Physics NCERT Exemplar Solutions Class 12th Chapter Eight Electromagnetic Waves

1. An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (figure). The wire is set into motion along its length with a uniform velocity v = v z_k . Calculate the pointing vector S = $\frac{1}{{\mathit{\mu }}_0}$ (E $\times \mathit{B}$ )

2. Sea water at frequency v = 4×10⁸ Hz has permittivity ε≈80 ${\mathit{\epsilon }}_0$ , permeability µ≈ ${\mathit{\mu }}_0$ and resistivity ρ = 0.25 m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V (t)= V₀sin(2 $\mathit{\pi }\mathit{v}\mathit{t}$ ). What fraction of the conduction current density is the displacement current density?

Explanation- suppose the distance between the plates is d and applied voltage is V_t= V₀2

Then electric field is E= $\frac{V}{d}$ sin(2 $\pi vt$ )

J_c= $\frac{1}{\rho }\frac{V_0}{d}sin\left(2\pi vt\right)$

  =  $J_0^{c}sin2\pi vt$

$J_0^c$   = $\frac{V_0}{\rho d}$

J_d= $\epsilon \frac{dE}{dt}$ = $\frac{V_0}{\rho d}$

   = $\epsilon \frac{2\pi v{V}_0}{d}$ cos(2 $\pi vt$ )

   = J₀^d cos 2 $\pi vt$

J₀^d= $\frac{2\pi V_{\epsilon }{V}_0}{d}$

$\frac{J_0^d}{J_0^c}$ = 2  $\pi V_{\epsilon }\rho$ = 2  $\pi 80{\epsilon }_0$ v  $\times$ 0.25 = 4  $\pi {\epsilon }_0$ v  $\times 10$ =4/9

3. A long straight cable of length l is placed symmetrically along z-axis and has radius a(<0

The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t) = Io sin (2πνt) 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 E(s,t)=  μ_oI_ovcos (2πνt) In (s/a)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 Id.

(iii) Compare the conduction current I_o with the displacement current I₀^d .

Explanation- E(s,t)= ${\mu }_o{I}_o\cos \left(2\pi vt\right)In\left(\frac{s}{a}\right)k$

Now displacement current J_d=e_{o $\frac{dE}{dt}=\frac{{\epsilon }_{o}d}{dt}\left[{\mu }_o{I}_o\cos \left(2\pi vt\right)In\left(\frac{s}{a}\right)k\right]$}

= $\frac{{\mu }_o{\epsilon }_o{I}_{o}vd}{dt}\left[cos2v\pi tIn\left(\frac{s}{a}\right)k\right]$

= $\frac{v^2}{c^2}2\pi I_{o}sin2\pi vtIn\left(\frac{a}{s}\right)k$

${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$ = $\frac{2\pi I_0}{{\lambda }_2}$ ina/s sin2 $\pi vtk$

I_d= $\int J_{d}sdsd\theta ={\int }_0^1{J}_{s}sds{\int }_0^{2\pi }d\theta$

  =( $\frac{2\pi }{\lambda }$ )^{2 $I_o{\int }_0^a\frac{a}{s}sdssin\pi vt$}

= $\frac{a^2}{2}{\left(\frac{2\pi }{\lambda }\right)}^2{I}_{o}sin2\pi vt{\int }_0^{a}In\left(\frac{a}{s}\right).d({\frac{s}{a})}^2$

After solving this we get

I_d= $\frac{a^2}{4}({\frac{2\pi }{\lambda })}^2{I}_{0}sin2\pi vt$

I_d= ${\left(\frac{2\pi a}{2\lambda }\right)}^2{I}_{o}sin2\pi vt=I_{od}sin2\pi vt$

I_od= ( ${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$

$\frac{I_{od}}{I_o}=({\frac{a\pi }{\lambda })}^2$

4. A plane EM wave travelling in vacuum along z direction is given by E= Eosin(kz-wt)i and B=Bosin(kz-wt)j

(i) Evaluate $\int E.dl$over the rectangular loop 1234 shown in figure

(ii) Evaluate $\int B.ds$over the surface bounded by loop 1234.

(iii) Use equation $\int E.dl=-\frac{d\phi }{dt}toprovethat\frac{E_o}{B_0}=c$

(iv) By using similar process and the equation $\int B.dl={\mu }_{o}I+{\epsilon }_o\frac{d\varnothing }{dt}provethatc=\frac{1}{\text{exception 25:}}$

Explanation-(i) $\left(i\right)\int E.dl$ = ${\int }_1^{2}E.dl+{\int }_2^{3}E.dl+{\int }_3^{4}E.dl+{\int }_4^{1}E.dl$

= E_oh[sin(kz₂-wt)-sin(kz₁-gwt)]

(ii) $\int B.ds=\int B.dscos0={\int }_{z1}^{z2}{B}_0\mathrm{s}\mathrm{i}\mathrm{n}(kz-wt)hdz$

= $\frac{-B_{o}h}{k}\cos \left(kz2-wt\right)\cos \left(kz1-wt\right)$

(iv) $\oint E.dl=\frac{-d\varnothing }{dt}$ =- $\oint B.ds$

=Eoh[sin(kz2-wt)-sin(kz1-wt)]

=- $\frac{d}{dt}[\frac{B_{yh}}{k}\cos \left(kz2-wt\right)-\mathrm{c}\mathrm{o}\mathrm{s}(kz1-wt\left)\right]$

Eo=Bow/k=Byc

Eo/Bo=c