Even though an electric field E exer ts a force $q \mathrm{E}$ on a charged particle yet electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.
Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure.
An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the $z$-axis (figure). The wire is set into motion along its length with a uniform velocity $v=v \hat{\mathbf{k}}_z$. Calculate the pointing vector $\mathbf{S}=\frac{1}{\alpha_0}(\mathbf{E} \times \mathbf{B})$
$$\begin{aligned} \text{Given,}\quad\mathbf{E} & =\frac{\lambda \hat{\mathbf{e}}_S}{2 \pi \varepsilon_0 a} \hat{\mathbf{j}} \\ \mathbf{B} & =\frac{\propto_0 \mathrm{i}}{2 \pi a} \hat{\mathbf{i}}=\frac{\propto_0 \lambda V}{2 \pi a} \hat{\mathbf{i}} \quad\left[\because I=\lambda_V\right]\\ \therefore\quad\mathbf{S} & =\frac{1}{\propto_0}[\mathbf{E} \times \mathbf{B}]=\frac{1}{\propto_0}\left[\frac{\lambda_{\hat{\mathbf{j}}}}{2 \pi \varepsilon_0 a} \times \frac{\propto_0}{2 \pi a} \lambda \sqrt{\hat{\mathbf{i}}}\right] \\ & =\frac{\lambda^2 V}{4 \pi^2 \varepsilon_0 a^2}(\hat{\mathbf{j}} \times \hat{\mathbf{i}})=-\frac{\lambda^2 V}{4 \pi^2 \varepsilon_0 \mathrm{a}^2} \hat{\mathbf{k}} \end{aligned}$$
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
$$\begin{aligned} B_0 & =E_0 \frac{\omega \propto_0 \varepsilon_0}{k} \\ \Rightarrow\quad \frac{E_0}{B_0} \frac{\omega}{k} & =\frac{1}{\propto_0 \varepsilon_0} \\ \text{But}\quad\frac{E_0}{B_0} & =c \text { and } \omega=c k \\ \Rightarrow\quad c \cdot c & =\frac{1}{\alpha_0 \varepsilon_0}, \text { therefore } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \end{aligned}$$