$=\frac{4\pi \times 12\times {10}^{-14}}{2[425{]}^{3/2}\times {10}^{-6}}$

$=\frac{4\pi \times 12\times {10}^{-14}}{2(20.6{)}^3\times {10}^{-6}}$

$\frac{E}{B}=\frac{9\times {10}^9\times 3.14\times {10}^{-6}\times 2\times (20.6{)}^3\times {10}^{-6}}{25\times {10}^{-4}\times 4\pi \times {10}^{-14}\times 12}$

$=\frac{9\times 3.14\times 2\times (20.6{)}^3}{25\times 4\pi \times 12}$

$=1.31\times {10}^{15}$