Exercises — Electromagnetic Waves

1. 14 marksNCERT Exercises, Chapter 8

   A capacitor made of two circular plates each of radius 12 cm, separated by 5.0 cm, is being charged by an external source with a constant charging current of 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff's junction rule valid at each plate? Explain.

   Hint. Displacement current is defined precisely so that the total current is continuous through the capacitor.

   (a) The plate area is $A=\pi r^2=\pi (0.12{)}^2=4.52\times 10^{-2}$ m${}^2$, so:

   $C=\frac{{\epsilon }_{0}A}{d}=\frac{8.854\times 10^{-12}\times 4.52\times 10^{-2}}{0.05}\approx 8.0\times 10^{-12}\text{F}=8.0\text{pF}$

   Since $Q=CV$ and the current is $\frac{dQ}{dt}$:

   $\frac{dV}{dt}=\frac{I}{C}=\frac{0.15}{8.0\times 10^{-12}}\approx 1.9\times 10^{10}\text{V/s}$

   (b) The displacement current equals the conduction current in magnitude:

   $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}=0.15\text{A}$

   (c) Yes, Kirchhoff's junction rule remains valid, provided the displacement current is included alongside the conduction current.

   Conduction current arrives at one plate but cannot cross the gap. Maxwell's insight was that the changing electric field in the gap constitutes a displacement current of exactly the same magnitude, so the total current is continuous everywhere around the circuit and the junction rule holds at each plate.

   ✦ (a) C = 8.0 pF, dV/dt = 1.9 x 10^10 V/s (b) Displacement current = 0.15 A (c) Yes, provided displacement current is included, since it exactly continues the conduction current across the gap

2. 24 marksNCERT Exercises, Chapter 8

   A parallel plate capacitor of circular plates of radius $R=6.0$ cm has capacitance $C=100$ pF. It is connected to a 230 V ac supply of angular frequency 300 rad s${}^{-1}$. (a) What is the rms conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

   Hint. Use the capacitive reactance for (a); for (c) note that only the fraction of displacement current within radius r contributes.

   (a) The capacitive reactance is:

   $X_C=\frac{1}{\omega C}=\frac{1}{300\times 100\times 10^{-12}}=3.33\times 10^7\Omega$

   $I_{\text{rms}}=\frac{230}{3.33\times 10^7}\approx 6.9\times 10^{-6}\text{A}=6.9\mu \text{A}$

   (b) Yes. The displacement current between the plates is exactly equal to the conduction current in the leads at every instant, which is what keeps the total current continuous.

   (c) Applying the Ampere-Maxwell law to a circle of radius $r=3.0$ cm between the plates, only the displacement current passing through that circle contributes, a fraction $r^2/R^2$ of the total:

   $B=\frac{{\mu }_0{I}_{0}r}{2\pi R^2}$

   with the amplitude $I_0=\sqrt{2}\times 6.9\times 10^{-6}=9.76\times 10^{-6}$ A:

   $B=\frac{4\pi \times 10^{-7}\times 9.76\times 10^{-6}\times 0.03}{2\pi \times (0.06{)}^2}\approx 1.6\times 10^{-11}\text{T}$

   Only part of the displacement current is enclosed because the field is spread uniformly over the full plate area.

   ✦ (a) I_rms = 6.9 microampere (b) Yes, they are equal at every instant (c) B = 1.6 x 10^-11 T

3. 32 marksNCERT Exercises, Chapter 8

   What physical quantity is the same for X-rays of wavelength $10^{-10}$ m, red light of wavelength 6800 Angstrom and radiowaves of wavelength 500 m?

   Hint. All three are electromagnetic waves travelling in the same medium.

   All three are electromagnetic waves, and in vacuum every electromagnetic wave travels at the same speed regardless of its wavelength.

   The quantity that is identical for all three is therefore the speed:

   $c=3\times 10^8{\text{m s}}^{-1}$

   Their wavelengths differ by twelve orders of magnitude, and their frequencies differ correspondingly, since $c=\nu \lambda$ is fixed. It is precisely because the product stays constant that a longer wavelength must mean a lower frequency.

   ✦ The speed in vacuum, 3 x 10^8 m/s

4. 43 marksNCERT Exercises, Chapter 8

   A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

   Hint. Electromagnetic waves are transverse, and E, B and the propagation direction form a mutually perpendicular set.

   Electromagnetic waves are transverse, so both field vectors are perpendicular to the direction of propagation. They are also perpendicular to each other.

   With the wave travelling along $z$, both $\vec{E}$ and $\vec{B}$ therefore lie in the $x$-$y$ plane, at right angles to one another. The triad $\vec{E}$, $\vec{B}$ and the propagation direction forms a right-handed set, with $\vec{E}\times \vec{B}$ pointing along the direction of travel.

   The wavelength follows from $c=\nu \lambda$:

   $\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{30\times 10^6}=10\text{m}$

   ✦ E and B are mutually perpendicular and both lie in the x-y plane, perpendicular to the z-direction of travel; wavelength = 10 m

5. 52 marksNCERT Exercises, Chapter 8

   A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

   Hint. Apply c = nu lambda at each end, noting that the higher frequency gives the shorter wavelength.

   Applying $\lambda =\frac{c}{\nu }$ at each end of the band:

   At $7.5$ MHz: $\lambda =\frac{3\times 10^8}{7.5\times 10^6}=40$ m

   At $12$ MHz: $\lambda =\frac{3\times 10^8}{12\times 10^6}=25$ m

   So the wavelength band is 25 m to 40 m.

   Note the order reverses: because wavelength is inversely proportional to frequency, the highest frequency corresponds to the shortest wavelength. Reporting the band as 40 m to 25 m in the same order as the frequencies would misstate which end is which.

   ✦ The wavelength band is 25 m to 40 m

6. 62 marksNCERT Exercises, Chapter 8

   A charged particle oscillates about its mean equilibrium position with a frequency of $10^9$ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

   Hint. An accelerating charge radiates at the frequency at which it oscillates.

   An oscillating charge radiates electromagnetic waves, and the frequency of the radiation is exactly the frequency at which the charge itself oscillates.

   The oscillating charge produces an oscillating electric field, which in turn generates an oscillating magnetic field, and the two regenerate each other and propagate outwards — all at the same frequency as the source.

   Therefore the frequency of the electromagnetic waves is:

   $\nu =10^9\text{Hz}$

   This identity between source and radiation frequency is the principle on which every radio transmitter operates.

   The frequencies match because each oscillation of the charge launches exactly one cycle of the radiated wave.

   ✦ 10^9 Hz, the same as the oscillation frequency of the charge

7. 72 marksNCERT Exercises, Chapter 8

   The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0=510$ nT. What is the amplitude of the electric field part of the wave?

   Hint. The ratio of the field amplitudes in vacuum is the speed of light.

   In an electromagnetic wave travelling through vacuum the amplitudes of the two fields are locked together by:

   $\frac{E_0}{B_0}=c$

   Therefore:

   $E_0=c{B}_0=3\times 10^8\times 510\times 10^{-9}$

   $E_0=153\text{N/C}$

   Because $c$ is so large, the electric field amplitude is numerically far greater than the magnetic one in SI units. This is why the electric field dominates the interaction of light with matter, even though the two carry equal energy densities.

   ✦ E0 = 153 N/C

8. 84 marksNCERT Exercises, Chapter 8

   Suppose the electric field amplitude of an electromagnetic wave is $E_0=120$ N/C and its frequency is $\nu =50.0$ MHz. (a) Determine $B_0$, $\omega$, $k$ and $\lambda$. (b) Find expressions for E and B.

   Hint. Work through the standard relations one at a time, then assemble the wave expressions.

   (a) Each quantity follows from a standard relation:

   $B_0=\frac{E_0}{c}=\frac{120}{3\times 10^8}=4.0\times 10^{-7}\text{T}=400\text{nT}$

   $\omega =2\pi \nu =2\pi \times 50.0\times 10^6\approx 3.14\times 10^8{\text{rad s}}^{-1}$

   $\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{50.0\times 10^6}=6.0\text{m}$

   $k=\frac{2\pi }{\lambda }=\frac{2\pi }{6.0}\approx 1.05{\text{rad m}}^{-1}$

   (b) Taking the wave to travel along $x$ with $\vec{E}$ along $y$ and $\vec{B}$ along $z$:

   $\vec{E}=\hat{j}(120\text{N/C})\sin \left[(1.05\text{rad/m})x-(3.14\times 10^8\text{rad/s})t\right]$

   $\vec{B}=\hat{k}(400\text{nT})\sin \left[(1.05\text{rad/m})x-(3.14\times 10^8\text{rad/s})t\right]$

   The two fields share the same phase, so they peak and vanish together — they are in step, not in quadrature.

   ✦ (a) B0 = 400 nT, omega = 3.14 x 10^8 rad/s, k = 1.05 rad/m, lambda = 6.0 m (b) E and B are in phase, perpendicular to each other and to the direction of travel

9. 94 marksNCERT Exercises, Chapter 8

   Use the formula $E=h\nu$ to obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies related to the sources of electromagnetic radiation?

   Hint. Compute a representative photon energy for each band, then compare with the energy scale of the process that emits it.

   Using $E=h\nu$ and converting to electronvolts by dividing by $1.6\times 10^{-19}$ gives representative values across the spectrum:

   Region	Frequency	Photon energy
   Radio waves	$10^6$ Hz	$\sim 4\times 10^{-9}$ eV
   Microwaves	$10^{10}$ Hz	$\sim 4\times 10^{-5}$ eV
   Infrared	$10^{13}$ Hz	$\sim 0.04$ eV
   Visible light	$10^{15}$ Hz	$\sim 2$ to $3$ eV
   Ultraviolet	$10^{16}$ Hz	$\sim 40$ eV
   X-rays	$10^{18}$ Hz	$\sim 4\times 10^3$ eV
   Gamma rays	$10^{20}$ Hz	$\sim 4\times 10^5$ eV

   The connection to the sources. Each band's photon energy matches the energy scale of the process that produces it, because a photon can only carry away what the source transition releases.

   Radio waves come from the collective oscillation of many electrons in an aerial, involving minute energy per photon. Infrared arises from molecular vibrations and rotations, whose energies are fractions of an electronvolt. Visible and ultraviolet light comes from outer electron transitions in atoms, of order a few to tens of electronvolts. X-rays come from inner shell electron transitions, which are far more tightly bound. Gamma rays originate in the nucleus, where binding energies are in the MeV range.

   So the spectrum is effectively a map of the energy scales of matter, from bulk oscillations up to nuclear transitions.

   ✦ Photon energy rises from about 10^-9 eV for radio waves to around 10^5 eV for gamma rays; each band matches the energy scale of its source, from electron oscillations in aerials through molecular vibrations and atomic transitions to nuclear processes

10. 105 marksNCERT Exercises, Chapter 8

    In a plane electromagnetic wave the electric field oscillates sinusoidally at a frequency of $2.0\times 10^{10}$ Hz and amplitude 48 V m${}^{-1}$. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals that of the B field.

    Hint. For part (c) express both densities in terms of E0 and use the relations E0 = c B0 and c squared = 1/(mu0 epsilon0).

    (a) From $c=\nu \lambda$:

    $\lambda =\frac{3\times 10^8}{2.0\times 10^{10}}=1.5\times 10^{-2}\text{m}=1.5\text{cm}$

    (b) The field amplitudes are related by $E_0=c{B}_0$:

    $B_0=\frac{48}{3\times 10^8}=1.6\times 10^{-7}\text{T}$

    (c) The average energy densities, using that the mean of ${\sin }^2$ over a cycle is $\frac{1}{2}$, are:

    $u_E=\frac{1}{4}{\epsilon }_0{E}_0^2,u_B=\frac{B_0^2}{4{\mu }_0}$

    Substituting $B_0=\frac{E_0}{c}$ into the magnetic expression:

    $u_B=\frac{E_0^2}{4{\mu }_0{c}^2}$

    But $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$, so ${\mu }_0{c}^2=\frac{1}{{\epsilon }_0}$ and therefore:

    $u_B=\frac{{\epsilon }_0{E}_0^2}{4}=u_E$

    The two are equal, which is why the energy of an electromagnetic wave is shared equally between its electric and magnetic components. Numerically both come to about $5.1\times 10^{-9}$ J m${}^{-3}$ here.

    ✦ (a) lambda = 1.5 cm (b) B0 = 1.6 x 10^-7 T (c) Both equal epsilon_0 E0 squared over 4, about 5.1 x 10^-9 J/m^3, since c squared = 1/(mu0 epsilon0)

Solutions written by the tuition.in editorial team and checked against leph108.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.