Question

A charge of 3.14 × 10−6 C 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

Explanation:

Step 1:

Given data in the question  

Charges of Magnitude, $q=3.14 \times 10^{-6} C$

Radius of the circle , $r=20 \mathrm{cm}=20 \times 10^{-2} \mathrm{m}$

Angular velocity of the circle,

$\omega=60 \mathrm{rad} / \mathrm{s}$

Time to Transition $1=\frac{2 \pi}{60}$

Step 2:

$\therefore \text { Current, } i=\frac{q}{t}=\frac{3.14 \times 10^{-6} \times 60}{2 \pi}=30 \times 10^{-6} \mathrm{A}$

The small element 1 and 2 of the ring respectively, E1 and E2 denote the electric area at a point on the axis at a distance of 5.00 cm from the centre.

E is the resulting electric area at a stage on the axis, at a distance of 5.00 cm from the center due to the whole ring.

Step 3:

The electric area at a stage on the axis is x distant from the centre

$E=\frac{x q}{4 \pi \varepsilon_{0}\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$

Step 4:

The magnetic field is given at a stage on the axis x distant from the centre

$B=\frac{\mu_{0}}{2} \frac{i r^{2}}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$

$\frac{E}{B}=\frac{\frac{x q}{4 \pi \varepsilon_{0}\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}}{\frac{\mu_{0}}{2} \frac{i r^{2}}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}}$

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

$=\frac{9 \times 3.14 \times 2 \times(20.6)^{3} \times 10^{15}}{25 \times 4 \pi \times 12}$

$=1.88 \times 10^{15} \mathrm{m} / \mathrm{s}$