Electromagnetic Waves NCERT Exemplar Solutions - Class 12

Quick reading. The cleanest shortcut for ``what part of the EM spectrum does a photon of energy E live in?'' is to convert energy to wavelength via λ(nm) = 1240/E(eV) and read off the band. The number 1240 is the constant hc expressed in eV nm — a unit combination engineered precisely for spectroscopy questions like this one.

1. Build the shortcut. Starting from E = hc/λ, substitute h = 4.136× 10^-15 eV s and c = 3× 10⁸ m/s = 3× 10¹⁷ nm/s: hc = (4.136× 10^-15)(3× 10¹⁷) = 1240 eV nm. Hence λ(nm) = 1240/E(eV) whenever E is in eV and λ in nm.
2. Use the shortcut: λ = 1240 eV nmE = 124011 nm = 112.7 nm.
3. Band lookup: the visible window is 400–700 nm; UV-A/B is roughly 200–400 nm; the deep UV (often called extreme or vacuum UV) is below ∼ 200 nm. At λ ≈ 113 nm we are well inside the ultraviolet — specifically the vacuum-UV regime, absorbed by atmospheric oxygen.
4. Cross-check via frequency: ν = c/λ = (3 × 10⁸)/(112.7 × 10^-9) = 2.66 × 10¹⁵ Hz, matching Step 2 of the main solution to within rounding.
5. Sanity on the band boundary: visible-red edge is ∼ 700 nm = 1.77 eV; visible-violet edge is ∼ 400 nm = 3.1 eV. Eleven eV is roughly 4× the violet edge — comfortably outside visible, and deep into ionising UV.

Why this matters. Bond-breaking energies for most diatomic molecules sit between ∼ 4 and 11 eV. That is exactly why ultraviolet radiation (and harder) is dangerous to biological molecules: each UV photon is energetic enough to cleave covalent bonds in DNA. The CO dissociation energy of 11 eV is, in fact, near the upper end of typical bond energies and explains why CO is photochemically stable in the lower atmosphere — only solar extreme-UV penetrating to the upper mesosphere can break it apart.

Option (c) ultraviolet.

Picture-first. Treat the perfect mirror like the fixed end of a string. A right-moving pulse hits the wall, comes back left, and the wall ``pulls down'' to keep the displacement zero there. Translate that intuition into the wave equation, then verify the boundary condition imposed by a perfect conductor: tangential E = 0 at the surface.

1. Direction reversal. The sign of ω t inside the cos flips when the wave changes its direction of travel. A wave proportional to cos(kz - ω t) moves in +ẑ; cos(kz + ω t) moves in -ẑ. So the reflected wave's spatial dependence becomes cos(kz + ω t).
2. Hard-reflector inversion. At a perfect conductor the tangential component of the total E vanishes at the boundary. So E_inc + E_r = 0 at z = a for all t. The only way two cosines of equal amplitude can cancel pointwise is for one to carry a π phase flip — multiply by -1.
3. Polarisation does not flip. The wall is optically inactive (no birefringence, no Faraday rotation). Therefore the polarisation direction î is preserved; only the sign and the direction of k̂ change.
4. Combine. The reflected wave reads E_r = -E₀îcos(kz+ω t). Plug in z = a: E_r(a,t) = -E₀îcos(ka+ω t), E_inc(a,t) = E₀îcos(ka-ω t). For these to cancel for all t we need cos(ka+ω t) = cos(ka-ω t), which is true at z = a only if sin(ka)sin(ω t) = 0 for all t. In the standard idealisation this is enforced by choosing the reference plane so that ka = 0 (or by interpreting the boundary at the antinode); the matching is then exact.
5. Magnetic-field check. The reflected B does not flip sign — only E does at a perfect conductor. Combined with the direction reversal of k̂, this preserves S = (1/₀) E×B in the -ẑ direction (returning energy back into the half-space).

Why this matters. The same π phase change underlies the dark central fringe in a Newton's-rings setup, the formation of standing waves in laser cavities and antinodes of B exactly at the surface of microwave cavities. In every case the conductor ``pulls'' the tangential E to zero by sourcing surface currents whose radiation cancels the incident field.

Option (c).

Strategic angle. Dimension-watch first. The unit of W/c is W/(m/s) = J/m = N (force), and integrating force over time gives momentum. So multiplying power by t and dividing by c is exactly the recipe — no need to compute energy as an intermediate step. This is the radiation-force route, complementary to the energy-then-momentum route in the main solution.

1. Total power on the surface: P = (20 W/cm²)(30 cm²) = 600 W. Note we kept the data in cm² because the cm² cancels cleanly with the intensity's W/cm².
2. Radiation force on a perfect absorber: F = Pc = 600 W3 × 10⁸ m/s = 2 × 10^-6 N. Unit check: W/(m/s) = (J/s)/(m/s) = J/m = N. The force is a steady 2 pushing the surface forward.
3. Impulse = momentum transferred: p = F t = (2 × 10^-6 N)(1800 s) = 3.6 × 10^-3 kg m/s. Express as 36× 10^-4 to match option (b)'s style.
4. Order-of-magnitude check. The total absorbed energy is U = Pt = (600 W)(1800 s) = 1.08 MJ — enough to boil ∼ 4 kg of water. Dividing by c = 3× 10⁸ m/s gives the same momentum, 3.6× 10^-3 kg m/s — about the linear momentum of a 1 g ball moving at 3.6 m/s. Substantial energy, tiny momentum: the 1/c in p = U/c is what makes radiation pressure perpetually small at terrestrial intensities.

Why this matters. Solar sails use exactly this mechanism: sunlight at 1361 W/m² on a perfectly reflecting sail of area A delivers a continuous force 2P/c that can accelerate a spacecraft over years. JAXA's IKAROS (2010) had a 200 m² sail giving ∼ 1.8 mN of thrust — small but free, and adding up to large Δ v over months of cruise.

Option (b): 36 × 10^-4 kg m/s.

Quick reading. Power scales as field squared; halve the power and the field falls by √2. That is the entire question. But the same proportionality has multiple consequences worth unpacking, because students routinely mis-apply I ∝ E² as E ∝ P.

1. Quadratic link. I = 12c₀ E₀² — intensity is quadratic in the field amplitude, not linear. So E₀² ∝ I.
2. Distance dependence cancels. At a fixed distance r the spherical area 4π r² is common to both bulbs, so the ratio of intensities is just the ratio of source powers: I₅₀/I₁₀₀ = 1/2.
3. Take the square root carefully. E₀² ∝ P at fixed r, so E₀ ∝ √P: E₅₀E₁₀₀ = √P₅₀P₁₀₀ = √12 = 1√2.
4. Distractor diagnosis. Option (a) E/2 corresponds to E ∝ P — i.e. forgetting the square root. Option (d) √2E is the inverted ratio (higher-power bulb's field). Option (b) 2E confuses intensity and field altogether. Only (c) survives.
5. Cross-check via formula. The radial intensity at distance r is I = P/(4π r²) and I = 12c₀ E₀². Equating: E₀ = √P2π r² c₀. Halving P (with r unchanged) multiplies E₀ by 1/√2, exactly.

Why this matters. The √P scaling is why doubling the power of a radio transmitter doesn't double its range — the field amplitude grows only by √2, and the range scales like the field's 1/r falloff. The same square-root crops up in mechanical waves (string-amplitude vs. power on a string), in optics (visibility of fringes vs. source brightness), and in acoustics (sound pressure ∝ √P).

Option (c): E/√2.

Picture-first. ``Energy flows in the direction E pushes a charge crossed with where the magnetic force then deflects it.'' That's the Poynting picture in one sentence. Three independent arguments converge on the same answer — order them so each reinforces the next.

1. Poynting vector. S = (1/₀) E×B. S points along the direction of energy flow for any EM field, and for a plane wave in vacuum that is the wave's direction of propagation.
2. Right-hand-rule test with concrete vectors. Take E = E₀î and B = B₀ĵ at an instant where both cosines are positive. Then E×B = E₀ B₀ (î×ĵ) = E₀ B₀ k̂, which is +ẑ. This is exactly the direction the wave's argument (kz - ω t) tells us to expect (phase fronts move in +ẑ). Consistency confirmed.
3. Eliminate B×E. The cross product is antisymmetric, so option (c) would put the energy flow in -ẑ — backwards. The convention picks the order E×B because Maxwell's equations produce that ordering naturally (Poynting's theorem starts from _t (u_E + u_B) + ∇·S = 0).
4. Quick rejection of (a) and (b). Both E and B are transverse to k̂ — they cannot themselves be the propagation direction. Transversality is a deep consequence of Maxwell's equations in vacuum (no sources ⇒ no longitudinal components).

Why this matters. This rule lets you instantly orient any EM-wave problem: pick the polarisation E, sketch B at 90^∘ in the plane perpendicular to k̂, and the wave travels along E×B. The same right-handed triad underlies the design of optical isolators, microwave waveguides and the polarisation logic of antenna arrays.

Option (d).

Strategic angle. Equipartition. In a plane EM wave the electric and magnetic ``springs'' carry equal average energy at every moment. This is not a numerical coincidence but a consequence of the wave equation linking E and B through E = cB.

1. Two energy-density formulae. u_E = 12₀ E² is the energy stored in the electric field per unit volume; u_B = B²/(2₀) is the same for the magnetic field. Both formulae are general (they apply to any electromagnetic field, not just waves).
2. EM-wave relation. For a plane wave in vacuum, E = c B at every instant — not just on average. Substitute into u_E: u_E = 12₀ (cB)² = 12₀ c² B².
3. Algebraic equality. Use c² = 1/(₀₀), i.e. ₀ c² = 1/₀: u_E = 12·1₀· B² = B²2₀ = u_B. So u_E = u_B at every point of space and every instant of time — perfect equipartition, no averaging needed.
4. Intensity ratio. Intensity is I = c u_total = c(u_E + u_B) = 2c u_E. Each contributes exactly half. So the ratio of the two contributions to I is 1:1.
5. Why options (a), (b), (d) are wrong. They treat u_E = 12₀ E² and u_B = B²/(2₀) as if E and B were independent variables. But they are not — the wave equation chains them through E = cB. Forgetting that link is the only way to manufacture ratios like c:1 or c²:1.

Why this matters. This is the EM analogue of equipartition between kinetic and potential energy in a simple harmonic oscillator; the wave equation is just two coupled first-order PDEs (one for E, one for B) that act as two coupled oscillators. The energy sloshes between electric and magnetic forms exactly the way kinetic and potential energy slosh in a pendulum — except here the sloshing happens twice per period because both u_E and u_B oscillate as cos².

Option (c): 1:1.

Strategic angle. Use power conservation across a sphere. Total radiated power P is fixed by the source, so the average power crossing any sphere of radius r around the antenna must equal P. That single constraint forces E₀∝ 1/r in the far field.

1. Power-conservation argument. Set P = I(r) · 4π r² (intensity times sphere area). P is set by the source and does not change with r; the area grows as r². Hence I(r) = P4π r², I∝ 1r².
2. Field-from-intensity. Intensity is proportional to E₀² in any plane wave: I = 12c₀ E₀². So E₀² ∝ 1/r² and therefore E₀ ∝ 1r.
3. Contrast with the near-field. For a static (or quasi-static) dipole moment p, the field falls as E∝ p/r³. This is the r≪ λ regime, dominated by the electrostatic potential. The 1/r form takes over only at r≫ λ, where the term carrying p̈ (acceleration) dominates the multipole expansion.
4. Quantitative crossover. The crossover from near to far field occurs at r∼ λ/(2π). Beyond that, the 1/r radiation field dwarfs the 1/r³ static tail, and beyond a few wavelengths the static tail is negligible.
5. This 1/r amplitude (not 1/r²) is what makes long-range radio possible. If E fell as 1/r², the signal would be undetectable beyond a few wavelengths of the antenna — radio communication would be impossible.

Why this matters. Antenna design is a long argument about how to maximise the 1/r far-field while shaping its angular distribution. Yagi, dish, and phased-array antennas all chase the same 1/r envelope. The same 1/r behaviour governs sound waves, gravitational waves, and any radiation phenomenon obeying a wave equation in three dimensions — it is geometry, not electromagnetism specifically.

Option (c): 1/r.

Strategic angle. Two independent yes/no checks: (i) is the B in option (b) really k̂×E/c? (ii) is the E here a fixed-direction oscillation? Each test is a one-line cross-product or a one-line phase comparison.

1. Cross-product check using the right-hand rule. Recall the cyclic identities: î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ. For the wave moving along +k̂: k̂×(E₁î+E₂ĵ) = E₁ĵ - E₂î, and dividing by c gives B = (E₁ĵ-E₂î)/c times the wave envelope. Confirms option (b).
2. Reject option (a). Its B is ∝ E₁î - E₂ĵ — the swap of î↔ĵ gone wrong. The Poynting vector E×B with option (a)'s B would not point along k̂, violating the propagation direction. Reject.
3. Polarisation analysis. E(z,t) = (E₁î+E₂ĵ)cos(kz-ω t). The direction vector is the constant unit ê = (E₁î+E₂ĵ)/√E₁²+E₂² — it never rotates with time. The tip of E therefore traces a straight line along ê, which is the textbook definition of plane (linear) polarisation. Confirms option (d).
4. Why option (c) is wrong. Circular polarisation requires the two Cartesian components of E to have the same amplitude and a 90^∘ phase difference, i.e. E = E₀[îcos(kz-ω t) ± ĵsin(kz-ω t)]. Here both components share the same cos(kz-ω t) — no phase difference at all, so the wave is linearly polarised.
5. Geometry of E. The direction of ê in the xy-plane is at angle arctan(E₂/E₁) from î, and B is perpendicular to it in the same plane (rotated by 90^∘ toward the propagation direction by right-hand rule), again constant in direction. Both fields oscillate along fixed directions while propagating in +k̂.

Why this matters. Polarisation is the cleanest way to test whether you've correctly identified the relative phase of two orthogonal field components in a wave. Polaroid filters, LCD screens, and 3D-glasses all rely on the linear-vs-circular distinction being sharp.

Options (b) and (d).

Strategic angle. Two checks per option: (i) does it respect transversality? (ii) does it sit in the right place in the E=c B×k̂ ring of identities?

1. Transversality kills (d) instantly. The condition k̂×E=0 forces E∥k̂ — a longitudinal electric field. An EM wave in vacuum is strictly transverse (this is what Maxwell's source-free equations enforce). So (d) is false. Accept (c) (which is just the transversality statement written as dot products).
2. Faraday's law for a plane wave. Substituting E = E₀cos(k·r - ω t) into ∇×E = -_tB gives k×E₀ = ω B₀ ⇒ B = 1ck̂×E, using ω/k = c in vacuum. Confirms option (a).
3. Read the same relation in reverse. Cross both sides of B = (1/c)k̂×E with k̂ from the right: B×k̂ = 1c(k̂×E)×k̂ = 1cE, where the last equality uses the BAC–CAB rule together with k̂·E=0. Therefore E = c B×k̂. Accept (b).
4. Re-cast option (a) form. The Exemplar writes B = (1/c)k̂×E = (1/ω)k̂×E. The second equality is only correct if we read k̂ as the full wave-vector k (not the unit vector) — then k×E/ω = k̂× E/c. The Exemplar's slightly loose notation hides this subtlety, but the physical content is the same. Accept (a).
5. Summary identity wheel (worth memorising): B = 1ck̂×E, E = -c k̂×B = c B×k̂, k̂·E = k̂·B = 0. Three relations, one underlying right-handed triad.

Why this matters. Memorising the small ring of EM-wave identities (B = k̂×E/c and its inverses) trivialises a large fraction of EM-wave MCQs and is the starting point for any antenna or waveguide derivation.

Options (a), (b), (c).

Picture-first. For propagation along x̂, sweep your hand through the yz-plane: any pair of perpendicular directions in that plane gives a valid (E,B). Pull two filters through the option list: forbid longitudinal components, then enforce k̂·(E×B) > 0.

1. First filter — no longitudinal components. Propagation is along x̂, so neither field may have an x̂-component. Option (a) has E_x (longitudinal E, forbidden by Gauss's law in vacuum). Option (c) has B_x (longitudinal B, forbidden by ∇·B=0 on a plane wave). Strike both.
2. Second filter — right-handed triad. For propagation along +x̂, we need E×B ∥ +x̂. Test (b): ŷ×ẑ = x̂. Passes. Test (d): ẑ×ŷ = -x̂ — but the option only specifies axes, not signs. With B along -ŷ the cross product becomes +x̂. So (d) represents the polarisation state ẑ rotated 90^∘ from (b)'s state ŷ, with matched sign convention. Both describe valid waves.
3. Polarisation interpretation. (b) is linearly polarised along ŷ; (d) is linearly polarised along ẑ. These are the two orthogonal polarisation basis states of an x-propagating wave; any other polarisation (including circular and elliptical) is a complex superposition of these two.
4. Field-amplitude link still holds. Even though directions differ between (b) and (d), the amplitudes must satisfy |E₀|/|B₀| = c for both options, just as in any plane wave. The question is asking only about axes, not magnitudes.
5. Why this question type matters. Multiple-correct problems on polarisation rely entirely on a clear count of how many independent transverse directions there are. In 3D for a wave moving along one axis, the transverse plane is 2D — so exactly two independent linear polarisations.

Why this matters. The number of independent polarisation states of a transverse wave in 3D equals the dimensionality of the plane perpendicular to k̂, which is 2. The same count shows up everywhere: two helicity states of the photon, two gravitational-wave polarisations (+ and ×), two linearly independent shear-wave polarisations in elastic media.

Options (b) and (d).

Strategic angle. Three independent facts about a 1 GHz radiator: frequency, wavelength, band. Build each from the charge-acceleration argument and the simple kinematics of a wave.

1. Frequency of radiation. A non-relativistically oscillating charge q(t) radiates at the same temporal frequency as its mechanical motion (Larmor radiation, Fourier analysis of q̈). So _rad = _osc = 10⁹ Hz. Confirms (a); refutes (b)'s factor of two (a common confusion with the squared intensity).
2. Wavelength from c=λ. λ = cν = 3 × 10⁸ m/s 10⁹ Hz = 0.3 m = 30 cm. Confirms option (c).
3. Spectrum locator. EM bands (very approximate):
   • Radio: ν 10⁹ Hz, λ 0.3 m;
   • Microwave: 10⁹ ν 10¹¹ Hz;
   • IR, visible, UV, X-ray, γ above that.
   1 GHz sits right on the radio/microwave boundary; the NCERT Exemplar classifies it as ``radio''. Confirms (d).
4. Antenna-length cross-check. A half-wave dipole at λ = 30 cm has length 15 cm, and a quarter-wave monopole is ∼ 7.5 cm — both match the physical size of mobile-phone antennas, confirming we are in the right band by another route.
5. Order-of-magnitude sanity. Visible-light photons have ν∼ 5× 10¹⁴ Hz, five orders of magnitude higher; 1 GHz is therefore very far from visible — consistent with the band being radio/microwave.

Why this matters. Mobile-phone bands (900, 1800, 2400 MHz) all live in exactly this regime. The fact that the antenna length is comparable to λ/4 ≈ 7.5 cm explains why mobile antennas are short. Wi-Fi at 2.4 GHz means λ = 12.5 cm — again of the order of an antenna's physical size.

Options (a), (c), (d).

Strategic angle. The only question is: does the charge have nonzero acceleration? The Larmor formula bakes the answer in cleanly, and an order-of-magnitude estimate shows just how sensitive the radiation is to a itself.

1. Filter all four options by a.
   • Rest: v = 0, a = 0.
   • Constant velocity: a = 0.
   • Circular orbit: a = v²/r ≠ 0 (centripetal).
   • Free fall in E: a = qE/m ≠ 0.
2. Apply the Larmor formula. P_rad = q² a²6π₀ c³. Radiation ⇔ nonzero a. So (b) and (d) radiate; (a) and (c) do not.
3. Why ``constant velocity'' is non-radiating, even though B exists. A uniformly moving charge produces a co-moving electric and magnetic field, but in the charge's rest frame both reduce to a pure Coulomb electrostatic field — no radiation, by Lorentz covariance. Only an acceleration leaves a non-removable retarded-field tail.
4. Numerical taste for (b)/(d). Take a 10 keV electron in a 1 m-radius orbit. Its v≈ 6× 10⁷ m/s, so a = v²/r ≈ 3.6× 10¹⁵ m/s² and P ≈ (1.6× 10^-19)²(3.6× 10¹⁵)² 6π(8.85× 10^-12)(3× 10⁸)³ ≈ 7× 10^-26 W. Tiny — but multiply by 10¹² electrons in a synchrotron and you get μW to mW of useful synchrotron light.
5. Distractor diagnosis. Option (a) is the classic trap: ``moving charges produce magnetic fields, don't magnetic fields radiate?'' No — a steady B does not radiate. Only changing fields radiate, and changes require acceleration of the source.

Why this matters. This is why classical atomic models fail: an electron orbiting a nucleus would radiate continuously (it's accelerating!) and spiral into the nucleus in ∼ 10^-11 seconds. The resolution required quantum mechanics, where stationary states have time-independent probability currents and therefore no Larmor radiation. Synchrotron light sources and bremsstrahlung X-ray tubes exploit the same Larmor formula on purpose.

Options (b) and (d).

Strategic angle. Two extremes and an inequality. Absorber: I/c. Reflector: 2I/c. Real surface: somewhere in between. Photons make the physics transparent, but the macroscopic energy–momentum argument is equally complete and pre-photon.

1. Photon picture. Each photon carries momentum hν/c when moving forward. Absorption stops the photon dead: surface gains hν/c per photon. Reflection reverses the photon: change in photon momentum is -2hν/c, so surface gains +2hν/c — twice the absorption case.
2. Convert to pressure. The photon flux on area A is Φ = I A/(hν) photons per second. Force is Φ times the per-photon impulse: F_abs = IAhν·hνc = IAc, F_ref = 2IAc. Divide by A for pressure: p_abs = I/c, p_ref = 2I/c. Confirms (a), refutes (b), confirms (c).
3. Pre-photon, classical derivation. Energy delivered in Δ t to area A: U = I AΔ t. Momentum carried by an EM wave is p = U/c. For absorption, all of this momentum is delivered: p_abs = U/(cAΔ t) = I/c. For reflection, the wave's own momentum reverses, so the surface receives 2U/c: p_ref = 2I/c.
4. Real surface inequality (d). For a partial reflector with reflectance R∈[0,1] and absorptance α=1-R, p = Ic(2R + α) = Ic(1 + R). At R=0 (perfect absorber): p = I/c. At R=1 (perfect reflector): p = 2I/c. For any real surface 0 < R < 1, I/c < p < 2I/c. Confirms (d).
5. Order-of-magnitude reality check. Sunlight on Earth: I = 1361 W/m². Perfectly reflecting: p ≈ 2× 1361/(3× 10⁸) ≈ 9 — about 10^-10 of atmospheric pressure. Tiny but real, and large enough to lift dust grains off small asteroids over time.

Why this matters. The IKAROS solar-sail probe (JAXA, 2010) used a thin aluminised polymer film closer to a perfect reflector than absorber to maximise this 2I/c thrust. The same physics drives comet-tail dynamics, Poynting–Robertson drag on dust orbiting the Sun, and Arthur Ashkin's optical tweezers.

Options (a), (c), (d).

Quick reading. Antenna picks up the component of E along its length. Rotate the antenna and you change that projection. Three things are happening simultaneously: polarisation of the broadcast, projection along the antenna, and the receiver's sensitivity to induced EMF.

1. Polarisation of the wave. A broadcast antenna driven along a fixed axis radiates an EM wave whose E vector lies along that same axis (the dipole's geometric plane). So the received wave is linearly polarised.
2. Antenna response. A straight conductor of length pointing along ̂ has an induced EMF E = ₀ E· d = E cosθ, where θ is the angle between E and ̂. At θ = 0 the EMF is maximum; at θ = π/2 it is zero.
3. Power received scales as cos²θ because the delivered power is E²/R where R is the antenna impedance. So rotating the antenna by, say, 45^∘ halves the received power, while 90^∘ drops it to zero in principle.
4. Why orientation matters in practice. AM broadcasts are vertically polarised; FM and TV often horizontally polarised; satellite uplinks sometimes circular. Pocket transistor radios with internal ferrite-rod antennas are magnetically coupled — and the rod's orientation must match B instead of E. Same physics, different field component.
5. Concept linkage to Malus' law. The cosine-projection rule for an antenna is the radio analogue of Malus' law for optical polarisers: I_out = I_incos²θ. Both reflect the projection of a polarisation vector onto a ``preferred axis''.

Why this matters. Polarisation matching is the entire reason that satellite TV dishes have a feed-horn whose polariser must be rotated to the correct angle when first installing. Mismatched polarisation can cost 20–30 dB of signal strength — the difference between a usable channel and pure noise.

Signal ∝ cosθ; orient antenna along E.

Strategic angle. Resonance + dipole coupling. Two ingredients must be present at the same time: a permanent molecular dipole, and an oscillating field at a frequency near the molecular relaxation rate.

1. Water has a large permanent dipole. The H-O-H bond angle is 104.5^∘, leaving a net dipole moment of ∼ 6.2× 10^-30 C m (≈ 1.85 debye) per molecule. Most other kitchen materials (glass, ceramic, many plastics) are non-polar or have much smaller dipole moments, so they couple weakly to the oven's E-field.
2. Torque on a dipole in an oscillating field. τ = p×E, which tries to align the dipole with E. Because E reverses every half-period T/2 = 1/(2ν) ∼ 2× 10^-10 s, the dipoles rotate back and forth at 2.45 GHz.
3. Microwave frequency (ν = 2.45 GHz) is tuned near the rotational/dipolar relaxation peak of liquid water at room temperature. ``Tuned near'' (not exactly at): engineers deliberately shift the oven frequency off the peak so the heating depth is uniform — exact resonance would heat only the surface where intensity is largest.
4. Energy-to-heat conversion happens through dielectric loss. As each water molecule rotates, it collides with its neighbours, transferring rotational kinetic energy into translational kinetic energy of the bulk — i.e. heat. The loss tangent tanδ of liquid water is large at microwave frequencies and small at radio/optical frequencies, which is why microwave heating is so efficient.
5. Why ice doesn't heat well. In ice, the water dipoles are frozen into a lattice — they cannot rotate freely, so the dielectric loss collapses by a factor of ∼ 100. That is why microwave ovens have a defrost cycle that pulses the magnetron: heat melts a thin liquid layer, which then absorbs microwaves rapidly and melts further layers.
6. Penetration depth. The microwave field decays into the food over ∼ 1–2 cm for typical water-rich foods at 2.45 GHz — small enough for surface heating, large enough for uniformity in a thin steak.

Why this matters. A microwave heats the food, not the plate. That is also why microwaves can be uneven: parts of the food with less water (a dry biscuit centre) heat far more slowly. The same dipolar-relaxation physics underlies industrial-scale dielectric drying of wood and paper, microwave-assisted chemistry, and THz spectroscopy of liquids.

Water's permanent dipole couples resonantly to the 2.45 GHz field, converting EM energy to heat.

Quick reading. ``Displacement current between plates equals the conduction current in the wires charging them.'' That is the content of I_d = dq/dt. The full derivation only needs to compute d_E/dt once and notice that the ₀ cancels.

1. Field between plates. The uniform field is E = σ/₀ = q/(₀ A). Multiply by area A to get the flux that threads any loop wholly between the plates: _E = q/₀.
2. Displacement current. I_d = ₀ d_Edt = ₀·1₀ dqdt = dqdt. The ₀ factors cancel — that is the structural identity ``I_d in the gap = I_c in the wire''.
3. Differentiate the given q(t) = q₀cos(2ν t). I_d = dqdt = -q₀ (2ν)sin(2ν t) = -2ν q₀sin(2ν t). Peak value |I_d|_max = 2ν q₀ = ω q₀.
4. Phase check. q but I_d∝ -sin: I_d leads q by π/2. This 90^∘ lead is exactly the AC-circuit result that ``current leads voltage in a capacitor'', reproduced here as a microscopic claim about the displacement-current density.
5. Dimensional check. [ν q₀] = (1/s) · C = A. So I_d comes out in amperes, the right unit for current. (Without Maxwell's ₀ d_E/dt prescription, the unit balance would fail in any non-vacuum problem.)

Why this matters. Without Maxwell's I_d, Amp`ere's law would be violated in the gap between charging-capacitor plates — and EM waves wouldn't exist. This is the historical step that turned electromagnetism from a set of empirical laws into a coherent field theory: the addition of _{0 t}E to the ₀J source of ∇×B.

I_d = -2ν q₀sin(2ν t).

Quick reading. Capacitor passes high frequencies, blocks low ones — straight from X_C = 1/(ω C). The displacement current is the gap-physics manifestation of this AC-circuit fact.

1. Capacitive reactance. The capacitor's reactance is X_C = 1/(ω C) = 1/(2ν C). Lower ν means larger X_C — the capacitor opposes the AC more strongly.
2. Peak conduction current through the wires charging the plate: I₀ = V₀/X_C = 2ν C V₀. Linear in ν.
3. Displacement current in the gap equals conduction current in the wires (Maxwell): I_d,max = 2ν C V₀. So I_d is set by the same formula.
4. Effect of decreasing ν. ν↓ ⇒ X_C↑ ⇒ I_d↓, linearly. At ν = 0 (DC) the capacitor blocks current entirely — and I_d = 0, as it should be (no time-varying field, no displacement current).
5. Numerical taste. Take C = 1 , V₀ = 10 V. At ν = 50 Hz (mains): I_d = 2π(50)(10^-6)(10) ≈ 3.1 mA. At ν = 5 Hz: I_d = 0.31 mA. A tenfold drop in frequency, a tenfold drop in current — visible to a milliammeter, exactly the linear scaling.
6. Equivalent statement. The displacement-current density J_d = _{0 t}E is itself ∝ ν for sinusoidal E, so the ``I_d ∝ ν'' result is just _t ∝ iω spelled out for a sinusoid.

Why this matters. This is why coupling capacitors in audio electronics work as high-pass filters: low-frequency components (hum at 50/60 Hz, slow drift) are attenuated, high-frequency components (audio signal at 100 Hz–20 kHz) pass. The same logic determines the crossover frequency in a loudspeaker network.

I_d,max ∝ ν; decreasing ν decreases I_d proportionally.

Strategic angle. Single plug-in: read B₀, drop into I = B₀² c/(2₀). Two cross-checks confirm the answer through independent routes — the E₀ form, and a numerical sanity check against solar irradiance.

1. Read off B₀. The wave is B_z(t) = 12× 10^-8sin(⋯) T, so the magnetic-field amplitude is B₀ = 1.2× 10^-7 T. Wave-vector k = 1.2× 10⁷ m^-1 and angular frequency ω = 3.6× 10¹⁵ rad/s identify the wave as visible-to-UV: λ = 2π/k ≈ 523 nm (green light).
2. Apply the magnetic intensity formula. I_av = 12B₀² c₀ = (1.2× 10^-7)² (3× 10⁸) 2 (4π× 10^-7). Compute numerator: (1.2)² = 1.44; (10^-7)² = 10^-14; 1.44× 10^-14× 3× 10⁸ = 4.32× 10^-6. Denominator: 8π× 10^-7 ≈ 2.513× 10^-6. Quotient: ≈ 1.72 W/m².
3. Cross-check via E₀ route. E₀ = c B₀ = (3× 10⁸)(1.2× 10^-7) = 36 V/m. Then aligned I_av &= 12c₀ E₀²
   &= 12(3× 10⁸)(8.85× 10^-12)(36)²
   &= 12(2.655× 10^-3)(1296)
   &≈ 1.72 W/m². aligned Same answer, confirming algebraic consistency.
4. Order-of-magnitude check. The Sun delivers about 1361 W/m² at Earth's orbit (the solar constant); the filtered beam here is roughly 1.72/1361 ≈ 1/800 of that — plausible for a beam from a floodlight after passing through a coloured filter.
5. Energy interpretation. For a 1 m² detector, the beam delivers 1.72 J per second — enough to register on a thermopile but invisible to a calorimeter.

Why this matters. Spectral radiometry (measuring the intensity of a single wavelength) underlies solar-cell calibration, laser-power-meter standards and astronomical photometry. The same plug-in formula gives all of them.

I_av ≈ 1.72 W/m².

Picture-first. Two in-phase cosines multiplied give a cos², which lifts the curve so it never goes negative. The ``never negative'' feature is the physical statement that energy flows in one direction (along k̂) for a wave moving in one direction — never reverses.

1. Sub the wave forms into the Poynting vector S = E×B/₀. With E = E₀îcos(ω t - kz) and B = B₀ĵcos(ω t - kz) (same phase): S(t) = E₀ B₀₀cos²(ω t - kz).
2. Graph features:
   • Always non-negative (cos²≥ 0): energy flow is always along +k̂.
   • Peaks at S_max = E₀ B₀/₀ whenever ω t - kz = nπ (every half-period of the field).
   • Zeros at ω t - kz = (n+12)π — the instants when both E and B pass through zero.
   • Period T/2 = π/ω: twice the field's temporal frequency because cos² has half the period of cos.
3. Time-average. ²= 1/2, so S= E₀ B₀2₀ = I, the wave's intensity. Drawn as a dashed horizontal line, this passes exactly through the midpoint of the cos² lobes.
4. Identity trick. Using cos²θ = (1 + cos 2θ)/2 we can re-write S(t) = E₀ B₀2₀[1 + cos(2ω t - 2kz)], i.e. a DC component (the mean intensity) plus a sinusoidal ripple at 2ω. The doubled frequency makes cos²-shaped graphs an immediate fingerprint of of two in-phase sinusoids.
5. Why a slow detector reads only the average. Visible light at ω∼ 10¹⁵ rad/s has T/2 ∼ 3× 10^-15 s — billions of times faster than the response of any photodetector or human eye. So we see the plateau S, not the underlying lobes.

Why this matters. Detectors (eyes, photodiodes, thermopiles) are too slow to resolve a single S(t) lobe; they measure the time-average, which is the intensity. The cos²-and-mean picture is the same one that governs LRC-circuit power dissipation, sound intensity, and squared photodiode signals in homodyne detection.

S-vs-t is a cos² curve of height E₀ B₀/₀ and period T/2 with mean E₀ B₀/(2₀).

Strategic angle. Light pushes. Identify the property, estimate the force, and pull out a comet-tail example to anchor the idea in everyday astronomy.

1. Identify the property. The laser holds a tiny ball against gravity. So the laser must be exerting an upward push on it. That push, divided by the ball's cross-sectional area, is the radiation pressure. The microscopic origin: each photon in the beam carries momentum hν/c, and on reflection/absorption transfers some or all of that to the ball.
2. Quick order of magnitude. For a ball of mass m∼ 10 and area A∼ 10^-6 m², weight is mg = 10^-7 N. Required pressure: p = F/A = 0.1 Pa. From p = 2I/c (perfect reflector), needed intensity is I = pc/2 ≈ (0.1)(3× 10⁸)/2 ≈ 1.5× 10⁷ W/m² — comfortably within a focused continuous-wave laser, so the demonstration is feasible (and indeed C. V. Raman performed it in the 1920s).
3. Comet-tail example. A comet near the Sun emits dust and gas. Solar photons, streaming radially outward, push the dust particles via radiation pressure on a timescale of days, building a tail that points away from the Sun regardless of the comet's velocity. The dust tail's curvature encodes the ratio of solar radiation pressure to solar gravity acting on each grain.
4. More examples for variety.
   • Solar sails (JAXA's IKAROS, 2010; LightSail-2, 2019) — propellantless spacecraft propulsion.
   • Optical tweezers (Ashkin, Nobel 2018) — trap living cells in a focused laser.
   • Atom-cooling traps (Doppler cooling) — slow atoms by photon-momentum kicks.
   • Crookes radiometer (folklore answer; technically a gas-kinetics effect, but often cited).
5. Concept link. Radiation pressure is the macroscopic face of the photon-momentum relation p = hν/c. Without this momentum, E×B in vacuum would carry only energy — no Newton-second-law balance against a gravitational pull would be possible.

Why this matters. Modern ``optical tweezers'' (Arthur Ashkin's Nobel work) use exactly this principle to trap and move microscopic particles, including single cells, single viruses, and even single atoms (in magneto-optical traps). The same radiation pressure governs the orbital decay of micron-scale dust grains around the Sun (Poynting–Robertson drag) and sets the upper mass limit for stars.

Radiation pressure; comet tails point away from the Sun.

Strategic angle. Amp`ere–Maxwell on a coaxial loop with no conduction current — exactly the situation in which Maxwell's correction reveals itself most cleanly. Symmetry does the work; only the displacement-current term contributes.

1. Symmetry setup. Take a circular loop of radius r coaxial with the capacitor's central axis, lying between the plates. Cylindrical symmetry forces B tangent to the loop with constant magnitude B on the loop.
2. Left-hand side (circulation): ∮B· d = B (2π r).
3. Right-hand side. Inside the gap there is no conduction current (I_c = 0). The only contribution is the displacement current I_d = ₀ d_Edt = ₀ π r² dEdt (flux through the loop's bounding disk; E uniform between plates). In a medium of relative permeability _r the Amp`ere–Maxwell coefficient is _0r.
4. Equate & solve. B (2π r) = _0r·₀π r² dEdt ⇒ B = _0r0 r2 dEdt. The Exemplar's form absorbs ₀ into the medium constant: B = (_0r r/2) (dE/dt).
5. Direction by right-hand rule. If E points downward (positive plate above) and dE/dt > 0 (capacitor charging more positively), then B circulates anticlockwise as seen from above — exactly the direction of a magnetic field induced by an upward-pointing conduction current. The displacement-current ``stands in'' for an equivalent conduction-current pattern.
6. Linear-in-r inside, 1/r outside. For r > R (plate radius), the area of the loop enclosing displacement current is the full plate area π R², so B(r > R) = _0r0 R²/(2r)· dE/dt — the usual 1/r Amp`erian falloff. The crossover at r = R matches both forms.

Why this matters. The same B-from-changing-E mechanism is what closes the loop in Maxwell's equations and allows free EM waves: a changing E sources a B, which by Faraday's law sources another E, and the leap-frog process propagates at speed c.

B = 12_0r0 r (dE/dt), equivalently (_0r r/2) (dE/dt).

Strategic angle. Match each application to its band, sort by wavelength from shortest to longest, then quote one extra application per band. The trick is being decisive about which band actually does each job — every band has a ``cousin'' that almost-but-not-quite works.

1. Satellite communication: microwave (₁). Bands like C (4–8 GHz), Ku (12–18 GHz), Ka (26–40 GHz) correspond to λ ∼ 1 cm to 7 cm — microwave through atmospheric windows that transmit cleanly without rain attenuation (mostly).
2. Germicidal water purifier: UV-C (₂). Wavelength ∼ 254 nm (mercury-lamp peak) damages DNA/RNA, killing bacteria and viruses. UV-A/B are less effective; visible light cannot disrupt covalent bonds.
3. Underground oil-leak detection: X-rays (₃). Penetrating radiation reveals density variations in pipe walls and surrounding soil. Wavelength ∼ 10^-10 m. (Some textbooks substitute gamma-rays here; both occupy the ionising end of the spectrum.)
4. Fog visibility on runways: infrared (₄). IR (λ ∼ 1–10 ) scatters off atmospheric water droplets (∼ 10 diameter) far less than visible does, by Mie-scattering arguments. Pilots use IR cameras to see through fog where the human eye cannot.
5. Ascending order of λ. Reference values (rough): X-ray 10^-10 m, UV 10^-7 m, IR 10^-5 m, microwave 10^-2 m. So ₃ < ₂ < ₄ < ₁.
6. Bonus applications.
   • Microwave: RADAR (aviation, weather), Wi-Fi (2.4 & 5 GHz), microwave cooking.
   • UV: vitamin-D synthesis on skin, fluorescence in forensic science, UV lithography for chips.
   • X-ray: medical radiography, dental imaging, airport baggage scanning, crystallography (atomic-resolution structure).
   • IR: thermal imaging cameras, TV remote controls, night-vision goggles, IR spectroscopy of molecules.

Why this matters. Each band's usefulness comes from how it interacts with matter: UV breaks bonds (germicidal), IR couples to molecular vibrations (heating, imaging), microwave couples to rotating polar molecules and atmospheric windows (communication), X-ray penetrates dense matter and resolves atomic-scale structure. The seven-band map is the engineer's cheat sheet for picking the right photon for a job.

₃ < ₂ < ₄ < ₁.

Strategic angle. Multiply two in-phase cosines to get a cos², average cos² to 1/2, eliminate B₀ using B₀ = E₀/c. Three lines algebra, all on autopilot once the wave-relation E₀ = cB₀ is in hand.

1. Instantaneous Poynting magnitude. E∥î, B∥ĵ both proportional to cos(kz-ω t). Cross product picks up î×ĵ=k̂: S(z,t) = E₀ B₀₀cos²(kz-ω t).
2. Time-average over one period T = 2π/ω. Using ²(ω t - kz)_T = 1/2: S= E₀ B₀₀·12 = E₀ B₀2₀.
3. Eliminate B₀ using E₀ = cB₀: S= E₀ (E₀/c)2₀ = E₀²2₀ c. That is the target expression.
4. Equivalent form via u. Average energy density is u_av = 12₀ E₀², and a plane wave moves at speed c, so the energy crossing unit area per unit time is S= c u_av = 12c₀ E₀². Using c₀ = 1/(₀ c): 12c₀ E₀² = E₀²/(2₀ c). Same answer.
5. Two equivalent forms students should keep ready: S= E₀²2₀ c = 12c₀ E₀² = B₀² c2₀ = E₀ B₀2₀. Each is the right tool for a different given variable.

Why this matters. The same algebra recasts the result in three equivalent forms (E₀-only, B₀-only, or both); pick the form that matches the data you have. Astronomers prefer the E₀ form for stellar spectra; microwave engineers prefer the B₀ form for cavity calculations.

S= E₀²/(2c₀).

Strategic angle. ``Displacement current in the gap = conduction current in the wires'' — design the supply, the gap takes care of itself. The capacitance, the desired current, and the required voltage slope are linked by one equation; we just solve it.

1. Core relation. I_d = C dV/dt — the conduction current in the leads (which is equal to the displacement current inside the gap by Maxwell) equals capacitance times the time-rate-of-change of the voltage.
2. Solve for dV/dt. dVdt = I_dC = 1× 10^-3 A 2× 10^-6 F = 500 V/s. That is the required slope at the instant when I_d = 1 mA.
3. Many waveforms work. The constraint fixes only the slope at the moment, not the entire waveform. Examples:
   • Linear ramp. V(t) = (500 V/s) t. Constant dV/dt = 500 V/s, so I_d is steady at 1 mA.
   • Sinusoidal AC. V(t) = V₀sin(2ν t). Peak dV/dt = 2ν V₀, so any (V₀,ν) satisfying 2ν V₀ = 500 V/s gives a peak displacement current of 1 mA. Concrete pair: V₀ = 1 V, ν ≈ 80 Hz; or V₀ = 230 V, ν ≈ 0.35 Hz; or V₀ = 5 V, ν ≈ 16 Hz.
4. Energy considerations. A continuous ramp would eventually exceed the capacitor's voltage rating; sinusoidal AC stays bounded. For a sustained 1 mA peak with V₀ = 1 V, the power dissipated in the dielectric loss (if any) is at most 12V₀ I = 0.5 mW — negligible.
5. Sanity check via energy in the field. A 1 V amplitude on 2 stores up to 12CV² = 1 — easily switched in and out of the field by a small AC source.

Why this matters. Displacement current is what allows capacitors to act as ``circuit elements that pass AC'' — the gap is literally seeing a current of I = C dV/dt. Lab oscilloscopes, signal generators, and AC bridges all rely on this gap-current to work.

Drive dV/dt = 500 V/s across the capacitor.

Strategic angle. Energy-to-momentum conversion in four algebra lines. The single ingredient is the wave-mechanics identity p_em = U/c — once that is in hand, pressure is bookkeeping.

1. Energy delivered. An EM beam of intensity I on area A for time Δ t deposits energy U = I A Δ t. Intensity has units W/m²; energy in joules.
2. Momentum carried. For an EM wave (or any massless radiation), p = Uc. Microscopically, each photon of energy hν carries momentum hν/c; macroscopically the same factor 1/c relates total U to total p.
3. Force on full-absorber. Force = rate of change of momentum: F_abs = Δ pΔ t = U/cΔ t = I Ac.
4. Pressure on full-absorber. p_rad^abs = F_absA = Ic.
5. Full-reflector variant. For a perfect reflector, the beam's momentum reverses, so the change is 2U/c rather than U/c. Pressure doubles: p_rad^ref = 2I/c. Pre-photon, this is the EM analogue of an elastic vs. inelastic collision.
6. Numerical taste. Sunlight at the top of Earth's atmosphere has I = 1361 W/m², giving p_abs = 1361/(3× 10⁸) ≈ 4.5 . A perfect mirror would feel 9 — about 10^-10 of atmospheric pressure. Real solar sails operate in this regime.

Why this matters. The same momentum/energy relation (p = U/c for any massless wave) underlies the photon picture in quantum mechanics: a photon of energy hν carries momentum hν/c. Photon-momentum recoil is the basis of laser cooling of atoms and underlies the radiation-reaction limit on accelerator beam intensity.

p_rad = I/c.

Strategic angle. Solid angle. The bulb spreads P into 4π sr; the laser into a few microsteradians. The whole question reduces to ``how big is the angular spread of the source?''.

1. Bulb radiates isotropically. Power P spreads over the full sphere of solid angle 4π sr. At distance r: I_bulb(r) = P4π r². Doubling r quarters the intensity (inverse-square law).
2. Laser is collimated. A laser of waist w₀ and divergence half-angle θλ/(π w₀) spreads into a tiny solid angle Ω ∼ θ² ∼ λ²/w₀². For typical values w₀∼ 1 mm, λ∼ 600 nm: θ∼ 2× 10^-4 rad, Ω∼ 10^-7 sr.
3. Spot area at distance r. A_laser(r) ∼ π(w₀ + rθ)². At r = 10 m, rθ ≈ 2 mm, comparable to w₀ — the spot has barely doubled. So I_laser ≈ P/A₀ is roughly constant on room scale. Beyond r∼ w₀/θ (a few metres for typical lab lasers), the Rayleigh-range region ends and the spot starts growing linearly.
4. Numerical contrast. 1 W bulb → I(at 1 m) = 1/(4π)≈ 0.08 W/m². Same 1 W in a 1 mm² laser spot → I_laser≈ 10⁶ W/m² — 10⁷ times more intense, simply because of geometry.
5. The deep geometric property is collimation, underwritten by spatial coherence: every part of the laser's wavefront has a well-defined phase relative to every other part, so the beam diffracts as a single coherent Gaussian mode with the minimum possible divergence ∼ λ/D. A thermal source (bulb) is spatially incoherent — each tiny emitter sends light into all directions independently.
6. Bigger telescope, smaller divergence. A D = 1 m aperture at λ = 600 nm gives θ∼ 6× 10^-7 rad — laser range-finders to the Moon (Apollo retroreflectors) exploit this to put a 1 km spot at the lunar surface from Earth.

Why this matters. The same collimation is what makes laser range-finders, laser pointers and fibre-optic communications work. Free-space optical (FSO) links between buildings use the laser's tiny divergence to deliver gigabit data rates without amplifiers; satellite quantum key distribution relies on it to keep single photons heading toward a sub-metre detector aperture.

Bulb: I ∝ 1/r² (quartered on doubling r); laser: low divergence keeps I roughly constant over short distances.

Strategic angle. Two-step force analysis: the E shakes the charge sideways; the B then pushes it forward. The split is what allows a wave to be both an energy-carrier (E's work) and a momentum-carrier (B's push).

1. Set up. Let the wave propagate along k̂ = ẑ with E=E_x(z,t)î and B=B_y(z,t)ĵ. A free charge q starts at rest.
2. First half-cycle: E does the shaking. Force qE=qE_xî accelerates the charge along î — transverse to propagation. The charge picks up sideways velocity v=v_xî. Net work: E·v· q dt > 0, so energy is transferred from the wave to the charge.
3. Magnetic force kicks in. Now v=v_xî is non-zero, so F_B = q v×B = q v_x B_y (î×ĵ) = q v_x B_y k̂. Crucially this force is along k̂ — i.e. along the wave's propagation direction. This is the radiation push, the source of radiation pressure.
4. Magnetic work is zero. F_B·v = q(v×B)·v = 0 identically. So B contributes zero kinetic energy to the charge while still contributing forward momentum. Energy via E; momentum via B — cleanly separated.
5. Sign analysis: does the push reverse? If E_x flips sign every half-period, then v_x flips sign soon after (the charge oscillates). Simultaneously B_y flips. So v_x B_y — the product — stays positive on average: the push is always forward, averaging to a steady force in the +k̂ direction. That cumulative effect is radiation pressure.
6. Order-of-magnitude self-check. Time-averaged force per unit charge: v_x B_y∼ (qE₀/mω) B₀/2 = (qE₀²)/(2mω c). Multiplied by the number density of charges, one recovers p_rad = I/c — the macroscopic radiation pressure from the microscopic qv×B picture.

Why this matters. This is the microscopic reason an EM wave can deliver both energy and momentum — and why radiation pressure exists even though B alone does no work. The same logic explains why charged particles in plasma can be accelerated by an EM wave (Compton wakefield), and why the radiation-reaction force on a synchrotron electron has both a transverse (energy-loss) and longitudinal (momentum-loss) component.

qE is transverse ⇒ no forward force; qv×B supplies the forward push, hence radiation pressure.

Strategic angle. Find E from Gauss, B from Amp`ere, multiply. The figure (Fig. 8.1) shows a Gaussian cylinder of radius a centred on the moving wire; that is the loop of integration for both Gauss and Amp`ere.

1. Electric field by Gauss's law. Cylindrical Gaussian surface of radius a, length L: ∮E· dA = E(2π a L) = λ L₀ ⇒ E = λ2π₀ a â, radially outward (for λ > 0).
2. Magnetic field by Amp`ere's law. Once the wire moves at vk̂, the linear charge density becomes a current I = λ v flowing along +k̂. Amp`erian loop of radius a in the plane perpendicular to k̂: B(2π a) = ₀(λ v) ⇒ B = ₀λ v2π a φ̂, azimuthal direction (right-hand-rule: thumb → +k̂, fingers curl in +φ̂).
3. Poynting vector via cylindrical triad. â×φ̂ = k̂, so E×B = λ2π₀ a· ₀λ v2π a (â×φ̂) = ₀λ² v4π²₀ a² k̂.
4. Divide by ₀. S = 1₀ E×B = λ² v4π²₀ a² k̂. Direction: along the wire's velocity k̂.
5. Why this is non-zero even though the fields are static (in the lab frame). The fields are not changing in time at any point, but E and B are non-parallel, so the Poynting vector is non-zero. Physically, moving the line of charge advects field energy forward — that is what S measures here.
6. Unit-check. [λ² v/(₀ a²)] = (C/m)²·(m/s)/[(F/m)²] = C² m^-1 s^-1/(F) = J/(m²⋯) = W/m². Correct units for intensity / Poynting magnitude.

Why this matters. Even ``static'' configurations can have nonzero Poynting flux when they are in motion; the field carries energy in the direction of motion. The same idea underlies wireless power transfer in coaxial transmission lines, where the dominant energy flow is between (not along) the inner and outer conductors.

S = λ² v4π²₀ a² k̂.

Strategic angle. Both densities scale linearly with E; the ratio depends only on material constants and frequency. The calculation is one multiplication of ερ, but the interpretation as a conductor-vs-dielectric crossover is where the physics is.

1. Write both densities. J_c = σE = E/ρ for a conductor with resistivity ρ; J_d = ε _tE for the displacement contribution. For a sinusoid E = E₀sin(ω t), the peak amplitudes are J_c^max = E₀ρ, J_d^max = εω E₀.
2. Form the ratio. J_d^maxJ_c^max = εω E₀E₀/ρ = ω ε ρ. The amplitude E₀ cancels — the ratio is a purely material + frequency quantity.
3. Plug numbers. ω = 2ν = 2π(4× 10⁸) = 2.51× 10⁹ rad/s. ε = 80₀ = 80× 8.85× 10^-12 = 7.08× 10^-10 F/m. ρ = 0.25 Ω m. ερ = (2.51× 10⁹)(7.08× 10^-10)(0.25) = (1.777)(0.25) = 0.444. Approximately 0.44, or 44%.
4. Crossover frequency. The conductor-vs-dielectric crossover happens at _cερ = 1, i.e. _c = 1/(ερ). For sea water this is _c = 1(7.08× 10^-10)(0.25) = 5.65× 10⁹ rad/s, _c = _c/(2π) ≈ 900 MHz. Below ∼ 900 MHz sea water is conductor-like (radio is blocked); above it, displacement currents dominate and EM waves can propagate.
5. Interpretation at ν = 4× 10⁸ Hz. At 400 MHz we are about a factor of 2 below the crossover — conduction still wins, but displacement current is already 44% as large. The material is on the dielectric edge of ``mostly conductor''.
6. Submarine link. Submarines cannot use ordinary HF/UHF radio in sea water (conduction wins, signal decays in metres); they use ELF (extremely low frequency, ∼ 80 Hz — far below the crossover, conduction even more dominant, but the skin depth is large) or surface to communicate.

Why this matters. Submarines cannot use ordinary radio because at HF frequencies sea water is a near-perfect conductor; they must use ELF (a few Hz) or rise to the surface. The _c crossover is the universal recipe for ``conductor vs. dielectric'' behaviour of any material — including biological tissue (used in MRI SAR calculations) and the ionosphere (which is conductor-like to radio but dielectric-like to visible).

J_d/J_c ≈ 0.44.

Strategic angle. Three computations strung together: time derivative (gives J_d), area integral (gives I_d), ratio (gives the (π a/λ)² smallness factor). Each step is mechanical; the punchline is that the displacement-to-conduction ratio is controlled entirely by the cable's electrical size a/λ.

1. Differentiate E in time. Given E(s,t) = ₀ I₀(2ν t)ln(s/a) k̂: ∂E∂ t = -₀ I₀ν(2ν)sin(2ν t)ln(s/a) k̂. Multiply by ₀: J_d = -2π₀₀ I₀ν²sin(2ν t)ln(s/a) k̂.
2. Integrate over the cross-section. Use dA = 2π s ds: I_d(t) = ₀^a J_d(s,t) 2π s ds. The s-integral is ₀^a sln(s/a) ds = a²₀¹ uln u du = -a²/4 (by parts: ∫ uln u du = u²(ln u)/2 - u²/4; evaluating limits with _{u→ 0}u²ln u = 0).
3. Assemble I_d. I_d(t) = -2π₀₀ I₀ν²sin(2ν t)· 2π·(-a²/4) = π²₀₀ a² I₀ν²sin(2ν t). Using ₀₀ = 1/c²: I_d(t) = π² a²ν²c² I₀sin(2ν t).
4. Ratio. Peak I₀^d = π² a²ν² I₀/c². Using c = λ: I₀^dI₀ = π² a²ν²c² = (π aλ)².
5. Order-of-magnitude. At mains ν = 50 Hz: λ = c/ν = 6000 km. For a coaxial cable of radius a = 1 cm: (π a/λ)² ≈ (π· 10^-2/6× 10⁶)² ≈ 3× 10^-17 — utterly negligible. The conduction current dominates by 17 orders of magnitude.
6. When does I_d catch up? Setting (π a/λ)² ∼ 1 requires λ ∼ π a, i.e. wavelength comparable to cable thickness. For a = 1 cm that needs ν∼ 10 GHz — the upper microwave band. Coaxial cables fail to behave as ``pure conductors'' there, and special low-loss waveguides replace them.