JEE Main chapters · Class 12 · Physics

Electromagnetic Waves: JEE Main Physics formulas and notes

Electromagnetic Waves is a short chapter that gives 1 question in most JEE Mains papers, and sometimes 2: the displacement current, the order and use of the spectrum, the relation E₀ = cB₀ with directions, the energy density and intensity, or radiation pressure. The formulas are few and each question is a one-step substitution if you know the direction rules and keep E₀ and rms values apart. About 4 hours is enough.

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Quick revision

• Iᵈ = ε₀ dΦᴇ/dt = C dV/dt; B = μ₀Iᵈr/2πR² inside, μ₀Iᵈ/2πr outside
• Maxwell: Gauss (E), Gauss (B) = 0, Faraday, Ampère–Maxwell with ε₀ dΦᴇ/dt
• c = 1/√(μ₀ε₀) = fλ = ω/k = E₀/B₀; E₀ = cB₀; v = c/n, n = √(μᵣεᵣ)
• E⃗ × B⃗ is the direction of travel; E and B are in phase; f never changes
• Order (long λ to short): radio, microwave, IR, visible, UV, X-ray, γ
• u = ½ε₀E² + B²/2μ₀ = ε₀E²; average ½ε₀E₀²
• I = ½ε₀cE₀² = cB₀²/2μ₀ = uc; point source I = P/4πr²
• p = U/c; absorber F = IA/c; reflector F = 2IA/c
• Photon energy E = 1240/λ(nm) eV

Formulas with solved examples

Displacement current and Maxwell

Displacement current

• Iᵈ = ε₀ (dΦᴇ)/(dt), Φᴇ = EA the electric flux
• for a charging capacitor Iᵈ = ε₀A (dE)/(dt) = C (dV)/(dt) = I(conduction)
• it is a changing electric field, not a flow of charge
• exists in the gap between the plates (no charge crosses it)
• total current I + Iᵈ is continuous, so Ampère’s law is made consistent

How to apply. Find the flux Φᴇ = EA (or E = V/d) as a function of time and differentiate. Between the plates of a charging capacitor the displacement current is equal to the conduction current in the wires.

Example. Capacitor of 100 pF with V changing at 5 × 10⁸ V/s: Iᵈ = C dV/dt = 100 × 10⁻¹² × 5 × 10⁸ = 0.05 A. Plates of radius 0.1 m with E changing at 10¹² V/m s: Iᵈ = 8.85 × 10⁻¹² × π(0.1)² × 10¹² = 0.28 A.

Magnetic field of the displacement current

• Ampère–Maxwell law: ∮B⃗·dl⃗ = μ₀(I + Iᵈ) = μ₀I + μ₀ε₀ (dΦᴇ)/(dt)
• between circular plates of radius R carrying Iᵈ: inside (r < R) B = (μ₀Iᵈr)/(2πR²), outside (r > R) B = (μ₀Iᵈ)/(2πr)
• B is the same as that of a wire carrying I at r > R
• a changing electric field produces a magnetic field

How to apply. Take a circle of radius r coaxial with the plates. For r < R only the fraction πr²/πR² of Iᵈ passes through it; for r > R all of it does. Then B × 2πr = μ₀ × (current through the circle).

Example. Iᵈ = 2 A, R = 0.1 m: at r = 0.04 m, B = 2 × 10⁻⁷ × 2 × 0.04 ÷ 0.1² = 1.6 μT. At r = 0.2 m, B = 2 × 10⁻⁷ × 2 ÷ 0.2 = 2 μT.

Maxwell’s four equations

• Gauss (electric): ∮E⃗·dA⃗ = (q)/(ε₀)
• Gauss (magnetic): ∮B⃗·dA⃗ = 0 (no monopoles)
• Faraday: ∮E⃗·dl⃗ = −(dΦʙ)/(dt)
• Ampère–Maxwell: ∮B⃗·dl⃗ = μ₀I + μ₀ε₀(dΦᴇ)/(dt)
• together with F = q(E + v × B) they describe all of electromagnetism
• from them c = (1)/(√(μ₀ε₀)) = 3 × 10⁸ m/s

How to apply. Know which law each equation states, and which one Maxwell added (the displacement term). The last two say: a changing B makes an E, a changing E makes a B; this is why waves exist.

Example. 1/√(μ₀ε₀) = 1 ÷ √(4π × 10⁻⁷ × 8.85 × 10⁻¹²) = 1 ÷ √(1.112 × 10⁻¹⁷) = 3.0 × 10⁸ m/s. The term Maxwell added is the second part of the Ampère–Maxwell law.

Nature of EM waves

Nature and production

• produced by accelerated (oscillating) charges, not by a charge moving with uniform velocity
• transverse: E⃗, B⃗ and the direction of travel are mutually perpendicular, E⃗ × B⃗ along c⃗
• E and B oscillate in phase (maximum together)
• no medium needed
• uncharged: not deflected by electric or magnetic fields
• carry energy and momentum
• obey superposition: show reflection, refraction, interference, diffraction, polarisation
• oscillator frequency f = (1)/(2π√(LC)) for an LC circuit

How to apply. Check the direction with E⃗ × B⃗: if E is along +y and the wave travels along +x then B is along +z. For a source circuit, the frequency of the wave equals the frequency of the charge oscillation.

Example. LC circuit with L = 1 μH and C = 100 pF: f = 1 ÷ (2π√(10⁻⁶ × 10⁻¹⁰)) = 1 ÷ (2π × 10⁻⁸) = 1.59 × 10⁷ Hz and λ = c/f = 18.9 m.

Wave equation, c = fλ and E₀ = cB₀

• E = E₀ sin(kx − ωt), B = B₀ sin(kx − ωt)
• k = (2π)/(λ), ω = 2πf, c = fλ = (ω)/(k)
• (E)/(B) = (E₀)/(B₀) = c at every instant and point (in vacuum)
• E₀ = cB₀
• c = (1)/(√(μ₀ε₀))
• the wave moves along +x for kx − ωt, along −x for kx + ωt

How to apply. Read k and ω from the equation: c = ω/k, λ = 2π/k, f = ω/2π. Then B₀ = E₀/c, with B along the third axis (E⃗ × B⃗ = direction of travel).

Example. E = 45 sin(2πx − 6π × 10⁸ t) V/m, along y: c = ω/k = 6π × 10⁸ ÷ 2π = 3 × 10⁸ m/s, λ = 1 m, f = 3 × 10⁸ Hz, B₀ = 45 ÷ 3 × 10⁸ = 1.5 × 10⁻⁷ T, along z.

EM waves in a medium

• v = (1)/(√(με)) = (c)/(√(μᵣεᵣ)) = (c)/(n), n = √(μᵣεᵣ)
• for a non-magnetic medium n = √εᵣ (dielectric constant K = εᵣ)
• frequency is fixed by the source and does not change; λ(medium) = (λ₀)/(n)
• (E₀)/(B₀) = v in the medium
• intensity and energy depend on ε (u = ½εE²)

How to apply. Find n from εᵣ and μᵣ, then v = c/n and λ = λ₀/n. Frequency stays the same when the wave enters a medium.

Example. A medium with εᵣ = 4 and μᵣ = 1: n = 2, v = 1.5 × 10⁸ m/s. A 600 nm wave becomes 300 nm and keeps its frequency of 5 × 10¹⁴ Hz.

Electromagnetic spectrum

Electromagnetic spectrum: order and ranges

• increasing frequency (decreasing λ): radio, microwave, infrared, visible, ultraviolet, X-rays, γ-rays
• radio: λ > 0.1 m
• microwave: 1 mm to 0.1 m (about 1 GHz to 300 GHz)
• infrared: 700 nm to 1 mm
• visible: 400 nm (violet) to 700 nm (red)
• ultraviolet: 1 nm to 400 nm
• X-rays: about 10⁻³ nm to 1 nm
• γ-rays: below about 10⁻³ nm (10⁻¹² m)
• all travel at c in vacuum; the boundaries overlap and are not sharp

How to apply. Convert with λ = c/f. Know the order from long to short λ: radio, microwave, IR, visible, UV, X-ray, γ. Violet has a shorter λ than red.

Example. A microwave oven works at 2.45 GHz: λ = 3 × 10⁸ ÷ 2.45 × 10⁹ = 0.122 m. Green light of 5.5 × 10¹⁴ Hz has λ = 545 nm, between 400 nm and 700 nm.

Spectrum: sources and uses

• radio: oscillating circuits; AM, FM, TV, mobile signals
• microwave: klystron or magnetron; radar, microwave oven, satellite links
• infrared: hot bodies, molecular vibrations; heating, remote controls, night vision, greenhouse effect
• visible: atoms and hot bodies; vision
• ultraviolet: very hot bodies, Sun, arc lamps; sterilisation, LASIK, absorbed by ozone and by glass
• X-rays: sudden stopping of fast electrons, inner-shell transitions; medical imaging and crystal structure
• γ-rays: nuclei; cancer therapy, killing germs, studying nuclei

How to apply. Match the application to the property: microwave resonates with water molecules, infrared is the heat radiation, UV causes fluorescence and sterilises, X-rays pass through soft tissue, γ-rays penetrate the most.

Example. Heat felt from a hot iron is infrared; the Earth’s warmth is kept by the atmosphere (greenhouse effect) because it absorbs infrared. Ozone absorbs the harmful UV-B from the Sun.

Photon energy across the spectrum

• E = hf = (hc)/(λ) = (1240)/(λ(nm)) eV
• h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J
• energy rises with frequency: radio is lowest, γ-ray highest
• visible photons have about 1.8 eV to 3.1 eV
• frequency is unchanged when light changes medium, so its photon energy is unchanged

How to apply. Take λ in nm in the shortcut 1240/λ to get the energy in eV. A smaller wavelength means a larger photon energy.

Example. Light of λ = 620 nm: E = 1240 ÷ 620 = 2.0 eV. An X-ray of 0.1 nm has 12 400 eV, about 6200 times more.

Energy, intensity and pressure

Energy density

• u = ½ε₀E² + (B²)/(2μ₀)
• for an EM wave the electric and the magnetic part are equal: uᴇ = uʙ so u = ε₀E² = (B²)/(μ₀)
• average u = ½ε₀E₀² = (B₀²)/(2μ₀)
• energy in a volume V is uV
• E₀ = cB₀ gives ½ε₀E₀² = B₀²/2μ₀

How to apply. Use the rms or the peak value consistently: the average energy density is half of ε₀E₀² because sin² averages to 1/2. Each field carries half of the total.

Example. E₀ = 30 V/m: average u = ½ × 8.85 × 10⁻¹² × 30² = 3.98 × 10⁻⁹ J/m³. The magnetic part is half of it, 2.0 × 10⁻⁹ J/m³.

Intensity and Poynting vector

• S⃗ = (1)/(μ₀) E⃗ × B⃗ (W/m², along the direction of travel)
• I = ⟨S⟩ = u c = ½ε₀cE₀² = (E₀²)/(2μ₀c) = (cB₀²)/(2μ₀)
• I ∝ E₀² ∝ B₀²
• isotropic point source of power P: I = (P)/(4πr²), so E₀ ∝ (1)/(r)
• energy crossing an area A in time t is IAt
• E(rms) = E₀/√2

How to apply. Find I from the power and the area (I = P/4πr² for a point source in all directions), then E₀ = √(2I/ε₀c) and B₀ = E₀/c. For a beam of constant cross-section I is the same all along it.

Example. E₀ = 100 V/m: I = ½ × 8.85 × 10⁻¹² × 3 × 10⁸ × 10⁴ = 13.3 W/m². A 50 W isotropic source at 2 m: I = 50 ÷ (4π × 4) = 0.995 W/m², E₀ = √(2 × 0.995 ÷ 2.655 × 10⁻³) = 27.4 V/m, B₀ = 9.1 × 10⁻⁸ T.

Momentum and radiation pressure

• momentum carried by energy U: p = (U)/(c)
• force and pressure on a surface of area A normal to the beam: perfect absorber F = (IA)/(c), P = (I)/(c)
• perfect reflector (normal incidence) F = (2IA)/(c), P = (2I)/(c)
• at angle θ to the normal: absorber P = (I cos²θ)/(c), reflector P = (2I cos²θ)/(c)
• radiation pressure is very small (the Sun gives about 4.5 × 10⁻⁶ Pa at the Earth)

How to apply. Decide whether the surface absorbs (momentum change U/c) or reflects (change 2U/c). Then pressure = (change in momentum per second per area) = I/c or 2I/c.

Example. I = 500 W/m² on a 2 m² absorber: F = 500 × 2 ÷ 3 × 10⁸ = 3.33 × 10⁻⁶ N. If the surface is a perfect mirror, F = 6.67 × 10⁻⁶ N. A light pulse of 6 J absorbed carries p = 6 ÷ 3 × 10⁸ = 2 × 10⁻⁸ kg m/s.

Laser beam and sail: typical numbers

• I = (P)/(A) for a beam of power P and area A
• E₀ = √(2I)/(ε₀c) = √(2Iμ₀c)
• force on an absorbing target F = P/c, on a reflecting one 2P/c
• a laser of 5 mW in 1 mm² gives I = 5000 W/m²
• the force depends on the power and not on the area

How to apply. From power and spot area get I. Then E₀ = 27.4√I (SI, since √(2/ε₀c) = 27.4) and B₀ = E₀/c.

Example. Beam power 5 mW, spot area 1 mm²: I = 5 × 10⁻³ ÷ 10⁻⁶ = 5000 W/m², E₀ = 27.4 × √5000 = 1940 V/m. Force on an absorber = 5 × 10⁻³ ÷ 3 × 10⁸ = 1.7 × 10⁻¹¹ N.

Most asked question types

How do I solve “Displacement current in a capacitor” questions in JEE Main?

A capacitor is charged by a changing voltage or a current. Find the displacement current, show it equals the conduction current, or find the magnetic field between the plates. 1. Iᵈ = ε₀ dΦᴇ/dt = C dV/dt. 2. The displacement current is the same as the conduction current in the wires. 3. For B use Ampère–Maxwell with the part of Iᵈ inside the loop. Example: C = 100 pF, dV/dt = 5 × 10⁸ V/s: Iᵈ = 0.05 A.

How do I solve “Maxwell’s equations and the displacement term” questions in JEE Main?

A statement or a list of equations is given. Name the law, say which equation was added by Maxwell, or say what a changing field produces. 1. Match each equation to its law. 2. The term ε₀ dΦᴇ/dt in Ampère’s law is Maxwell’s. 3. A changing E produces B and a changing B produces E. Example: The equation containing μ₀ε₀ dΦᴇ/dt is the Ampère–Maxwell law.

How do I solve “Directions of E, B and the wave” questions in JEE Main?

The direction of E or B and of travel are given for an EM wave. Find the remaining one or the equation of the wave. 1. E⃗ × B⃗ gives the direction of travel. 2. E and B are in phase and perpendicular. 3. Use E₀ = cB₀ for the amplitude. Example: E along +y and travel along +x give B along +z.

How do I solve “Wave equation: find E₀, B₀, λ, f, k, ω” questions in JEE Main?

The equation of an EM wave is given, such as E = E₀ sin(kx − ωt). Find the amplitude of B, the wavelength, the frequency, the speed or the direction. 1. Compare with E₀ sin(kx − ωt): read k and ω. 2. λ = 2π/k, f = ω/2π, c = ω/k. 3. B₀ = E₀/c; the sign gives the direction. Example: E = 45 sin(2πx − 6π × 10⁸ t): λ = 1 m, f = 3 × 10⁸ Hz, B₀ = 1.5 × 10⁻⁷ T.

How do I solve “Speed and wavelength in a medium” questions in JEE Main?

An EM wave passes into a medium of given εᵣ and μᵣ or n. Find the speed, the wavelength or the frequency. 1. n = √(μᵣεᵣ), v = c/n. 2. Frequency is unchanged. 3. λ = λ₀/n. Example: εᵣ = 4, μᵣ = 1: v = 1.5 × 10⁸ m/s.

How do I solve “Parts of the electromagnetic spectrum” questions in JEE Main?

Names of waves are given with wavelengths, frequencies, sources or uses. Arrange them in order, or match source with use. 1. Order by frequency: radio, microwave, IR, visible, UV, X-ray, γ. 2. Use λ = c/f to compare. 3. Match the use to the property of the wave. Example: Microwave oven: 2.45 GHz, λ = 12 cm.

How do I solve “Energy density and the electric and magnetic parts” questions in JEE Main?

The amplitude of E or B is given. Find the energy density, the share of the electric or magnetic part, or the energy in a volume. 1. u = ½ε₀E₀² (average) and the electric and magnetic parts are equal. 2. Use B₀ = E₀/c to switch between E and B. 3. Energy = u × volume. Example: E₀ = 30 V/m: u = 3.98 × 10⁻⁹ J/m³.

How do I solve “Intensity of a source and the field amplitude” questions in JEE Main?

A bulb, laser or antenna of given power is observed at some distance. Find the intensity, E₀ or B₀, or the energy received. 1. I = P/4πr² for a point source (P/A for a beam). 2. E₀ = √(2I/ε₀c), B₀ = E₀/c. 3. Energy = I × A × t. Example: 50 W source at 2 m: I = 0.995 W/m², E₀ = 27.4 V/m.

How do I solve “Radiation pressure and force” questions in JEE Main?

Light falls on an absorbing or reflecting surface. Find the pressure, the force or the momentum transferred. 1. Absorber: P = I/c; reflector: 2I/c. 2. Force = pressure × area; momentum = U/c. 3. For a slanted surface use I cos²θ in the pressure. Example: 500 W/m² on 2 m² absorber: F = 3.33 × 10⁻⁶ N.

Common myths and the truth

• Myth: Displacement current is the current of charges moving across the gap of a capacitor. Truth: No charge crosses the gap; the displacement current comes from the changing electric field and produces the magnetic field in the same way a current does.
• Myth: EM waves need a medium, like sound. Truth: They travel in vacuum at c; the medium only slows them to c/n.
• Myth: E and B are 90 degrees out of phase in an EM wave. Truth: They are perpendicular in space but in phase in time: both are largest at the same instant.
• Myth: When light enters glass its frequency decreases. Truth: The frequency is set by the source and stays the same; the speed and the wavelength fall by the factor n.
• Myth: X-rays and γ-rays are different in kind. Truth: Both are EM waves; they differ in origin (electrons and atoms for X-rays, nuclei for γ-rays), and their ranges overlap.
• Myth: Infrared rays are the only heat rays. Truth: Every EM wave that is absorbed warms the body; infrared is the radiation emitted by warm bodies and absorbed most easily by matter.
• Myth: The magnetic energy in an EM wave is much smaller than the electric energy. Truth: They are equal at every point, uᴇ = uʙ, because E = cB.
• Myth: The radiation pressure of light is large enough to feel. Truth: It is about 10⁻⁶ Pa for sunlight; it matters only for very intense beams, comet tails and space sails.

Mistakes to avoid

• The average intensity uses E₀²/2; do not forget the ½ (or use the rms value).
• Displacement current is not a movement of charge; there are no charges crossing the gap.
• E₀ = cB₀ holds in vacuum; in a medium E₀/B₀ = v = c/n.
• Frequency is constant when the wave goes from one medium to another; wavelength is not.
• Radiation pressure for a reflector is double that of an absorber (normal incidence).
• A uniformly moving charge does not radiate; an accelerated charge does.
• Do not confuse the order of the spectrum: in terms of frequency, infrared is below visible and ultraviolet above.
• I = P/4πr² makes E₀ fall as 1/r, not as 1/r².

Simulations for Electromagnetic Waves

• Radiation from a dipole antenna — A half-wave dipole radiates most strongly at right angles to its length and not at all along it.
• Displacement current between plates — Charge and discharge a capacitor, and connect a changing electric field with current continuity.
• Light entering a medium (3D fields) — Watch E and B oscillate at right angles as light travels into glass. Rotate the polarisation plane.
• Electromagnetic wave — See perpendicular electric and magnetic fields oscillate as a wave travels.
• Photon energy across the spectrum — Pick a frequency and read the energy carried by each photon.
• Explore the EM spectrum — Slide across the spectrum: the source glows with that radiation and shows its wavelength and frequency.

More: Class 12 Physics simulations · all JEE Main chapters.