Question

A uniformly charged ring of radius $0.1m$ rotates at a frequency of ${10}^{4}rps$ about its axis. Find the ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance $0.2m$ from the centre. (Use speed of light $c=3\times {10}^{8}m/s,{\pi }^2=10$)

Answer 1

Energy density of $\overrightarrow{E}=\frac{1}{2}{\epsilon }_0{\left(E\right)}^2$

where $\overrightarrow{E}=\frac{KxQ}{{(x^2+a^2)}^{3/2}}$

where $a:$ radius $=0.1m$

$x:$ point of measurement $=0.2m$

Now, Energy density of $\overrightarrow{B}=\frac{1}{2}\frac{B^2}{{\mu }_0}$

$\therefore \frac{Energy\overrightarrow{E}}{Energy\overrightarrow{B}}=\frac{{\epsilon }_0{E}^2}{B^2}{\mu }_0$

$\therefore {\mu }_0{\epsilon }_0=\frac{1}{C^2}$

$\therefore \frac{E_{\overrightarrow{E}}}{E_{\overrightarrow{B}}}=\frac{1}{C^2}\frac{{\overrightarrow{E}}^2}{{\left(\overrightarrow{B}\right)}^2}⟶\left(1\right)$

$\frac{\overrightarrow{E}}{\overrightarrow{B}}=\frac{KxQ}{{(x^2+a^2)}^{3/2}}÷\frac{{\mu }_{0}Q{a}^{3}w}{4\pi a{(a^2+x^2)}^{3/2}}$

$=\frac{xQ}{{\epsilon }_0}÷\frac{{\mu }_{0}Q{a}^{3}w}{a}$

$=\frac{xQ}{{\epsilon }_0}\times \frac{1}{{\mu }_{0}Q{a}^{2}w}=\frac{x}{a^2\left({\epsilon }_0{\mu }_0\right)w}$

where $w:$ angular frequency $=2r\left({10}^4\right)$

$\therefore$ Substituting all the values.

$=\frac{0.2C^2}{{\left(0.1\right)}^{2}2\pi {10}^4}⟶\left(2\right)$

Substituting (2) in (1)

$\frac{E_{\overrightarrow{E}}}{E_{\overrightarrow{B}}}=\frac{1}{C^2}{\left\{\frac{0.2}{{\left(0.1\right)}^2}\frac{C^2}{2\pi {10}^4}\right\}}^2$

$={\left(\frac{20\times C}{2r{10}^4}\right)}^2$

$={\left(\frac{10C}{\pi {10}^4}\right)}^2$

$=\frac{{(3\times {10}^8)}^2}{10\times {10}^6}=9\times {10}^9$

$\therefore$ Ratio $=9\times {10}^9$ is the correct answer.