Show that the magnetic field $B$ at a point in between the plates of a parallel plate capacitor during charging is $\frac{\propto_0 \varepsilon_0 r}{2} \frac{d E}{d t}$ (symbols having usual meaning).
Consider the figure ginen below to prove that the magneti field $B$ at a point in between the plater of a paravel- plate copocior during charging is $\frac{\varepsilon_0 \propto_0 r}{2} \frac{d E}{d t}$
Let $I_d$ be the displacement current in the region between two plates of parallel plate capacitor, in the figure.
The magnetic field induction at a point in a region between two plates of capacitor at a perpendicular distance $r$ from the axis of plates is
$$\begin{array}{rlr} B & =\frac{\propto_0 2 I_d}{4 \pi r}=\frac{\propto_0}{2 \pi r} I_d=\frac{\propto_0}{2 \pi r} \times \varepsilon_0 \frac{d \phi_E}{d t} & {\left[\because I_d=\frac{E_0 d \phi_E}{d t}\right]} \\ & =\frac{\propto_0 \varepsilon_0}{2 \pi r} \frac{d}{d t}\left(E \pi r^2\right)=\frac{\propto_0 \varepsilon_0}{2 \pi r} \pi r^2 \frac{d E}{d t} & {\left[\because \phi_E=E \pi r^2\right]} \\ B & =\frac{\propto_0 \varepsilon_0 r}{2} \frac{d E}{d t} & \end{array}$$
Electromagnetic waves with wavelength
(i) $\lambda_1$ is used in satellite communication.
(ii) $\lambda_2$ is used to kill germs in water purifies.
(iii) $\lambda_3$ is used to detect leakage of oil in underground pipelines.
(iv) $\lambda_4$ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.
(a) (i) Microwave is used in satellite communications.
So, $\lambda_1$ is the wavelength of microwave.
(ii) Ultraviolet rays are used to kill germs in water purifier. So, $\lambda_2$ is the wavelength of UV rays.
(iii) $X$-rays are used to detect leakage of oil in underground pipelines. So, $\lambda_3$ is the wavelength of $X$-rays.
(iv) Infrared is used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.
(b) Wavelength of X-rays < wavelength of UV $<$ wavelength of infrared $<$ wavelength of microwave.
$$\Rightarrow \quad \lambda_3<\lambda_2<\lambda_4<\lambda_1 $$
(c) (i) Microwave is used in radar.
(ii) UV is used in LASIK eye surgery.
(iii) X-ray is used to detect a fracture in bones.
(iv) Infrared is used in optical communication.
Show that average value of radiant flux density $S$ over a single period $T$ is given by $S=\frac{1}{2 c \propto_0} E_0^2$.
Radiant flux density $S=\frac{1}{\propto_0}(\mathbf{E} \times \mathbf{B})=C^2 \varepsilon_0(\mathbf{E} \times \mathbf{B})$
$$\left[\because c=\frac{1}{\sqrt{\propto_0 \varepsilon_0}}\right]$$
Suppose electromagnetic waves be propagating along $x$-axis. The electric field vector of electromagnetic wave be along $y$-axis and magnetic field vector be along $z$-axis. Therefore,
$$\begin{aligned} \mathbf{E}_0 & =\mathbf{E}_0 \cos (k x-\omega t) \\ \text{and}\quad\mathbf{B} & =\mathbf{B}_0 \cos (k x-\omega t) \\ \mathbf{E} \times \mathbf{B} & =\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \\ \mathbf{S} & =\mathbf{c}^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\ & =c^2 \varepsilon_0\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \end{aligned}$$
Average value of the magnitude of radiant flux density over complete cycle is
$$\begin{array}{rlrl} S_{\mathrm{av}} & =c^2 \varepsilon_{\mathrm{o}}\left|\mathbf{E}_0 \times \mathbf{B}_0\right| \frac{1}{T} \int_0^T \cos ^2(k x-\omega t) d t \\ & =c^2 \varepsilon_0 E_0 B_0 \times \frac{1}{T} \times \frac{T}{2} \quad\left[\because \int_0^T \cos ^2(k x-\omega t) d t=\frac{T}{2}\right] \\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{c^2}{2} \varepsilon_0 E_0\left(\frac{E_0}{c}\right) \quad\left[\text { As, } C=\frac{E_0}{B_0}\right]\\ & =\frac{c}{2} \varepsilon_0 E_0^2=\frac{c}{2} \times \frac{1}{c^2 \alpha_0} E_0^2 \quad {\left[C=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{c^2 \alpha_0}\right]}\\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{E_0^2}{2 \alpha_0 c} \end{array}$$
Hence proved.
You are given a $2 \propto F$ parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
Given, capacitance of capacitor $C=2 \propto F$,
Displacement current $I_d=1 \mathrm{~mA}$
Charge $$q=C V$$
$$\begin{array}{rlr} I_d d t & =C d V & {[\because q=i t]} \\ \text{or} \quad I_d & =C \frac{d V}{d t} \\ 1 \times 10^{-3} & =2 \times 10^{-6} \times \frac{d V}{d t} \\ \text{or}\quad\frac{d V}{d t} & =\frac{1}{2} \times 10^{+3}=500 \mathrm{~V} / \mathrm{s} & \end{array}$$
So, by applying a varying potential difference of 500 V/s, we would produce a displacement current of desired value.
Show that the radiation pressure exerted by an EM wave of intensity $I$ on a surface kept in vacuum is $\frac{I}{C}$.
$$\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}$$
Force is the rate of change of momentum i.e.,
$$\begin{aligned} \text{i.e.,}\quad F & =\frac{d p}{d t} \\ \text{Energy in time dt,}\quad U & =p \cdot C \text { or } p=\frac{U}{C} \\ \therefore\quad\text { Pressure } & =\frac{1}{A} \cdot \frac{U}{C \cdot d t} \\ \text { Pressure } & =\frac{I}{C} \quad\left[\because I=\text { Intensity }=\frac{U}{A \cdot d t}\right] \end{aligned}$$