NCERT Exemplar Solutions Class 12 Physics Electromagnetic Waves — Very Short, Short and Long Answer Questions Q14–32

NCERT Exemplar Problems, Class 12 Physics — Electromagnetic Waves, Very Short Answer Questions 14 to 20, Short Answer Questions 21 to 27 and Long Answer Questions 28 to 32. All 19 questions solved. The book numbers them 8.14 to 8.32.

Three ideas carry this page. A changing electric field acts like a current: the displacement current $i_D = \varepsilon_0\,d\Phi_E/dt$ completes Ampere’s law, and it is what lets a wave sustain itself. The Poynting vector $\mathbf S = \frac{1}{\mu_0}\mathbf E\times\mathbf B$ gives the energy flow, and its time average is the intensity. A wave that carries energy $U$ also carries momentum $U/c$, so it exerts radiation pressure.

Key insight. Write the fields as functions of $kz-\omega t$, then do the calculus honestly. Faraday’s law on a loop gives $E_0 = cB_0$, Ampere-Maxwell on another loop gives $c = 1/\sqrt{\mu_0\varepsilon_0}$, and the averages in Questions 23 and 32 all come from $\langle\sin^2\rangle = \langle\cos^2\rangle = \tfrac12$.

Very Short Answer Questions

Question 14

Why is the orientation of the portable radio with respect to broadcasting station important?

Solution. The radio waves from a broadcasting station are plane polarised: the electric field oscillates in one fixed direction. The telescopic rod of the radio acts as an antenna, and the wave drives a current in it only through the component of $\mathbf E$ along the rod. The signal is strongest when the rod is parallel to the electric field of the wave and weakest when it is perpendicular. So the radio must be turned until the rod lines up with $\mathbf E$.

Question 15

Why does microwave oven heats up a food item containing water molecules most efficiently?

Solution. Water molecules are polar and have natural rotational frequencies that lie in the microwave region. The oven’s microwaves are chosen to match one of these resonant frequencies, so the molecules absorb energy strongly from the oscillating field. The absorbed energy turns into kinetic energy of the molecules, which is heat, and it is spread through the food by collisions.

Question 16

The charge on a parallel plate capacitor varies as $q = q_0\cos 2\pi\nu t$. The plates are very large and close together (area $= A$, separation $= d$). Neglecting the edge effects, find the displacement current through the capacitor?

Solution. The displacement current in the gap equals the conduction current in the wires, because the field between the plates is $E = q/\varepsilon_0A$ and $i_D = \varepsilon_0\,A\,dE/dt = dq/dt$. So $$i_D = \frac{dq}{dt} = -2\pi\nu q_0\sin 2\pi\nu t.$$

Question 17

A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?

Solution. The reactance of the capacitor is $X_C = 1/\omega C$. When the frequency falls, $X_C$ rises, so for a given source voltage the conduction current in the circuit falls. The displacement current between the plates always equals the conduction current, $i_D = i_C$, so it falls too. Equivalently, $i_D = C\,dV/dt$ and a slower oscillation has a smaller $dV/dt$.

Question 18

The magnetic field of a beam emerging from a filter facing a floodlight is given by $B_0 = 12\times10^{-8}\sin(1.20\times10^{7}z – 3.60\times10^{15}t)$ T. What is the average intensity of the beam?

Solution. Compare with $B_0\sin(kz-\omega t)$: the amplitude is $B_0 = 12\times10^{-8}$ T, with $k = 1.20\times10^{7}\ \text{m}^{-1}$ and $\omega = 3.60\times10^{15}\ \text{s}^{-1}$. The wave speed is $\omega/k = 3\times10^{8}$ m/s, as it should be. The electric amplitude is $$E_0 = cB_0 = 3\times10^{8}\times12\times10^{-8} = 36\ \text{V/m}.$$ The average intensity is $$I_{av} = \frac{cB_0^2}{2\mu_0} = \frac{3\times10^{8}\times(12\times10^{-8})^2}{2\times4\pi\times10^{-7}} \approx 1.72\ \text{W/m}^2.$$ For interest, $\lambda = 2\pi/k \approx 524$ nm and $\nu = \omega/2\pi \approx 5.7\times10^{14}$ Hz, which is green light.

Question 19

Poynting vectors $\mathbf S$ is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propagation. Mathematically, it is given by $\mathbf S = \dfrac{1}{\mu_0}\mathbf E\times\mathbf B$. Show the nature of $S$ vs $t$ graph.

Solution. Take a wave along $z$ with $\mathbf E = E_0\sin(kz-\omega t)\,\hat{\mathbf i}$ and $\mathbf B = B_0\sin(kz-\omega t)\,\hat{\mathbf j}$. Since $\hat{\mathbf i}\times\hat{\mathbf j} = \hat{\mathbf k}$, $$S = \frac{E_0B_0}{\mu_0}\sin^2(kz-\omega t).$$ At a fixed point, $\sin^2$ is never negative, so $S$ is always along $+z$. It rises from $0$ to the maximum $E_0B_0/\mu_0$ and falls back to $0$, repeating with period $T/2 = \pi/\omega$, half the period of the wave. The graph is a train of equal humps resting on the time axis, never going below it. Its average is $E_0B_0/2\mu_0$.

Question 20

Professor C.V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.

Solution. He was showing that electromagnetic waves carry momentum. The beam is absorbed or reflected by the ball, the momentum it carries is transferred, and the resulting radiation force balances the weight of the light ball. Another example is the tail of a comet, which always points away from the Sun because the pressure of sunlight pushes the dust and gas outwards.

Short Answer Questions

Question 21

Show that the magnetic field $B$ at a point in between the plates of a parallel-plate capacitor during charging is $\dfrac{\varepsilon_0\mu_r}{2}\dfrac{dE}{dt}$ (symbols having usual meaning).

Solution. The printed $\varepsilon_0\mu_r$ is a misprint for $\mu_0 r$, where $r$ is the distance from the axis. Take a circle of radius $r$ between the plates, with its centre on the axis. There is no conduction current in the gap, so the Ampere-Maxwell law reads $$\oint\mathbf B\cdot d\mathbf l = \mu_0\varepsilon_0\frac{d\Phi_E}{dt}.$$ By symmetry $B$ is tangential and constant on the circle, so the left side is $B\cdot2\pi r$. The electric flux through the circle is $\Phi_E = E\,\pi r^2$. Hence $$B\,2\pi r = \mu_0\varepsilon_0\,\pi r^2\frac{dE}{dt}\quad\Rightarrow\quad B = \frac{\mu_0\varepsilon_0\, r}{2}\frac{dE}{dt}.$$

Question 22

Electromagnetic waves with wavelength (i) $\lambda_1$ is used in satellite communication. (ii) $\lambda_2$ is used to kill germs in water purifiers. (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.

Solution.

(a) $\lambda_1$ is microwaves, $\lambda_2$ is ultraviolet, $\lambda_3$ is X-rays and $\lambda_4$ is infrared.

(b) The spectrum in order of increasing wavelength runs X-rays, ultraviolet, infrared, microwaves, so $$\lambda_3 < \lambda_2 < \lambda_4 < \lambda_1.$$

(c) One more use of each: microwaves in radar; ultraviolet in LASIK eye surgery; X-rays in imaging bone fractures; infrared in remote controls or optical-fibre communication.

Question 23

Show that average value of radiant flux density ‘$S$’ over a single period ‘$T$’ is given by $S = \dfrac{1}{2c\mu_0}E_0^2$.

Solution. For a plane wave $B = E/c$, so the instantaneous Poynting magnitude is $$S = \frac{1}{\mu_0}EB = \frac{E_0^2}{c\mu_0}\cos^2(kz-\omega t).$$ Over one period $\cos^2$ averages to $\tfrac12$, since $\cos^2\theta = \tfrac12(1+\cos2\theta)$ and the $\cos 2\theta$ term integrates to zero over a period. Therefore $$S_{av} = \frac{E_0^2}{2c\mu_0}.$$

Question 24

You are given a $2\ \mu\text{F}$ parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?

Solution. The displacement current equals the conduction current in the leads, $i_D = i_C = dq/dt = C\,dV/dt$. So the voltage across the capacitor must change at the rate $$\frac{dV}{dt} = \frac{i_D}{C} = \frac{1\times10^{-3}}{2\times10^{-6}} = 500\ \text{V/s}.$$

Question 25

Show that the radiation pressure exerted by an EM wave of intensity $I$ on a surface kept in vacuum is $I/c$.

Solution. Let a wave of intensity $I$ fall normally on an area $A$ and be completely absorbed. In time $\Delta t$ the energy delivered is $U = IA\,\Delta t$. A wave carrying energy $U$ carries momentum $p = U/c$, so the momentum transferred is $$p = \frac{IA\,\Delta t}{c}.$$ The force is the rate of change of momentum, $F = p/\Delta t = IA/c$, and the pressure is $$P = \frac{F}{A} = \frac{I}{c}.$$

Question 26

What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant. What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

Solution. A bulb radiates in all directions, so at distance $r$ its power is spread over a sphere of area $4\pi r^2$ and $I\propto 1/r^2$. Doubling the distance cuts the intensity to one quarter.

A laser beam is highly collimated: its rays are very nearly parallel and its divergence is negligible. The beam therefore keeps the same cross-sectional area as it travels, so the same power crosses the same area and the intensity stays constant. That is the geometrical property the bulb lacks.

Question 27

Even though an electric field $\mathbf E$ exerts a force $q\mathbf E$ on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.

Solution. The electric force on a charge in the wave oscillates in direction, so over a full cycle it averages to zero and gives no steady push. What the electric field does is set the charges moving along $\mathbf E$ and do work on them, which is how the wave transfers energy.

Once a charge moves with velocity $\mathbf v$ along $\mathbf E$, the magnetic field of the wave exerts $q\mathbf v\times\mathbf B$ on it. Since $\mathbf v$ is along $\mathbf E$, this force points along $\mathbf E\times\mathbf B$, the direction of propagation, and it keeps the same sign through the cycle because $\mathbf v$ and $\mathbf B$ reverse together. This steady magnetic force is the radiation pressure.

Long Answer Questions

Question 28

An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the $z$-axis (Fig. 8.1). The wire is set into motion along its length with a uniform velocity $\mathbf v = v\,\hat{\mathbf k}_z$. Calculate the Poynting vector $\mathbf S = \dfrac{1}{\mu_0}(\mathbf E\times\mathbf B)$.

Fig. 8.1 shows the wire along the $z$-axis, a cylinder of radius $a$ drawn round it, and the velocity $V$ pointing up.

Solution. Look at a point at distance $a$ from the wire, say $(0,a,0)$.

The electric field of the line charge is radial: $$\mathbf E = \frac{\lambda}{2\pi\varepsilon_0 a}\,\hat{\mathbf j}.$$ The moving charge is a current $I = \lambda v$ along $+z$. By the right-hand rule its field circles the wire, and at $(0,a,0)$ the circling direction $\hat{\boldsymbol\phi}$ is $-\hat{\mathbf i}$: $$\mathbf B = \frac{\mu_0\lambda v}{2\pi a}\,(-\hat{\mathbf i}).$$ Then, using $\hat{\mathbf j}\times(-\hat{\mathbf i}) = +\hat{\mathbf k}$, $$\mathbf S = \frac{1}{\mu_0}\,\mathbf E\times\mathbf B = \frac{1}{\mu_0}\cdot\frac{\lambda}{2\pi\varepsilon_0 a}\cdot\frac{\mu_0\lambda v}{2\pi a}\,\hat{\mathbf k} = \frac{\lambda^2 v}{4\pi^2\varepsilon_0 a^2}\,\hat{\mathbf k}.$$ Energy therefore flows along the wire, in the direction of its motion, and falls as $1/a^2$.

Printed key error. The key gives $-\hat{\mathbf k}$, which takes $\mathbf B$ along $+\hat{\mathbf i}$ at this point. The right-hand rule for a current along $+z$ gives $\mathbf B$ along $-\hat{\mathbf i}$ at $(0,a,0)$, so $\mathbf E\times\mathbf B$ is along $+\hat{\mathbf k}$. The magnitude in the key is right.

Question 29

Sea water at frequency $\nu = 4\times10^{8}$ Hz has permittivity $\varepsilon \approx 80\,\varepsilon_0$, permeability $\mu \approx \mu_0$ and resistivity $\rho = 0.25\ \Omega$-m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t) = V_0\sin(2\pi\nu t)$. What fraction of the conduction current density is the displacement current density?

Solution. In the water the field is $E = E_0\sin 2\pi\nu t$ between the plates. The conduction current density is $$J_c = \frac{E}{\rho} = \frac{E_0}{\rho}\sin 2\pi\nu t.$$ The displacement current density is $$J_d = \varepsilon\frac{dE}{dt} = 2\pi\nu\,\varepsilon E_0\cos 2\pi\nu t.$$ The ratio of the amplitudes is $$\frac{J_d}{J_c} = 2\pi\nu\,\varepsilon\rho = 2\pi\times4\times10^{8}\times80\times8.85\times10^{-12}\times0.25 \approx 0.445.$$ So the displacement current density is about $0.44$ of the conduction current density. The two are $90^\circ$ out of phase (cosine against sine), so this is a ratio of amplitudes, not of instantaneous values.

Question 30

A long straight cable of length $l$ is placed symmetrically along $z$-axis and has radius $a\ (\ll l)$. The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I(t) = I_0\sin(2\pi\nu t)$ 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 $\mathbf E(s,t) = \mu_0I_0\nu\cos(2\pi\nu t)\ln\left(\dfrac{s}{a}\right)\hat{\mathbf k}$. (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 $I_0^d$.

Solution. (i) The displacement current density is $\mathbf J_d = \varepsilon_0\,\partial\mathbf E/\partial t$: $$\mathbf J_d = -\varepsilon_0\mu_0I_0\nu\,(2\pi\nu)\sin(2\pi\nu t)\ln\frac{s}{a}\,\hat{\mathbf k} = \frac{2\pi I_0\nu^2}{c^2}\ln\frac{a}{s}\,\sin(2\pi\nu t)\,\hat{\mathbf k}.$$ With $\lambda = c/\nu$ this is $$\mathbf J_d = \frac{2\pi I_0}{\lambda^2}\ln\frac{a}{s}\,\sin(2\pi\nu t)\,\hat{\mathbf k}.$$

(ii) Integrate over rings of area $2\pi s\,ds$: $$I_d = \int_0^a J_d\,2\pi s\,ds = \frac{4\pi^2I_0}{\lambda^2}\sin(2\pi\nu t)\int_0^a s\ln\frac{a}{s}\,ds.$$ The integral is $\int_0^a s\ln(a/s)\,ds = a^2/4$ (it is the negative of $\int_0^a s\ln(s/a)\,ds = -a^2/4$). So $$I_d = \left(\frac{\pi a}{\lambda}\right)^2 I_0\sin(2\pi\nu t).$$

(iii) The amplitudes compare as $$\frac{I_0^d}{I_0} = \left(\frac{\pi a}{\lambda}\right)^2 .$$ For a cable much narrower than the wavelength, $a\ll\lambda$, this is far smaller than $1$, so the displacement current is negligible beside the conduction current.

Question 31

A plane EM wave travelling in vacuum along $z$ direction is given by $\mathbf E = E_0\sin(kz-\omega t)\,\hat{\mathbf i}$ and $\mathbf B = B_0\sin(kz-\omega t)\,\hat{\mathbf j}$. (i) Evaluate $\oint\mathbf E\cdot d\mathbf l$ over the rectangular loop 1234 shown in Fig 8.2. (ii) Evaluate $\int\mathbf B\cdot d\mathbf s$ over the surface bounded by loop 1234. (iii) Use equation $\oint\mathbf E\cdot d\mathbf l = -\dfrac{d\phi_B}{dt}$ to prove $\dfrac{E_0}{B_0} = c$. (iv) By using similar process and the equation $\oint\mathbf B\cdot d\mathbf l = \mu_0I + \varepsilon_0\dfrac{d\phi_E}{dt}$, prove that $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$.

Fig. 8.2 shows the loop in the $xz$ plane: corners 1 and 2 on the $z$-axis at $z_1$ and $z_2$, and corners 4 and 3 at height $h$ along $x$ above them.

Solution. (i) Go round 1, 2, 3, 4. Sides 1-2 and 3-4 run along $z$, perpendicular to $\mathbf E$, so they contribute nothing. Side 2-3 runs along $+x$ at $z_2$ and side 4-1 runs along $-x$ at $z_1$: $$\oint\mathbf E\cdot d\mathbf l = E_0h\left[\sin(kz_2-\omega t)-\sin(kz_1-\omega t)\right].$$

(ii) The loop lies in the $xz$ plane, so its normal is along $y$, the direction of $\mathbf B$. Use strips of area $h\,dz$: $$\int\mathbf B\cdot d\mathbf s = \int_{z_1}^{z_2}B_0\sin(kz-\omega t)\,h\,dz = \frac{B_0h}{k}\left[\cos(kz_1-\omega t)-\cos(kz_2-\omega t)\right].$$

(iii) Differentiate (ii) with respect to $t$: $$\frac{d\phi_B}{dt} = \frac{B_0h\omega}{k}\left[\sin(kz_1-\omega t)-\sin(kz_2-\omega t)\right].$$ Then $-\dfrac{d\phi_B}{dt} = \dfrac{B_0h\omega}{k}\left[\sin(kz_2-\omega t)-\sin(kz_1-\omega t)\right]$. Equating this with (i), the brackets cancel and $$E_0 = \frac{B_0\,\omega}{k}\ \Rightarrow\ \frac{E_0}{B_0} = \frac{\omega}{k} = c.$$

(iv) Now take a similar rectangular loop in the $yz$ plane, with sides of length $h$ along $y$ at $z_1$ and $z_2$. The normal is along $x$, the direction of $\mathbf E$, and there is no conduction current in vacuum, so $I = 0$. The same working as above, with $\mathbf B$ along $y$ and the flux of $\mathbf E$ through the loop, gives $$\oint\mathbf B\cdot d\mathbf l = B_0h\left[\sin(kz_1-\omega t)-\sin(kz_2-\omega t)\right],$$ $$\varepsilon_0\frac{d\phi_E}{dt} = \varepsilon_0\frac{E_0h\omega}{k}\left[\sin(kz_1-\omega t)-\sin(kz_2-\omega t)\right].$$ The Ampere-Maxwell law then gives $B_0 = \mu_0\varepsilon_0E_0\,\omega/k$. Using $\omega/k = c$ and $E_0 = cB_0$ from part (iii), $$B_0 = \mu_0\varepsilon_0\,c\,(cB_0)\ \Rightarrow\ c^2 = \frac{1}{\mu_0\varepsilon_0}\ \Rightarrow\ c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}.$$

Question 32

A plane EM wave travelling along $z$ direction is described by $\mathbf E = E_0\sin(kz-\omega t)\,\hat{\mathbf i}$ and $\mathbf B = B_0\sin(kz-\omega t)\,\hat{\mathbf j}$. Show that (i) The average energy density of the wave is given by $u_{av} = \dfrac14\varepsilon_0E_0^2 + \dfrac14\dfrac{B_0^2}{\mu_0}$. (ii) The time averaged intensity of the wave is given by $I_{av} = \dfrac12 c\varepsilon_0E_0^2$.

Solution. (i) The energy density is the sum of the electric and magnetic parts, $$u = \tfrac12\varepsilon_0E^2 + \frac{B^2}{2\mu_0} = \tfrac12\varepsilon_0E_0^2\sin^2(kz-\omega t) + \frac{B_0^2}{2\mu_0}\sin^2(kz-\omega t).$$ The time average of $\sin^2$ over a period is $\tfrac12$, so $$u_{av} = \tfrac14\varepsilon_0E_0^2 + \tfrac14\frac{B_0^2}{\mu_0}.$$

(ii) In vacuum $B_0 = E_0/c$ and $c^2 = 1/\mu_0\varepsilon_0$, so $$\frac{B_0^2}{\mu_0} = \frac{E_0^2}{c^2\mu_0} = \varepsilon_0E_0^2.$$ The two terms in $u_{av}$ are therefore equal and $u_{av} = \tfrac12\varepsilon_0E_0^2$. Energy flows at speed $c$, so the intensity is energy density times speed: $$I_{av} = u_{av}\,c = \tfrac12\,c\,\varepsilon_0E_0^2.$$