Refer to Prob. 7.16, to which the correct answer was

$\mathit{E}\mathit{(}\mathit{s}\mathit{,}\mathit{t}\mathit{)}\mathit{=}\frac{{\mathit{\mu }}_{\mathit{0}}{\mathit{I}}_{\mathit{0}}\mathit{\omega }}{\mathit{2}\mathit{\tau }\mathit{\tau }}\mathit{s}\mathit{i}\mathit{n}\mathit{\left(}\mathit{\omega }\mathit{t}\mathit{\right)}\mathit{I}\mathit{n}\mathit{\left(}\frac{\mathit{a}}{\mathit{s}}\mathit{\right)}\hat{\mathit{z}}$

(a) Find the displacement current density ${\mathit{J}}_{\mathbf{d}}$·

(b) Integrate it to get the total displacement current,

${\mathit{I}}_{\mathbf{d}}\mathbf{=}\mathbf{\int }{\mathit{J}}_{\mathbf{d}}\mathbf{.}\mathit{d}\mathit{a}$

Compare ${\mathit{I}}_{\mathbf{d}}$ and I. (What's their ratio?) If the outer cylinder were, say, 2 mm in diameter, how high would the frequency have to be, for${\mathit{I}}_{\mathbf{d}}$ to be 1% of I ? [This problem is designed to indicate why Faraday never discovered displacement currents, and why it is ordinarily safe to ignore them unless the frequency is extremely high.]

The electric field for a current carrying straight wire is,$E\left(s,t\right)=\frac{{\mu }_0{\in }_0\omega }{2\pi }\sin \left(\omega t\right)\mathrm{In}\left(\frac{a}{s}\right)\hat{z}$.

The diameter of the outer cylinder is, $\mathrm{d}=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}$.

The ratio $I_d$of and Iis,$\frac{I_d}{I}=1\%=\frac{1}{100}$.

Consider the electric field inside the conducting material of a capacitor changes, then a certain amount of current is produced. The current produced in the conductor is described as the ‘displacement current’.

The formula for the displacement current $\left(I_d\right)$is given by,

$I_d={\in }_0\frac{d{\Phi }_E}{dt}$

Here, ${\Phi }_E$ represents the electric flux and${\in }_0$ is the permittivity of vacuum.\

The given expression for the electric field having current flowing down the straight wire of radius and length is given by,

$E\left(s,t\right)=\frac{{\mu }_0{I}_0\omega }{2\pi }\sin \left(\omega t\right)\mathrm{In}\left(\frac{a}{s}\right)\hat{z}$

Then the formula for the displacement current density is given by,

$$\begin{gathered} J_d={\in }_0\frac{dE}{dt} \\ {J}_d={\in }_0\frac{d}{dt}\left(\frac{{\mu }_0{I}_0\omega }{2\pi }\sin \left(\omega t\right)In\left(\frac{a}{s}\right)\hat{z}\right) \end{gathered}$$

Solving it,

$$\begin{gathered} J_d={\in }_0\frac{{\mu }_0{I}_0\omega }{2\pi }\cos \left(\omega t\right)\mathrm{In}\left(\frac{a}{s}\right)\hat{z} \\ {J}_d={\in }_0\frac{{\mu }_0{\omega }^2}{2\pi }\left(I_0\cos \left(\omega t\right)\right)\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{s}}\right)\hat{\mathrm{z}} \end{gathered}$$

Putting, in expression,

$J_d=\frac{{\mu }_0{\in }_0{\omega }^{2}l}{2\pi }\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{s}}\right)\hat{\mathrm{z}}$

Hence, the displacement current density is $\frac{{\mu }_0{\in }_0{\omega }^{2}l}{2\pi }\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{s}}\right)\hat{\mathrm{z}}$.

Integrating the formula for the displacement current density $\left(J_d\right)$over a small area $\left(d_a\right)$to get the total displacement current $\left(I_d\right)$,

$$\begin{gathered} I_d=\int J_d.da \\ {I}_d=\int \frac{{\mu }_0{\in }_0{\omega }^{2}I}{2\pi }\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{s}}\right)\hat{\mathrm{z}}.\mathrm{da} \end{gathered}$$

Putting, $\hat{z}.da=2\pi s.ds$

Taking constant terms out of integral and integrating between 0 to a,

$I_d=\frac{{\mu }_0{\in }_0{\omega }^{2}I}{2\pi }{\int }_0^a\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{s}}\right)\left(2\pi s.ds\right)$

$$\begin{gathered} I_d={\mu }_0{\in }_0{\omega }^{2}I{\int }_0^a\left(\mathrm{In}a-\mathrm{In}s\right)\left(s.ds\right) \\ {I}_d={\mu }_0{\in }_0{\omega }^{2}I{\int }_0^a\left(s.\mathrm{In}a-s.\mathrm{In}s\right)ds \end{gathered}$$

Solving the integral,

$$\begin{gathered} I_d={\mu }_0{\in }_0{\omega }^{2}I{\left[\frac{s^2}{2}\left(\mathrm{In}a\right)-\frac{s^2}{2}\left(\mathrm{In}s\right)+\frac{s^2}{4}\right]}_0^a \\ {I}_d={\mu }_0{\in }_0{\omega }^{2}I\left[\frac{a^2}{2}\left(\mathrm{In}a\right)-\frac{a^2}{2}\left(\mathrm{In}a\right)+\frac{a^2}{4}\right] \\ {I}_d=\frac{{\mu }_0{\in }_0{\omega }^{2}I{a}^2}{4} \end{gathered}$$

Hence, the total displacement current is $\frac{{\mu }_0{\in }_0{\omega }^{2}I{a}^2}{4}$.

According to the question, the ratio of $I_d$ and I is given by,

$$\begin{gathered} \frac{I_d}{I}=\frac{\frac{{\mu }_0{\in }_0{\omega }^{2}I{a}^2}{4}}{I} \\ \frac{I_d}{I}=\frac{{\mu }_0{\in }_0{\omega }^2{a}^2}{4} \end{gathered}$$

It is known that, ${\mu }_0{\in }_0=\frac{1}{c^2}$, here c is speed of light. So,

$$\begin{gathered} \frac{I_d}{I}=\frac{{\omega }^2{a}^2}{4c^2} \\ \frac{I_d}{I}={\left(\frac{\omega a}{2c}\right)}^2 \end{gathered}$$

It is given that $\frac{I_d}{I}=\frac{1}{100}$, so,

$$\begin{gathered} \frac{1}{100}={\left(\frac{\omega a}{2c}\right)}^2 \\ \frac{1}{10}=\frac{\omega a}{2c} \\ \frac{\omega a}{2c}=\frac{1}{10} \\ \omega =\frac{2c}{10a} \end{gathered}$$

Putting the value of radius $a=\frac{2\times {10}^{-3}}{2}={10}^{-3}\mathrm{m}$, and speed of light

$$\begin{gathered} c=3\times {10}^8\frac{m}{s}, \\ \omega =\frac{2\times 3\times {10}^8{\displaystyle \frac{m}{s}}}{10\times {10}^{-3}\mathrm{m}} \\ \omega =0.6\times {10}^{11}{\mathrm{s}}^{-1} \\ \mathrm{\omega }=6\times {10}^{10}{\mathrm{s}}^{-1} \end{gathered}$$

Frequency in Hertz is given by,

$$\begin{gathered} v=\frac{\omega }{2\pi } \\ v=\frac{6\times {10}^{10}{s}^{-1}}{2\pi } \\ v=0.955\times {10}^{10}\mathrm{Hz} \end{gathered}$$

It can be written as,

$$\begin{gathered} v\approx {10}^{10}\mathrm{Hz} \\ \mathrm{v}\approx {10}^4\mathrm{MHz} \end{gathered}$$

Hence, the value of the frequency has to be ${10}^4\mathrm{MHz}$.