Sea water at frequency $v=4 \times 10^8 \mathrm{~Hz}$ has permittivity $\varepsilon \approx 80 \varepsilon_0$, permeability $\alpha \approx \alpha_0$ and resistivity $\rho=0.25 \mathrm{~m}$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t)=V_0 \sin (2 \pi v t)$. What fraction of the conduction current density is the displacement current density?
Suppose distance between the parallel plates is $d$ and applied voltage $V_{(t)}=V_0 2 \pi v t$.
Thus, electric field
$$\begin{aligned} & E=\frac{V_0}{d} \sin (2 \pi v t) \\ & \text { Now using Ohm's law, } \\ & J_c=\frac{1}{\rho} \frac{V_0}{d} \sin (2 \pi v t) \\ & \Rightarrow \quad=\frac{V_0}{\rho d} \sin (2 \pi v t)=J_0^c \sin 2 \pi v t \\ \text{Here,}\quad & J_0^c=\frac{V_0}{\rho d} \end{aligned}$$
$$\begin{aligned} &\text { Now the displacement current density is given as }\\ &\begin{aligned} J_d & =\varepsilon \frac{\delta E}{d t}=\frac{\varepsilon \delta}{d t} \quad\left[\frac{V_0}{d} \sin (2 \pi v t)\right]\\ & =\frac{\varepsilon 2 \pi v V_0}{d} \cos (2 \pi v t) \\ \Rightarrow \quad & =J_0^d \cos (2 \pi v t) \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { where, } \quad J_0^d & =\frac{2 \pi V \varepsilon V_0}{d} \\ \Rightarrow \quad \frac{J_0^d}{J_0^c} & =\frac{2 \pi v \varepsilon V_0}{d} \cdot \frac{\rho d}{V_0}=2 \pi \vee \varepsilon \rho \\ & =2 \pi \times 80 \varepsilon_0 \vee \times 0.25=4 \pi \varepsilon_0 \vee \times 10 \\ & =\frac{10 v}{9 \times 10^9}=\frac{4}{9} \end{aligned}$$
A long straight cable of length $l$ is placed symmetrically along $z$-axis and has radius $a(<< l)$. The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I(t)=I_0 \sin (2 \pi v t)$ flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance $s$ from the wire inside the cable is
$$\mathbf{E}(s, t)=\alpha_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}$$
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current $I^d$.
(iii) Compare the conduction current $I_0$ with the displacement current $I_0^d$.
(i) Given, the induced electric field at a distance $r$ from the wire inside the cable is
$$\mathbf{E}(s, t)=\propto_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}$$
Now, displacement current density,
$$J_d=\varepsilon_0 \frac{d \mathbf{E}}{d t}=\varepsilon_0 \frac{d}{d t}\left[\propto _0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}\right]$$
$$\begin{aligned} &\begin{aligned} & =\varepsilon_0 \alpha_0 I_0 v \frac{d}{d t}[\cos 2 \pi v t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \\ & =\frac{1}{c^2} I_0 v^2 2 \pi[-\sin 2 \pi v t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \left[\because l_4 \frac{s}{a}=-l_4 \frac{a}{s}\right]\\ & =\frac{v^2}{c^2} 2 \pi I_0 \sin 2 \pi v t \ln \left(\frac{a}{s}\right) \hat{\mathbf{k}} \\ & =\frac{1}{\lambda^2} 2 \pi I_0 \ln \left(\frac{a}{s}\right) \sin 2 \pi v t \hat{\mathbf{k}} \\ & =\frac{2 \pi I_0}{\lambda_2} \ln \frac{a}{s} \sin 2 \pi v t \hat{\mathbf{k}} \end{aligned}\\ \end{aligned}$$
$$\begin{aligned} \text{(ii)}\quad I_d & =\int J_d s d s d \theta=\int_{s-0}^a J_d s d s \int_0^{2 \pi} d \theta=\int_{s=0}^a J_d s d s \times 2 \pi \\ & =\int_{s-0}^a\left[\frac{2 \pi}{\lambda^2} I_0 \log _e\left(\frac{a}{s}\right) s d s \sin 2 \pi v t\right] \times 2 \pi \\ \Rightarrow\quad & =\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \int_{s=0}^a\left(\frac{a}{s}\right) s d s \sin 2 \pi v t \\ & =\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \int_{s-0}^a \ln \left(\frac{a}{s}\right) \frac{1}{2} d\left(s^2\right) \cdot \sin 2 \pi v t \\ & =\frac{a^2}{2}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right) \cdot d\left(\frac{s}{a}\right)^2 \\ & =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right)^2 \cdot d\left(\frac{s}{a}\right)^2 \\ & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s-0}^a \ln \left(\frac{s}{a}\right)^2 \cdot d\left(\frac{s}{a}\right)^2 \\ & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \times(-1) \quad \left[\because \int_{s=0}^a \ln \left(\frac{s}{a}\right)^2 d\left(\frac{s}{a}\right)^2=-1\right]\\ \therefore\quad I_d & =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \\ \Rightarrow\quad & =\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0 \sin 2 \pi v t \end{aligned}$$
(iii) The displacement current,
$$\begin{array}{ll} & I_d=\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0 \sin 2 \pi v t=I_{0 d} \sin 2 \pi v t \\ \text { Here, } & I_{0 d}=\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0=\left(\frac{a \pi}{\lambda}\right)^2 I_0 \\ \therefore & \frac{I_{0 d}}{I_0}=\left(\frac{a \pi}{\lambda}\right)^2 \end{array}$$
A plane EM wave travelling in vacuum along $z$-direction is given by $\mathbf{E}=E_0 \sin (k z-\omega t) \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \sin (k z-\omega t) \hat{\mathbf{j}}$
(i) Evaluate $\int \mathbf{E} \cdot \mathbf{d l}$ over the rectangular loop 1234 shown in figure.
(ii) Evaluate $\int \mathbf{B} \cdot \mathrm{ds}$ over the surface bounded by loop 1234.
(iii) Use equation $\int \mathbf{E} \cdot \mathbf{d l}=\frac{-d \phi_B}{d t}$ to prove $\frac{E_0}{B_0}=c$.
(iv) By using similar process and the equation $\int \mathbf{B} \cdot \mathbf{d l}=\alpha_0 I+\varepsilon_0 \frac{d \phi_E}{d t}$, prove that $c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}}$
(i) Consider the figure given below
During the propagation of electromagnetic wave a long $z$-axis, let electric field vector $(\mathbf{E})$ be along $x$-axis and magnetic field vector $\mathbf{B}$ along $y$-axis, i.e., $\mathbf{E}=E_0 \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$.
Line integral of $E$ over the closed rectangular path 1234 in $x$ - $z$ plane of the figure
$$\begin{aligned} \oint \mathrm{E} \cdot \mathrm{dl} & =\int_1^2 \mathrm{E} \cdot \mathrm{dl}+\int_2^3 \mathrm{E} \cdot \mathrm{dl}+\int_3^4 \mathrm{E} \cdot \mathrm{dl}+\int_4^1 \mathrm{E} \cdot \mathrm{dl} \\ & =\int_1^2 \mathrm{E} \cdot \mathrm{dl} \cos 90+\int_2^3 \mathrm{E} \cdot \mathrm{dl} \cos 0+\int_3^4 \mathrm{E} \cdot \mathrm{dl} \cos 90+\int_4^1 \mathrm{E} \cdot \mathrm{dl} \cos 180 \Upsilon \\ & =E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-g \omega t\right)\right] \end{aligned}$$
(ii) For evaluating $\int \mathrm{B} \cdot \mathrm{ds}$, let us consider the rectangle 1234 to be made of strips of are $d s=h d z \text { each. }$
$$\begin{aligned} & \qquad \begin{aligned} \int \mathrm{B} \cdot \mathrm{ds}=\int \mathrm{B} \cdot \mathrm{ds} \cos 0 & =\int \mathrm{B} \cdot \mathrm{ds}=\int_{z_1}^{z_2} B_0 \sin (k z-\omega t) h d z \\ & =\frac{-B_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \end{aligned} \end{aligned}$$
(iii) Given, $\oint \mathrm{E} \cdot \mathrm{dl}=\frac{-d \phi_B}{d t}=-\frac{d}{d t} \oint \mathrm{~B} \cdot \mathrm{ds}$ Putting the values from Eqs. (i) and (ii), we get $$ \begin{aligned} & E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\ & =\frac{-d}{d t}\left[\frac{B_y h}{k}\left\{\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right]\right. \\ & =\frac{B_y h}{k} \omega\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\ \Rightarrow \quad E_0 & =\frac{B_0 \omega}{k}=B_y c \left(\because \frac{\omega}{k}=c\right)\\ \Rightarrow \quad \frac{E_0}{B_0} & =c \end{aligned}$$
(iv) For evaluating $\boldsymbol{\rho B} \cdot \mathrm{dl}$, let us consider a loop 1234 in $y-z$ plane as shown in figure given below
$$\begin{aligned} \oint \mathbf{B} \cdot \mathbf{d l} & =\int_1^2 \mathbf{B} \cdot \mathbf{d l}+\int_2^3 \mathbf{B} \cdot \mathbf{d l}+\int_3^4 \mathbf{B} \cdot \mathbf{d l}+\int_4^1 \mathbf{B} \cdot \mathbf{d l} \\ & =\int_1^2 \mathbf{B} \cdot \mathbf{d l} \cos 0+\int_2^3 \mathbf{B} \cdot \mathbf{d l} \cos 90 \Upsilon+\int_3^4 \mathbf{B} \cdot \mathbf{d l} \cos 180 \Upsilon_{+}+\int_4^1 \mathbf{B} \cdot \mathbf{d l} \cos 90 \Upsilon \\ & =B_0 h\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right.\quad\text{.... (iii)} \end{aligned}$$
Now to evaluate $\phi_E=\int \mathbf{E} \cdot \mathrm{ds}$, let us consider the rectangle 1234 to be made of strips of area $h d_2$ each.
$$ \begin{aligned} & \phi_E=\int \mathrm{E} \cdot \mathrm{ds}=\int E d s \cos 0=\int E d s=\int_{z_1}^{z_2} E_0 \sin \left(k z_1-\omega t\right) h d z \\ & =-\frac{E_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \\ & \therefore \quad \frac{d \phi_E}{d t}=\frac{E_0 h \omega}{k}\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right] \quad\text{.... (iv)}\\ & \text { In } \quad \oint \mathrm{B} \cdot \mathrm{dl}=\alpha_0\left(I+\frac{\varepsilon_0 d \phi_E}{d t}\right), I=\text { conduction current } \\ & =0 \text { in vacuum } \\ & \therefore \quad \quad \quad \oint\mathrm{B} \cdot \mathrm{dl}=\alpha_0 \varepsilon \frac{d \phi_E}{d t} \end{aligned}$$
Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get
$$\begin{aligned} B_0 & =E_0 \frac{\omega \propto_0 \varepsilon_0}{k} \\ \Rightarrow\quad \frac{E_0}{B_0} \frac{\omega}{k} & =\frac{1}{\propto_0 \varepsilon_0} \\ \text{But}\quad\frac{E_0}{B_0} & =c \text { and } \omega=c k \\ \Rightarrow\quad c \cdot c & =\frac{1}{\alpha_0 \varepsilon_0}, \text { therefore } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \end{aligned}$$
A plane EM wave travelling along $z$-direction is described by $\mathbf{E}=E_0 \sin (k z-\omega t) \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \sin (k z-\omega t) \hat{\mathbf{j}}$. Show that
(i) the average energy density of the wave is given by
$$u_{\mathrm{av}}=\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{B_0^2}{\alpha_0}$$
(ii) the time averaged intensity of the wave is given by
$$I_{\mathrm{av}}=\frac{1}{2} c \varepsilon_o E_0^2 .$$
(i) The electromagnetic wave carry energy which is due to electric field vector and magnetic field vector. In electromagnetic wave, $E$ and $B$ vary from point to point and from moment to moment. Let $E$ and $B$ be their time averages.
The energy density due to electric field $E$ is
$$u_E=\frac{1}{2} \varepsilon_0 E^2$$
The energy density due to magnetic field $B$ is
$$u_B=\frac{1}{2} \frac{B^2}{\alpha_0}$$
Total average energy density of electromagnetic wave
$$u_{a v}=u_E+u_B=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\alpha_0}$$
Let the EM wave be propagating along $z$-direction. The electric field vector and magnetic field vector be represented by
$$\begin{aligned} & E=E_0 \sin (k z-\omega t) \\ & B=B_0 \sin (k z-\omega t) \end{aligned}$$
The time average value of $E^2$ over complete cycle $=\frac{E_0^2}{2}$
and time average value of $B^2$ over complete cycle $=\frac{B_0^2}{2}$
$$\begin{aligned} u_{\mathrm{av}} & =\frac{1}{2} \frac{\varepsilon_0 E_0^2}{2}+\frac{1}{2} \alpha_0\left(\frac{B_0^2}{2}\right) \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{B_0^2}{4 \alpha_0} \end{aligned}$$
$$\begin{aligned} &\text { (ii) We know that } E_0=c B_0 \text { and } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}}\\ &\begin{aligned} & \therefore \quad \frac{1}{4} \frac{B_0^2}{\alpha_0}=\frac{1}{4} \frac{E_0^2 / C^2}{\alpha_0}=\frac{E_0^2}{4 \propto_0} \times \propto_0 \varepsilon_0=\frac{1}{4} \varepsilon_0 E_0^2 \\ & \therefore \quad u_B=u_E \end{aligned} \end{aligned}$$
$$\begin{aligned} \text{Hence,}\quad U_{\mathrm{av}} & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{B_0^2}{\alpha_0} \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \varepsilon_0 E_0^2 \\ & =\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\alpha_0} \end{aligned}$$
$$\begin{aligned} &\text { Time average intensity of the wave }\\ &I_{\mathrm{av}}=U_{\mathrm{av}} C=\frac{1}{2} \varepsilon_0 E_0^2 C=\frac{1}{2} \varepsilon_0 E_0^2 \end{aligned}$$