NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves

Why Maxwell needed displacement current. Ampère's original law ∮ B· dl = ₀ I_enc ran into trouble at a charging capacitor: an Ampèrian loop around the wire encloses I_c, but the same loop with its surface deformed to pass between the plates encloses no conduction current at all. Same loop, two different surfaces, two answers - a paradox. Maxwell patched it by adding ₀ d_E/dt, so the law became ∮ B· dl = ₀(I_c + ₀ d_Edt). This single fix produced symmetry with Faraday's law and predicted electromagnetic waves travelling at c = 1/√₀₀ - light!

Alternative computation of dV/dt. Note that V = q/C, so dV/dt = (1/C) dq/dt = I_c/C - exactly what we used. Equivalently E = V/d, so the field between the plates also changes at dEdt = 1d·dVdt = 1.87510¹⁰0.05 = 3.7510¹¹ V m^-1 s^-1.

Common mistake. Students often forget that R is in centimetres and plug R = 12 directly into π R², giving an area of 452 m² and an absurd capacitance. Convert to metres first, every single time.

Real-world flavour. In a working AM radio antenna or RF circuit, charging currents drive capacitor-like structures at megahertz frequencies. The displacement current in the dielectric is precisely what radiates the electromagnetic wave outward; without Maxwell's correction, neither radio nor Wi-Fi would be theoretically possible.

Symmetry argument behind the r²/R² factor. The displacement current density between uniformly charged circular plates is uniform across the plate area, j_d = I_d/π R². For an Ampèrian circle of radius r < R, the enclosed displacement current is j_d · π r² = I_d r²/R². This is the magnetic-field analogue of the "uniform current density inside a wire" problem and gives the familiar linear-in-r field inside B ∝ r and 1/r outside.

Alternative method via d_E/dt. One can also compute B by writing the flux through the Ampèrian disc explicitly: _E = E π r² with E = q/₀π R². Then d_Edt = r²₀ R²dqdt = r² I_c0 R², and B2π r = ₀₀ d_E/dt gives the same answer. The two formulations are equivalent because ₀ d_E/dt is the displacement current.

Common mistake. Confusing rms and peak. The question asks for the amplitude (peak) of B, so we must use I₀ = √2 I_rms, not I_rms directly. Similarly, do not confuse the electric field amplitude E₀ (V/m) with the magnetic field amplitude B₀ (T) - they differ by a factor of c (see Q 8.7).

Real-world flavour. The magnetic field of order 10^-11 T is incredibly tiny - roughly a billion times weaker than the Earth's field. Yet inside a working radio transmitter, similar capacitor-like elements generate the wave that drives an antenna; once radiated, the same EM wave carries information across continents.

Why c is universal. Maxwell derived the wave equation for the EM field directly from his four equations and found that the wave speed depends only on the constants ₀ and ₀, not on the source or the observer. Einstein elevated this to the second postulate of special relativity: the speed of light in vacuum is the same in every inertial frame. As a consequence, nothing carrying energy or information can outrun c.

What changes from X-ray to radio. Across the EM spectrum, what changes are wavelength, frequency, photon energy E = hν and the way each wave interacts with matter. X-rays ionise atoms; visible light triggers chemical and biological responses (vision); radio waves drive electrons in antennas. But the speed of propagation never changes.

Common mistake. Saying "wavelength" or "frequency" is the same - they are not, and the calculation above shows they differ by 10+ orders of magnitude.

Real-world flavour. Optical fibres, GPS timing, and radar ranging all rely on the invariance and precise value of c. The metre itself is now defined by fixing c ≡ 299 792 458 m/s; we measure lengths by measuring time-of-flight of light.

Why EM waves are transverse. Maxwell's equation ∇· E = 0 in vacuum forces any plane-wave solution Er,t = E₀ e^{ik·r - ω t} to satisfy k· E₀ = 0 - that is, E must be perpendicular to the wave vector k. The same equation for B gives k·B₀ = 0. And Faraday's law links them: B = k× E/c. So the trio E, B, k forms a right-handed orthogonal set.

Alternate way to remember the geometry. Curl the fingers of your right hand from E toward B; the thumb gives the propagation direction. If E along +x and B along +y, then E× B = +z - wave goes in +z. Reverse either field, and the wave reverses.

Common mistake. Drawing E and B along the same axis. They are always perpendicular to each other. Also confusing "perpendicular to propagation" with "always horizontal" - there is no global vertical for an EM wave; only the propagation direction matters.

Real-world flavour. A wavelength of 10 m sits in the upper VHF / lower shortwave band - used historically for FM radio, amateur ("ham") radio, and aircraft communication. Antennas tuned for this band are typically a metre or two long because optimal dipole length is λ/2 or λ/4.

Where this sits in the spectrum. Frequencies of 7.5-12 MHz fall in the High Frequency (HF) shortwave band, used worldwide for international broadcasting (e.g. BBC World Service, AIR shortwave) and amateur radio. These wavelengths reflect off the ionosphere and so can travel thousands of kilometres beyond the horizon - that's how long-distance shortwave broadcasting works.

Alternative shortcut. Memorise: = 300/. Then 300/7.5 = 40 and 300/12 = 25 - instant answer, no scientific notation needed.

Common mistake. Swapping the endpoints. The smallest wavelength corresponds to the largest frequency, not the other way round.

Real-world flavour. Compare with the AM broadcast band 530 kHz-1610 kHz → λ ≈ 186-566 m and FM broadcast 88 MHz-108 MHz → λ ≈ 2.8-3.4 m. Shorter wavelengths need shorter antennas, which is why FM car antennas are short 75 cm = λ/4.

Why the frequencies match. When the source charge has displacement x(t) = x₀2π₀ t, its acceleration is a(t) = -2π₀²x(t), which oscillates at the same ₀. The radiation electric field at far distance, by the Liénard-Wiechert formulas, is proportional to the retarded acceleration. So the radiated wave's electric field also oscillates at ₀. Spectral purity follows from the purely sinusoidal motion of the source.

Alternative phrasing. Think of the charge as one end of an antenna driven at ₀. The current in the antenna oscillates at ₀; by Maxwell's equations the radiated EM wave carries the same period. Frequency is a temporal property; it cannot be created or destroyed in linear propagation.

Common mistake. Some students try to derive a different wave frequency from the oscillation amplitude or assume the wave frequency is doubled (perhaps confusing with intensity or energy, which do go as the square). It is the field that copies the source frequency directly.

Real-world flavour. 1 GHz is the territory of mobile-phone signals (older 2G/3G bands), microwave ovens (2.45 GHz nearby), GPS L-band (~1.5 GHz), and Wi-Fi (2.4 GHz, 5 GHz). An oscillating charge at this frequency, multiplied billions of times in a real antenna's electron sea, is exactly how a Wi-Fi router emits its signal.

Why the ratio is exactly c. A plane wave moves rigidly at speed c. In one period the wave advances by λ; the field at a given point completes one cycle. Faraday's law balances the rate at which B changes (in time) against the spatial twist of E. For a sinusoid this enforces a fixed ratio of amplitudes equal to the wave speed. The same ratio falls out of the Ampère-Maxwell law applied to the same plane wave.

Energy partition. Because E₀ = cB₀, the electric and magnetic energy densities u_E = 12₀ E², u_B = B²2₀ are exactly equal at every instant (using c² = 1/₀₀). Total energy is split 50-50 between the electric and magnetic field - see Q 8.10(c).

Common mistake - units confusion. Students mix up tesla (T) and gauss (G), or use B in nT but forget to convert. Plain SI: Ts^-1 = V/m, which makes the calculation dimensionally clean.

Real-world flavour. B₀ of order 500 nT is comparable to ambient natural radio noise; E₀ ≈ 153 V/m is a fairly strong field - about the strength found very close to a high-power radio transmitter. Free-space wave impedance Z₀ = E/H = ₀ c = 377 Ω where H = B/₀ is a universal constant used to design antennas.

Choosing the direction of propagation. The problem does not specify a direction, so any consistent choice is acceptable. The convention is: if E is along y and the wave moves along +x, then B is along +z; if you flip B to -z, the wave must move along -x to conserve the right-hand rule.

Alternative form using sin. Equally valid is E = y E₀sin(ω t - kx), B = z B₀sin(ω t - kx), which represents the same wave with a different choice of phase origin. NCERT typically uses cosine; many textbooks use sine - interchangeable so long as both fields use the same phase.

Common mistake - units confusion between B and E. E is in V/m (or N/C, which is the same), and B is in tesla. They differ by a factor of c, so naively comparing their magnitudes ("B is much smaller!") is meaningless without restoring the c factor that lurks behind the units.

Real-world flavour. Frequency 50 MHz, wavelength 6 m, sits in the low-VHF band - used historically for analogue TV channel 2, and now for some amateur radio. A horizontal half-wave dipole here would be ∼ 3 m long. The field amplitudes E₀ = 120 V/m, B₀ = 400 nT are far above natural background and would correspond to a powerful broadcast located close to the receiver.

Why eV instead of joules. Atomic and nuclear energy gaps are conveniently ∼ 1-10⁶ eV, giving compact numbers. Using joules buries the physics under 10^-19 prefactors. The shortcut EeV = 1240/ is worth memorising - JEE/NEET-level photoelectric, atomic-physics and X-ray problems all use it.

Alternative way to see the source-energy match. A characteristic energy Δ E in a source corresponds to an emission frequency ν = Δ E/h. Hydrogen's outer-electron transitions release a few eV → visible/UV (Balmer series). Inner-electron transitions in heavy atoms release keV → X-rays. Nuclear de-excitation releases MeV → gamma rays. Conversely, the photon you detect tells you the energy scale of its source.

Common mistake. Mixing up the inverse relation: students sometimes think higher frequency means lower photon energy. Remember E = hν - frequency and energy go up together; wavelength is what goes down.

Real-world flavour. Medical X-ray photons (~50 keV) penetrate soft tissue because they are far more energetic than the binding energies of light atoms (C, H, N, O), so they aren't absorbed strongly except by heavier atoms like calcium in bone. Gamma photons from cobalt-60 (~1.3 MeV) penetrate even further, used in cancer radiotherapy and industrial radiography.

Deep meaning of the equality. An EM wave carries equal amounts of energy in its electric and magnetic components - they are not independent reservoirs but two manifestations of one unified field, related by Lorentz transformation. In a different inertial frame, what is purely electric here may appear partly magnetic; but the total energy density (and the speed c) remain frame-consistent in vacuum.

Total time-averaged energy density and intensity. u= u_E+ u_B= 12₀ E₀² ≈ 1.02× 10^-8 J m^-3. The intensity (power per unit area) is then I = uc ≈ 3.1 W m^-2 - roughly that of bright office lighting at a desk surface.

Common mistake - units confusion between B and E. When students compute B²/2₀ directly, they often forget to square the 10^-7, or mix units of tesla with gauss. Stick to SI and double-check the exponents.

Real-world flavour. 20 GHz, 1.5 cm wavelength sits in the K-band microwave region - used by satellite TV downlinks, automotive radar, and certain radio-astronomy bands. Water absorbs strongly near 22 GHz (the famous water-vapour line), so atmospheric humidity matters in satellite link design. The microwave-oven frequency (2.45 GHz) lies an order of magnitude lower, where water still couples but the wavelength ∼ 12 cm is large enough to penetrate food uniformly.