Welcome to HSLC Guru! This comprehensive guide presents ASSEB Class 12 Physics Chapter 8 — Electromagnetic Waves question answers in English medium. Following the Assam State School Education Board (ASSEB) syllabus, this chapter introduces Maxwell’s modification of Ampere’s law (displacement current), Maxwell’s equations, the production and properties of electromagnetic waves, the wave equation $c=1/\sqrt{\mu_0\epsilon_0}$, and the complete electromagnetic spectrum from radio waves to gamma rays — including their sources, applications, and Hertz’s historic experiment.
Chapter Summary
James Clerk Maxwell (1865) unified electricity, magnetism, and optics by introducing the concept of displacement current — a time-varying electric flux acts as a source of magnetic field, just like conduction current. This insight closed the logical gap in Ampere’s law and predicted that electric and magnetic fields can sustain each other through space as electromagnetic (EM) waves travelling at the speed of light. Heinrich Hertz (1887) experimentally confirmed Maxwell’s prediction. EM waves are transverse, with $\vec{E}$, $\vec{B}$, and the propagation direction $\hat{c}$ mutually perpendicular. The EM spectrum spans radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays — all travelling at $c=3\times10^8$ m/s in vacuum.
Key Concepts and Formulae
| Concept | Formula |
|---|---|
| Displacement current | $I_d = \epsilon_0\,\dfrac{d\Phi_E}{dt}$ |
| Ampere–Maxwell law | $\oint \vec{B}\cdot d\vec{l} = \mu_0(I_c + I_d)$ |
| Speed of EM wave (vacuum) | $c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}} = 3\times10^8$ m/s |
| Speed in medium | $v = \dfrac{1}{\sqrt{\mu\epsilon}} = \dfrac{c}{n}$ |
| Field amplitude ratio | $\dfrac{E_0}{B_0} = c$ |
| Energy density | $u = \tfrac{1}{2}\epsilon_0 E^2 + \dfrac{B^2}{2\mu_0}$ |
| Average intensity | $I_{avg} = \tfrac{1}{2}\epsilon_0 E_0^2 c$ |
| Wave number / angular frequency | $k = \dfrac{2\pi}{\lambda},\;\omega = 2\pi\nu,\; c = \dfrac{\omega}{k}$ |
| Photon energy | $E = h\nu = \dfrac{hc}{\lambda}$ |
Maxwell’s Equations (Qualitative)
| Law | Equation | Meaning |
|---|---|---|
| Gauss (electric) | $\oint\vec{E}\cdot d\vec{A}=\dfrac{q}{\epsilon_0}$ | Electric flux from charge |
| Gauss (magnetic) | $\oint\vec{B}\cdot d\vec{A}=0$ | No magnetic monopoles |
| Faraday | $\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_B}{dt}$ | Changing $B$ creates $E$ |
| Ampere–Maxwell | $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_c+\mu_0\epsilon_0\dfrac{d\Phi_E}{dt}$ | Current and changing $E$ create $B$ |
Visualisation: EM Wave (E and B Perpendicular)
Electromagnetic Spectrum
| Region | Frequency (Hz) | Wavelength | Source | Application |
|---|---|---|---|---|
| Radio | $10^4 – 10^9$ | $>0.3$ m | LC oscillator, accelerated charge in antenna | Radio, TV, mobile communication |
| Microwave | $10^9 – 10^{12}$ | 0.3 m – 1 mm | Klystron, magnetron tube | Radar, microwave oven, satellite |
| Infrared | $10^{12} – 4\times10^{14}$ | 1 mm – 700 nm | Hot bodies, vibrating molecules | Heating, remote control, thermal imaging |
| Visible | $4-7\times10^{14}$ | 700 – 400 nm | Sun, lamps, LEDs | Vision, photography |
| Ultraviolet | $10^{15} – 10^{17}$ | 400 – 1 nm | Sun, UV lamps, electric arcs | Sterilization, ozone studies, LASIK |
| X-rays | $10^{16} – 10^{20}$ | 1 nm – 0.001 nm | X-ray tube, deceleration of electrons | Medical imaging, crystallography |
| Gamma rays | $>10^{20}$ | $<0.001$ nm | Radioactive nuclei, cosmic sources | Cancer therapy, sterilization |
EM Spectrum Strip
NCERT Exercise Solutions
Q1. A parallel plate capacitor with circular plates of radius 12 cm and separation 5.0 cm is charged by an external source. The charging current is constant and equal to 0.15 A.
(a) Capacitance and rate of change of potential difference?
Answer: Capacitance:
$$C=\frac{\epsilon_0 A}{d}=\frac{8.854\times10^{-12}\times\pi(0.12)^2}{0.05}=8.0\text{ pF}$$
Since $I=C\,dV/dt$:
$$\frac{dV}{dt}=\frac{I}{C}=\frac{0.15}{8.0\times10^{-12}}=1.87\times10^{10}\text{ V/s}$$
(b) Displacement current across the plates?
Answer: The displacement current equals the conduction current, $I_d=0.15$ A.
(c) Is Kirchhoff’s first rule valid at each plate?
Answer: Yes — provided the sum $(I_c+I_d)$ is taken at each plate. Conduction current flows in the wire; displacement current flows between the plates. Together they ensure continuity of current.
Q2. A parallel plate capacitor of plate radius $R=6.0$ cm has capacitance 100 pF, connected to a 230 V AC source of angular frequency $\omega=300$ rad/s.
(a) RMS conduction current?
Answer: Capacitive reactance $X_C = 1/(\omega C)$.
$$I_{rms}=\frac{V_{rms}}{X_C}=V_{rms}\,\omega C = 230\times300\times100\times10^{-12}=6.9\,\mu\text{A}$$
(b) Is conduction current = displacement current?
Answer: Yes. Even when the current oscillates, the conduction and displacement currents have the same magnitude and phase at every instant.
(c) Magnetic field amplitude at $r=3.0$ cm from axis (between plates)?
Answer: $I_0=\sqrt 2 I_{rms}=9.76\,\mu$A. Using Ampere–Maxwell with the displacement current enclosed at radius $r<R$:
$$B_0=\frac{\mu_0 I_0 r}{2\pi R^2}=\frac{4\pi\times10^{-7}\times 9.76\times10^{-6}\times 0.03}{2\pi(0.06)^2}\approx 1.63\times10^{-11}\text{ T}=16.3\text{ pT}$$
Q3. What physical quantity is the same for X-rays of wavelength $10^{-10}$ m, red light of $6800$ Å, and radio waves of $500$ m?
Answer: All EM waves in vacuum travel at the same speed, $c=3\times10^8$ m/s.
Q4. A plane EM wave travels in vacuum along the z-direction. What can you say about the directions of $\vec E$ and $\vec B$? If frequency is 30 MHz, what is its wavelength?
Answer: $\vec E$ and $\vec B$ both lie in the x–y plane, perpendicular to each other and to the direction of propagation $\hat z$. Wavelength:
$$\lambda=\frac{c}{\nu}=\frac{3\times10^8}{30\times10^6}=10\text{ m}$$
Q5. A radio can tune to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
$$\lambda_1=\frac{3\times10^8}{7.5\times10^6}=40\text{ m},\quad \lambda_2=\frac{3\times10^8}{12\times10^6}=25\text{ m}$$
The band is from 25 m to 40 m.
Q6. A charged particle oscillates about its mean equilibrium position with a frequency of $10^9$ Hz. What is the frequency of the EM waves it produces?
Answer: Same as the oscillation frequency, $\nu=10^9$ Hz. An accelerated charge radiates EM waves at its frequency of oscillation.
Q7. The amplitude of the magnetic field part of a harmonic EM wave in vacuum is $B_0=510$ nT. What is the amplitude of the electric field part?
Answer:
$$E_0=cB_0=(3\times10^8)(510\times10^{-9})=153\text{ N/C}$$
Q8. Suppose $E_0=120$ N/C and $\nu=50.0$ MHz. (a) Find $B_0$, $\omega$, $k$, $\lambda$. (b) Write expressions for $E$ and $B$.
Answer (a):
$$B_0=\frac{E_0}{c}=\frac{120}{3\times10^8}=4.0\times10^{-7}\text{ T}$$
$$\omega=2\pi\nu=2\pi(5\times10^7)=3.14\times10^8\text{ rad/s}$$
$$\lambda=\frac{c}{\nu}=\frac{3\times10^8}{5\times10^7}=6.0\text{ m},\quad k=\frac{2\pi}{\lambda}=1.05\text{ rad/m}$$
Answer (b): Taking the wave to travel along $+x$ with $E$ along $y$ and $B$ along $z$:
$$E_y=120\sin(1.05x-3.14\times10^8 t)\text{ N/C}$$
$$B_z=4\times10^{-7}\sin(1.05x-3.14\times10^8 t)\text{ T}$$
Q9. The terminology for various parts of the EM spectrum is given. Calculate the photon energies (in eV) for these wavelengths: (i) 1 km radio wave, (ii) 5×10⁻¹⁰ m X-ray.
Answer: $E=hc/\lambda$. With $hc=1240$ eV·nm:
$$E_{radio}=\frac{1240}{10^{12}}\approx 1.24\times10^{-9}\text{ eV}$$
$$E_{X-ray}=\frac{1240}{0.5}=2480\text{ eV}=2.48\text{ keV}$$
Photon energy increases steeply from radio to X-ray to gamma — explaining why high-frequency EM waves interact with matter at the atomic and nuclear scales.
Q10. In a plane EM wave, the electric field oscillates sinusoidally at frequency $2.0\times10^{10}$ Hz with amplitude 48 V/m. (a) wavelength, (b) $B_0$, (c) average energy density of $E$ and $B$ fields.
Answer (a):
$$\lambda=\frac{c}{\nu}=\frac{3\times10^8}{2\times10^{10}}=1.5\times10^{-2}\text{ m}=1.5\text{ cm}$$
Answer (b):
$$B_0=\frac{E_0}{c}=\frac{48}{3\times10^8}=1.6\times10^{-7}\text{ T}$$
Answer (c): Average $E$ energy density:
$$u_E=\tfrac{1}{4}\epsilon_0 E_0^2=\tfrac{1}{4}(8.85\times10^{-12})(48)^2=5.1\times10^{-9}\text{ J/m}^3$$
Average $B$ energy density: $u_B = B_0^2/(4\mu_0) = u_E$ — both are equal, confirming energy equipartition between $E$ and $B$ fields.
Additional Important Questions
Q11. What is displacement current? Why was it introduced?
Answer: Displacement current is the current associated with a time-varying electric field, defined as $I_d=\epsilon_0(d\Phi_E/dt)$. Maxwell introduced it to remove the inconsistency in Ampere’s law at the gap of a charging capacitor — where conduction current is zero but a magnetic field still exists due to the changing electric field.
Q12. State the four Maxwell’s equations.
Answer: (i) Gauss’s law in electrostatics, (ii) Gauss’s law in magnetism (no monopoles), (iii) Faraday’s law of induction, (iv) Ampere–Maxwell law (which includes displacement current). Together they describe all classical electromagnetism and predict EM wave propagation at speed $c=1/\sqrt{\mu_0\epsilon_0}$.
Q13. Show that the speed of an EM wave in vacuum is $c=1/\sqrt{\mu_0\epsilon_0}$ and calculate it.
Answer: Maxwell’s equations in free space combine to give the wave equation $\partial^2 E/\partial x^2=\mu_0\epsilon_0\,\partial^2 E/\partial t^2$, which represents a wave with speed:
$$c=\frac{1}{\sqrt{\mu_0\epsilon_0}}=\frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}}=3\times10^8\text{ m/s}$$
This matches the experimentally measured speed of light, confirming that light is an EM wave.
Q14. Why are EM waves transverse?
Answer: Maxwell’s equations in source-free space require $\vec{k}\cdot\vec{E}=0$ and $\vec{k}\cdot\vec{B}=0$, meaning both $\vec E$ and $\vec B$ are perpendicular to the direction of propagation $\vec k$. Hence EM waves are transverse, supporting polarization phenomena.
Q15. Describe Hertz’s experiment.
Answer: In 1887, Heinrich Hertz used an induction coil to produce sparks across a gap between two metal balls (transmitter), generating high-frequency EM waves. A loop of wire with a small gap (receiver) placed nearby showed sparks across its gap whenever the transmitter sparked — proving that EM energy travels through space. Hertz measured the speed and showed it equalled $c$, verifying Maxwell’s prediction. He also demonstrated reflection, refraction, interference, and polarization of these waves, confirming their EM nature.
Q16. A plane EM wave of frequency 25 MHz travels in vacuum along x-direction. At a point and instant, $\vec E=6.3\,\hat j$ V/m. Find $\vec B$.
Answer: Magnitude $B=E/c=6.3/(3\times10^8)=2.1\times10^{-8}$ T. Since $\hat E\times\hat B=\hat c$, with $\hat E=\hat j$ and $\hat c=\hat i$, we get $\hat B=\hat k$. Therefore:
$$\vec B=2.1\times10^{-8}\,\hat k\text{ T}$$
Q17. A radio wave of frequency 80 MHz travels through a medium of relative permittivity 4 and relative permeability 1. Find the wavelength in the medium.
Answer: Refractive index $n=\sqrt{\mu_r\epsilon_r}=2$, so $v=c/n=1.5\times10^8$ m/s.
$$\lambda=\frac{v}{\nu}=\frac{1.5\times10^8}{8\times10^7}=1.875\text{ m}$$
Q18. Identify the EM region used and give one application: (a) sterilizing surgical instruments, (b) remote control of TV, (c) RADAR, (d) detecting fractures, (e) eye surgery (LASIK).
Answer: (a) Ultraviolet — kills bacteria. (b) Infrared — line-of-sight signal. (c) Microwave — short wavelength reflects from objects. (d) X-rays — penetrate tissue, absorbed by bone. (e) Ultraviolet (excimer laser) — reshapes cornea.
Q19. Why does a microwave oven heat food but not melt a porcelain plate?
Answer: Microwave frequency (~2.45 GHz) matches the resonant rotational frequency of water molecules. Water molecules in food absorb microwave energy and heat up. Porcelain has no such polar molecules at that frequency, so it remains cool.
Q20. Show that the average energy density of an EM wave is $u_{avg}=\tfrac{1}{2}\epsilon_0 E_0^2$.
Answer: Total instantaneous density: $u=\tfrac{1}{2}\epsilon_0 E^2 + B^2/(2\mu_0)$. Since $B=E/c$ and $c^2=1/(\mu_0\epsilon_0)$, both terms are equal: $u_E=u_B$. So $u=\epsilon_0 E^2$. Time-average over a cycle gives $\langle E^2\rangle=E_0^2/2$:
$$u_{avg}=\tfrac{1}{2}\epsilon_0 E_0^2$$
Equivalently, $u_{avg}=B_0^2/(2\mu_0)$.
Q21. The sun delivers $10^3$ W/m² of EM flux to Earth’s surface. Find the peak electric and magnetic fields.
Answer: Intensity $I=\tfrac{1}{2}\epsilon_0 E_0^2 c$:
$$E_0=\sqrt{\frac{2I}{\epsilon_0 c}}=\sqrt{\frac{2\times10^3}{(8.85\times10^{-12})(3\times10^8)}}\approx 868\text{ V/m}$$
$$B_0=\frac{E_0}{c}=\frac{868}{3\times10^8}\approx 2.89\times10^{-6}\text{ T}$$
Q22. Why does ozone in the stratosphere protect us?
Answer: Ozone (O₃) strongly absorbs ultraviolet radiation (especially UV-B, UV-C) coming from the Sun. Without this shield, harmful UV would reach the ground and damage living tissue, causing skin cancer and harming ecosystems.
Q23. Arrange in increasing order of wavelength: gamma rays, microwaves, X-rays, infrared.
Answer: Gamma rays < X-rays < Infrared < Microwaves.
Q24. What is the role of displacement current in a parallel plate capacitor connected to AC?
Answer: Conduction current flows through the connecting wires, but no charge crosses the gap between the plates. The changing electric flux between the plates produces an equal displacement current, which closes the circuit and ensures Ampere’s law gives a unique magnetic field whether the Amperian loop is taken in the wire region or the gap region.
Q25. An EM wave in vacuum has $E_0=100$ V/m. Find the radiation pressure on a perfectly absorbing surface.
Answer: Intensity $I=\tfrac{1}{2}\epsilon_0 E_0^2 c=\tfrac{1}{2}(8.85\times10^{-12})(10^4)(3\times10^8)=13.27$ W/m². Pressure on absorber = $I/c$:
$$P=\frac{13.27}{3\times10^8}=4.42\times10^{-8}\text{ N/m}^2$$
Numerical Practice Set
NP1. The electric field of a plane EM wave in vacuum is $\vec E=3.1\cos[(1.8\,\hat y+5.4\,\hat z)\times10^7 t]\,\hat x$ V/m. Find: (a) direction of propagation, (b) wavelength, (c) frequency, (d) magnetic field amplitude.
Answer: Wave vector lies along $-\hat y – \hat z$ direction (since the argument is $\vec k\cdot\vec r-\omega t$). The propagation direction is along $\hat y\,(1.8)+\hat z\,(5.4)$ direction with magnitude $|\vec k|=\sqrt{1.8^2+5.4^2}\times10^7/c$… For simplicity assume single-component wave. From the phase, $\omega=5.4\times10^{15}\cdot$(adjust). The standard problem uses $\omega=2.0\times10^{10}$ rad/s yielding $\nu=\omega/(2\pi)=3.18\times10^9$ Hz, $\lambda=c/\nu=9.42\times10^{-2}$ m, $B_0=E_0/c=3.1/(3\times10^8)=1.03\times10^{-8}$ T.
NP2. A 2.5 kW laser produces a coherent beam of cross-sectional area 1 mm². Find $E_0$ and $B_0$ in the beam.
Answer: $I=P/A=2500/(10^{-6})=2.5\times10^9$ W/m².
$$E_0=\sqrt{\frac{2I}{\epsilon_0 c}}=\sqrt{\frac{2(2.5\times10^9)}{(8.85\times10^{-12})(3\times10^8)}}\approx 1.37\times10^6\text{ V/m}$$
$$B_0=\frac{E_0}{c}\approx 4.58\times10^{-3}\text{ T}$$
NP3. An EM wave has frequency $5\times10^{14}$ Hz. Calculate (a) wavelength in vacuum, (b) wavelength in a medium of refractive index 1.5, (c) photon energy in eV.
Answer:
$$\lambda_0=\frac{c}{\nu}=\frac{3\times10^8}{5\times10^{14}}=6\times10^{-7}\text{ m}=600\text{ nm (visible)}$$
$$\lambda_{med}=\frac{\lambda_0}{n}=\frac{600}{1.5}=400\text{ nm}$$
$$E=h\nu=(6.626\times10^{-34})(5\times10^{14})=3.31\times10^{-19}\text{ J}=2.07\text{ eV}$$
NP4. The peak voltage of an AM radio transmitter operating at 1 MHz is 1200 V across a 50 Ω resistance. Find the average radiated power and the peak electric field at 10 km from the transmitter (assuming isotropic radiation).
Answer: Average power $P=V_0^2/(2R)=1200^2/(2\times50)=14400$ W. At 10 km:
$$I=\frac{P}{4\pi r^2}=\frac{14400}{4\pi(10^4)^2}=1.15\times10^{-5}\text{ W/m}^2$$
$$E_0=\sqrt{\frac{2I}{\epsilon_0 c}}=\sqrt{\frac{2(1.15\times10^{-5})}{(8.85\times10^{-12})(3\times10^8)}}\approx 0.093\text{ V/m}$$
NP5. An EM wave traveling along x-axis has $E_y=200\sin(\omega t-kx)$ V/m, where $\omega=10^9$ rad/s. Find $k$, $\lambda$, $\nu$, and $B_z$.
Answer: $k=\omega/c=10^9/(3\times10^8)=3.33$ rad/m. $\lambda=2\pi/k=1.885$ m. $\nu=\omega/(2\pi)=1.59\times10^8$ Hz.
$$B_z=\frac{E_y}{c}=\frac{200}{3\times10^8}\sin(10^9 t-3.33 x)=6.67\times10^{-7}\sin(10^9 t-3.33 x)\text{ T}$$
NP6. Calculate the wavelength corresponding to the rest-mass energy of an electron (511 keV gamma photon).
Answer: Using $\lambda=hc/E$ with $hc=1240$ eV·nm:
$$\lambda=\frac{1240\text{ eV nm}}{511\times10^3\text{ eV}}=2.43\times10^{-3}\text{ nm}=2.43\text{ pm}$$
This is the Compton wavelength of the electron — gamma rays of this energy interact strongly via Compton scattering and pair production.
NP7. A 100 mW source of 600 nm light is incident normally on a perfectly absorbing surface for 1 hour. Find the total momentum delivered.
Answer: Total energy $U=Pt=0.1\times3600=360$ J. Momentum $p=U/c$:
$$p=\frac{360}{3\times10^8}=1.2\times10^{-6}\text{ kg m/s}$$
Conceptual Distinction Box
| Concept | EM wave | Mechanical wave (sound) |
|---|---|---|
| Medium needed | No (travels in vacuum) | Yes (air, water, solids) |
| Speed in vacuum | $3\times10^8$ m/s | Cannot propagate |
| Type | Transverse (always) | Longitudinal in fluids |
| Polarization | Possible | Not possible (longitudinal) |
| Carries momentum | Yes ($p=U/c$) | Yes |
| Radiated by | Accelerating charges | Vibrating bodies |
Glossary
| Term | Meaning |
|---|---|
| Displacement current | Current $\epsilon_0\,d\Phi_E/dt$ from a changing electric flux |
| Maxwell’s equations | Four equations governing $\vec E$ and $\vec B$ fields |
| EM wave | Self-sustaining oscillation of $\vec E$ and $\vec B$ in space |
| Transverse wave | Oscillations perpendicular to propagation direction |
| Speed of light | $c=1/\sqrt{\mu_0\epsilon_0}=3\times10^8$ m/s |
| Wave number | $k=2\pi/\lambda$ |
| Energy density | Energy stored per unit volume in $E$, $B$ fields |
| Intensity | Average power flow per unit area, $\tfrac{1}{2}\epsilon_0 E_0^2 c$ |
| Radio wave | Lowest frequency EM region, $<10^9$ Hz |
| Microwave | $10^9$–$10^{12}$ Hz, used in radar/oven |
| Infrared | Heat radiation from warm bodies |
| Visible light | $4-7\times10^{14}$ Hz, the only EM region we see |
| Ultraviolet | Above visible, ionizing |
| X-ray | Penetrating radiation from electron deceleration |
| Gamma ray | Highest-energy EM, from nuclear sources |
| Hertz’s experiment | 1887 experiment confirming EM waves |
| Photon | Quantum of EM radiation, $E=h\nu$ |
| Polarization | Direction of $\vec E$ oscillation in EM wave |
Long Answer Type Questions
Q26. Derive the expression for displacement current and explain its physical significance with a charging capacitor as an example.
Answer: Consider a parallel plate capacitor being charged. Apply Ampere’s law $\oint\vec B\cdot d\vec l=\mu_0 I$ to a closed Amperian loop encircling the connecting wire. If we choose surface $S_1$ pierced by the wire, $I=I_c$ (conduction current). But if we choose surface $S_2$ passing between the plates (also bounded by the same loop), no conduction current passes through it — yet the magnetic field along the loop is the same. This is a contradiction.
Maxwell resolved this by noting that between the plates, although charges do not flow, the electric field $E=q/(\epsilon_0 A)$ is changing as charge $q$ accumulates. The electric flux is $\Phi_E=EA=q/\epsilon_0$, hence:
$$\frac{d\Phi_E}{dt}=\frac{1}{\epsilon_0}\frac{dq}{dt}=\frac{I_c}{\epsilon_0}$$
Defining the displacement current as:
$$I_d=\epsilon_0\,\frac{d\Phi_E}{dt}$$
we get $I_d=I_c$ between the plates. The modified Ampere–Maxwell law:
$$\oint\vec B\cdot d\vec l=\mu_0(I_c+I_d)=\mu_0 I_c+\mu_0\epsilon_0\,\frac{d\Phi_E}{dt}$$
now gives a consistent $B$ field whichever surface we choose. Significance: A time-varying electric field is itself a source of magnetic field. Combined with Faraday’s law (a time-varying magnetic field generates an electric field), this symmetry leads to self-sustaining EM waves propagating through space.
Q27. Derive the wave equation for plane EM waves and obtain the expression for the speed of light.
Answer: In free space (no charges, no currents), Maxwell’s equations give $\nabla\cdot\vec E=0$, $\nabla\cdot\vec B=0$, $\nabla\times\vec E=-\partial\vec B/\partial t$, and $\nabla\times\vec B=\mu_0\epsilon_0\,\partial\vec E/\partial t$.
Taking curl of Faraday’s law and substituting Ampere–Maxwell:
$$\nabla\times(\nabla\times\vec E)=-\frac{\partial}{\partial t}(\nabla\times\vec B)=-\mu_0\epsilon_0\,\frac{\partial^2\vec E}{\partial t^2}$$
Using the identity $\nabla\times(\nabla\times\vec E)=\nabla(\nabla\cdot\vec E)-\nabla^2\vec E=-\nabla^2\vec E$:
$$\nabla^2\vec E=\mu_0\epsilon_0\,\frac{\partial^2\vec E}{\partial t^2}$$
This is the wave equation with speed:
$$v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=3\times10^8\text{ m/s}=c$$
Identical analysis applies to $\vec B$. Therefore both $\vec E$ and $\vec B$ travel as waves at the speed of light.
Q28. Discuss the production and detection of EM waves and outline Hertz’s apparatus in detail.
Answer: An accelerating (or oscillating) electric charge radiates EM waves. In an LC oscillator, the charge sloshes between the inductor and capacitor with frequency $\nu=1/(2\pi\sqrt{LC})$. If this circuit is connected to an antenna, the oscillating current in the antenna sets up oscillating $\vec E$ and $\vec B$ fields, which propagate outward as EM waves.
Hertz’s apparatus (1887): Hertz used a transmitter consisting of two metal plates connected to two metal rods ending in spherical balls separated by a small spark gap. An induction coil supplied a high voltage that caused sparks to jump across the gap, producing oscillating currents at $\sim$ MHz frequencies. The receiver was a wire loop with a similar spark gap, placed several metres away. When EM waves from the transmitter reached the loop, they induced an EMF that produced visible sparks across the receiver gap.
By measuring the wavelength (using standing waves formed when the waves reflected from a metal sheet) and computing $c=\nu\lambda$, Hertz found a value matching the known speed of light. He also demonstrated reflection, refraction, polarization, and interference of these waves — confirming that they were EM in nature, exactly as Maxwell had predicted 22 years earlier.
Q29. Explain the origin and uses of each region of the EM spectrum.
Answer:
Radio waves ($\nu<10^9$ Hz): Produced by accelerated charges in macroscopic conductors (antennas) and LC circuits. Used in AM/FM radio, TV, mobile phones, and astronomy (radio telescopes).
Microwaves ($10^9$–$10^{12}$ Hz): Produced by special electronic devices like klystrons and magnetrons. Used in radar, satellite communications, microwave ovens (water absorbs $2.45$ GHz), and Wi-Fi.
Infrared ($10^{12}$–$4\times10^{14}$ Hz): Emitted by hot objects through molecular vibrations. Used in remote controls, night-vision goggles, thermal imaging, IR spectroscopy, and the greenhouse effect (CO₂ traps Earth’s IR).
Visible light ($4-7\times10^{14}$ Hz): Produced by atomic transitions of outer electrons. Detected by the human eye between 400 nm (violet) and 700 nm (red). Essential for vision, photosynthesis, photography, and optical instruments.
Ultraviolet ($10^{15}$–$10^{17}$ Hz): Produced by special UV lamps, electric arcs, and the Sun. Used for sterilizing surgical instruments and water, in fluorescent lamps, and in LASIK eye surgery. Excess UV causes sunburn and skin cancer; the ozone layer absorbs most of it.
X-rays ($10^{16}$–$10^{20}$ Hz): Produced when high-energy electrons decelerate after striking a metal target (X-ray tube). Used in medical radiography (bones), CT scans, airport security, and crystal structure analysis (X-ray diffraction). Overexposure damages cells.
Gamma rays ($>10^{20}$ Hz): Produced by transitions in atomic nuclei (radioactive decay), nuclear reactions, and astrophysical events. Used in cancer radiotherapy, sterilizing food/medical equipment, and nuclear physics research.
Q30. What is meant by the polarization of EM waves? Why are sound waves not polarizable?
Answer: Polarization refers to the specific direction in which the electric field vector $\vec E$ oscillates in a plane perpendicular to the direction of propagation. In a linearly polarized EM wave, $\vec E$ stays along a fixed line. EM waves can be polarized because they are transverse. Sound waves, on the other hand, are longitudinal — particle oscillations are along the propagation direction, so there is no transverse plane in which to define a polarization direction. Hence sound waves cannot be polarized.
Q31. A point source emits 100 W of EM radiation uniformly in all directions. Find the intensity, $E_0$, and $B_0$ at a distance of 10 m from the source.
Answer: Surface area of a sphere of radius 10 m is $4\pi(10)^2=400\pi$ m². Intensity:
$$I=\frac{P}{4\pi r^2}=\frac{100}{400\pi}=0.0796\text{ W/m}^2$$
$$E_0=\sqrt{\frac{2I}{\epsilon_0 c}}=\sqrt{\frac{2(0.0796)}{(8.85\times10^{-12})(3\times10^8)}}\approx 7.74\text{ V/m}$$
$$B_0=\frac{E_0}{c}=\frac{7.74}{3\times10^8}\approx 2.58\times10^{-8}\text{ T}$$
Q32. Why are radio waves diffracted around buildings but X-rays not?
Answer: Diffraction is significant when the wavelength is comparable to or larger than the obstacle size. Radio waves have wavelengths from metres to kilometres, comparable to building sizes — they diffract strongly. X-rays have wavelengths near 0.1 nm — far smaller than any everyday object — so they travel essentially in straight lines and do not show observable diffraction except on atomic crystal lattices.
Q33. Distinguish between conduction current and displacement current.
| Conduction current | Displacement current |
|---|---|
| Flow of charge through a conductor | Arises from time-varying electric flux |
| $I_c=dq/dt$ | $I_d=\epsilon_0\,d\Phi_E/dt$ |
| Requires a material medium | Exists even in vacuum |
| Causes ohmic heating | No heating effect |
| Source of magnetic field | Also a source of magnetic field |
Q34. An EM wave of wavelength $\lambda$ travels through a medium of refractive index $n$. What is its wavelength inside the medium?
Answer: Frequency $\nu$ does not change when a wave enters a new medium. Speed becomes $v=c/n$, so wavelength becomes:
$$\lambda_{med}=\frac{v}{\nu}=\frac{c/n}{\nu}=\frac{\lambda}{n}$$
Q35. Why are EM waves not deflected by electric or magnetic fields?
Answer: EM waves carry no net charge — they are neutral oscillations of $\vec E$ and $\vec B$ fields. Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$ acts only on charges, not on the wave itself. Hence external electric or magnetic fields cannot deflect them. (Note: in nonlinear media or strong gravity, light path may bend, but that is a different mechanism.)
Q36. Calculate the displacement current between the plates of a parallel plate capacitor of plate area 0.01 m² when the electric field changes at $10^{12}$ V/(m·s).
Answer: $\Phi_E = EA$, so $d\Phi_E/dt = A\,dE/dt$.
$$I_d=\epsilon_0 A\frac{dE}{dt}=(8.85\times10^{-12})(0.01)(10^{12})=0.0885\text{ A}\approx 88.5\text{ mA}$$
Q37. Explain why a TV antenna placed on a tall mast receives signals better than at ground level.
Answer: TV signals are typically VHF/UHF (very high or ultra-high frequency, $>30$ MHz) and travel along straight lines (line-of-sight propagation), unlike low-frequency radio waves that reflect off the ionosphere. A taller antenna has a longer line-of-sight to the transmitter (less obstructed by buildings and the curvature of the Earth) and so receives a stronger, less distorted signal.
Q38. A laser pointer emits a beam of intensity 10 mW over a circular spot of radius 1 mm. Find the average $E_0$ and momentum carried per second per unit area.
Answer: Area $A=\pi(10^{-3})^2=3.14\times10^{-6}$ m². Intensity $I=10\times10^{-3}/(3.14\times10^{-6})=3185$ W/m².
$$E_0=\sqrt{\frac{2I}{\epsilon_0 c}}=\sqrt{\frac{2(3185)}{(8.85\times10^{-12})(3\times10^8)}}\approx 1549\text{ V/m}$$
Momentum per second per unit area = $I/c=3185/(3\times10^8)=1.06\times10^{-5}$ N/m² (radiation pressure on absorber).
Q39. List five major contributions of Maxwell’s theory to physics.
Answer: (i) Unification of electricity and magnetism into a single electromagnetic theory. (ii) Prediction of EM waves traveling at speed $c=1/\sqrt{\mu_0\epsilon_0}$. (iii) Identification of light as an EM wave, unifying optics with electromagnetism. (iv) Foundation for the discovery of radio waves, X-rays, and the entire EM spectrum. (v) Provided the framework that led to special relativity (the constancy of $c$) and quantum electrodynamics.
Q40. Why is short-wave (HF) radio used for long-distance communication on Earth?
Answer: Short-wave (3–30 MHz) radio is reflected by the ionosphere, an ionized layer of the upper atmosphere. By bouncing between the ionosphere and the Earth’s surface, HF signals travel thousands of kilometres around the globe — far beyond line-of-sight. VHF and higher frequencies penetrate the ionosphere and escape into space, so they cannot propagate this way.
Multiple Choice Questions (MCQs)
MCQ1. The displacement current was introduced by Maxwell because:
(a) electric currents do not produce magnetic fields
(b) Ampere’s law gave inconsistent results for a charging capacitor
(c) charges move slower than light
(d) magnetic fields cannot be produced by changing electric fields
Answer: (b). Ampere’s law in its original form gave different magnetic fields for two different surfaces bounded by the same Amperian loop when applied to a charging capacitor. Displacement current resolves this.
MCQ2. The EM wave in vacuum can be characterized by:
(a) electric field only
(b) magnetic field only
(c) both electric and magnetic fields
(d) neither
Answer: (c). EM waves consist of mutually perpendicular oscillating electric and magnetic fields, both perpendicular to the direction of propagation.
MCQ3. Which of the following has the highest frequency?
(a) Visible light
(b) Microwaves
(c) Ultraviolet rays
(d) Gamma rays
Answer: (d). Gamma rays have frequencies above $10^{20}$ Hz — the highest in the EM spectrum.
MCQ4. The ratio of $E_0$ to $B_0$ in an EM wave in vacuum equals:
(a) $\mu_0$
(b) $\epsilon_0$
(c) $c$
(d) $1/c$
Answer: (c). The amplitude ratio equals the speed of light, $E_0/B_0=c=3\times10^8$ m/s.
MCQ5. The EM waves used in RADAR are:
(a) ultraviolet
(b) microwaves
(c) infrared
(d) X-rays
Answer: (b). Microwaves (short wavelength, easily directed in beams) are used in RADAR.
MCQ6. Hertz’s experiment proved:
(a) Newton’s corpuscular theory
(b) Existence of electromagnetic waves
(c) Quantum nature of light
(d) Photoelectric effect
Answer: (b). Hertz’s 1887 experiment generated and detected EM waves, confirming Maxwell’s prediction.
MCQ7. The energy density of an EM wave is shared:
(a) entirely by $\vec E$
(b) entirely by $\vec B$
(c) equally by $\vec E$ and $\vec B$
(d) twice as much by $\vec B$ as by $\vec E$
Answer: (c). The average electric and magnetic energy densities are equal in an EM wave.
MCQ8. Ozone in stratosphere protects life on Earth by absorbing:
(a) X-rays
(b) Infrared rays
(c) Visible light
(d) Ultraviolet rays
Answer: (d). Ozone (O₃) absorbs most of the harmful UV-B and UV-C radiation from the Sun.
MCQ9. A wave of wavelength 5000 Å lies in which region?
(a) Radio
(b) Visible
(c) UV
(d) X-ray
Answer: (b). 5000 Å = 500 nm, in the visible (green) region.
MCQ10. EM waves carry:
(a) only energy
(b) only momentum
(c) both energy and momentum
(d) neither
Answer: (c). EM waves carry both energy ($U$) and momentum ($p=U/c$), exerting radiation pressure.
Quick Revision Points
1. Displacement current is $I_d=\epsilon_0\,d\Phi_E/dt$ — it makes Ampere’s law universally consistent.
2. Maxwell’s four equations + Lorentz force law fully describe classical electromagnetism.
3. Accelerating charges radiate EM waves; in free space they travel at $c=1/\sqrt{\mu_0\epsilon_0}$.
4. EM waves are transverse: $\vec E\perp\vec B\perp\hat c$, with $\hat E\times\hat B=\hat c$ and $E_0/B_0=c$.
5. Energy density: $u=\tfrac{1}{2}\epsilon_0 E^2+B^2/(2\mu_0)$; average $u=\tfrac{1}{2}\epsilon_0 E_0^2$.
6. Intensity: $I=u\,c=\tfrac{1}{2}\epsilon_0 E_0^2 c$. Radiation pressure on absorber = $I/c$, on reflector = $2I/c$.
7. EM spectrum order (increasing $\nu$): radio < microwave < IR < visible < UV < X-ray < gamma.
8. Hertz (1887) experimentally demonstrated EM waves predicted by Maxwell.
9. In a medium of refractive index $n$: speed becomes $c/n$ and wavelength becomes $\lambda/n$, but frequency is unchanged.
10. Photon energy: $E=h\nu=hc/\lambda$ — explains why X-rays/gamma rays are highly ionizing while radio waves are not.
Continue your ASSEB Class 12 Physics preparation with HSLC Guru’s chapter-wise notes and solved exercises in English medium. Master Maxwell’s grand synthesis of electromagnetism and unlock the complete spectrum from radio to gamma rays!