JEE Physics — Electromagnetic Waves Complete Chapter Guide

Chapter Overview & Weightage

Electromagnetic Waves is a compact but high-yield chapter for JEE Main. The theory is lighter than Electrostatics or Modern Physics, but examiners squeeze surprisingly tricky conceptual questions out of it. One well-prepared question here can be the margin between 99 and 99.5 percentile.

JEE Main Weightage Data

Year	Questions	Marks	Topics Tested
2024 (Jan + Apr)	1–2	4–8	Displacement current, EM spectrum order
2023	1	4	Properties of EM waves, energy density
2022	1–2	4–8	Maxwell’s equations concept, speed of EM waves
2021	1	4	Electromagnetic spectrum, wavelength ranges
2020	2	8	Displacement current, EM wave properties

Roughly 2–3% weightage in JEE Main. JEE Advanced rarely tests this chapter directly — it appears more as supporting context in optics or modern physics problems.

The chapter breaks into three clear areas: displacement current + Maxwell’s correction, properties of EM waves, and the electromagnetic spectrum. All three are equally important for JEE Main.

Key Concepts You Must Know

These are ordered by how frequently they appear in PYQs — prioritise accordingly.

Tier 1 — Almost Always Asked:

Displacement current: definition, formula, and why Maxwell introduced it
Speed of EM waves in vacuum: $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
EM waves are transverse — $\vec{E}$ , $\vec{B}$ , and direction of propagation are mutually perpendicular
Relation between $E$ and $B$ : $E = cB$ at every instant

Tier 2 — Frequently Asked:

Energy stored in EM waves: electric and magnetic energy densities are equal
Intensity of EM wave and its relation to amplitude
Poynting vector — direction and physical meaning
The complete EM spectrum: order by frequency and wavelength, sources, uses

Tier 3 — Occasionally Asked:

Maxwell’s four equations (conceptual identification, not solving them)
Momentum of EM waves and radiation pressure
Refractive index of medium and speed: $v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$

Important Formulas

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

When to use: Any question asking about current continuity in a capacitor circuit, or “what completes the circuit between capacitor plates.” This is Maxwell’s modification to Ampere’s law.

The total current in Ampere’s law becomes $I + I_d$ , making it universally valid.

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \text{ m/s}

When to use: Establishing the link between electromagnetism and optics. This result — that $c$ equals the speed of light — was Maxwell’s greatest triumph.

\frac{E_0}{B_0} = c \quad \Longrightarrow \quad \frac{E}{B} = c

When to use: Any numerical giving either $E_0$ or $B_0$ and asking for the other. The ratio equals $c$ always, not just at amplitude — it holds at every point, every instant.

u_E = \frac{1}{2}\varepsilon_0 E^2, \qquad u_B = \frac{B^2}{2\mu_0}, \qquad u_E = u_B

When to use: Questions asking which field stores more energy — the answer is always “equal.” Total energy density $u = \varepsilon_0 E^2$ .

I = \frac{1}{2} c \varepsilon_0 E_0^2 = \frac{c B_0^2}{2\mu_0}

\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0} \quad \text{(Poynting vector)}

When to use: Intensity problems, radiation pressure questions ( $P = I/c$ for complete absorption, $P = 2I/c$ for complete reflection).

v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}

When to use: When a medium is given with relative permittivity $\varepsilon_r$ and relative permeability $\mu_r$ . For most non-magnetic media, $\mu_r = 1$ , so $v = c/\sqrt{\varepsilon_r} = c/n$ .

The Electromagnetic Spectrum

This table is the single most tested piece of factual knowledge from this chapter:

Type	Wavelength Range	Frequency Range	Source	Key Use
Radio waves	>0.1 m	< $3 \times 10^9$ Hz	Oscillating circuits	AM/FM, radar
Microwaves	1 mm – 0.1 m	$3\times10^9$ – $3\times10^{11}$ Hz	Klystron, magnetron	Microwave ovens, satellite
Infrared	700 nm – 1 mm	$3\times10^{11}$ – $4\times10^{14}$ Hz	Hot objects, IR LEDs	Remote controls, thermal imaging
Visible	400 – 700 nm	$4\times10^{14}$ – $7\times10^{14}$ Hz	Hot bodies, atomic transitions	Vision
Ultraviolet	1 nm – 400 nm	$7\times10^{14}$ – $3\times10^{17}$ Hz	Sun, arc lamps	Sterilisation, vitamin D
X-rays	0.001 – 1 nm	$3\times10^{17}$ – $3\times10^{20}$ Hz	High-energy electrons on metal	Medical imaging, crystallography
Gamma rays	<0.001 nm	> $3\times10^{20}$ Hz	Nuclear decay	Cancer treatment, sterilisation

Memorise the order: Radio → Micro → IR → Visible → UV → X-ray → Gamma, going from lowest to highest frequency (and longest to shortest wavelength). JEE often asks you to arrange in order — the VIBGYOR part of visible is a subrange within this.

Solved Previous Year Questions

PYQ 1 — Displacement Current (JEE Main 2023, April Session)

Q: A parallel plate capacitor with plate area $A = 10^{-3}$ m² and plate separation $d = 1$ mm is connected to a $100$ V, $50$ Hz AC source. Find the displacement current between the plates.

Solution:

Displacement current equals the conduction current feeding the capacitor. The capacitor’s charge is $q = CV$ , so $I_d = \frac{dq}{dt} = C\frac{dV}{dt}$ .

Capacitance: $C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 10^{-3}}{10^{-3}} = 8.85 \times 10^{-12}$ F

Voltage: $V = 100\sin(2\pi \times 50 \times t)$

\frac{dV}{dt}\bigg|_{\max} = 100 \times 2\pi \times 50 = 10^4\pi \text{ V/s}

I_d^{\max} = C \cdot \frac{dV}{dt}\bigg|_{\max} = 8.85 \times 10^{-12} \times 10^4\pi \approx 2.78 \times 10^{-7} \text{ A}

The key physical insight: $I_d$ between the plates equals $I$ in the wires. There’s no actual charge crossing the gap — but the changing electric flux creates an equivalent current that keeps Ampere’s law consistent.

PYQ 2 — E and B in an EM Wave (JEE Main 2024, January Shift 2)

Q: In an EM wave, the electric field oscillates with amplitude $48$ V/m. What is the amplitude of the magnetic field?

Solution:

We use $B_0 = \frac{E_0}{c}$ directly.

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

That’s 160 nT — tiny compared to Earth’s magnetic field (~ $50\,\mu$ T), which is why we usually detect EM radiation via its electric component.

Students sometimes use $E = cB$ and rearrange as $B = cE$ , forgetting the direction of division. Remember: E is large, B is small, just like $c$ is large. So $B_0 = E_0 / c$ .

PYQ 3 — Energy Density (JEE Main 2022, June Session)

Q: At a point in space, the electric field of an EM wave is $E = 100\sin(\omega t)$ V/m. Find the average total energy density at that point.

Solution:

Average electric energy density: $\langle u_E \rangle = \frac{1}{2}\varepsilon_0 E_0^2 \cdot \frac{1}{2} = \frac{1}{4}\varepsilon_0 E_0^2$

Wait — let’s be careful. The energy density is $u_E = \frac{1}{2}\varepsilon_0 E^2$ . Since $E = E_0\sin(\omega t)$ , the time average of $\sin^2$ is $\frac{1}{2}$ :

\langle u_E \rangle = \frac{1}{2}\varepsilon_0 \langle E^2 \rangle = \frac{1}{2}\varepsilon_0 \cdot \frac{E_0^2}{2} = \frac{\varepsilon_0 E_0^2}{4}

Since $\langle u_E \rangle = \langle u_B \rangle$ , total average energy density:

\langle u \rangle = 2\langle u_E \rangle = \frac{\varepsilon_0 E_0^2}{2} = \frac{8.85 \times 10^{-12} \times 10^4}{2} \approx 4.4 \times 10^{-8} \text{ J/m}^3

The most common error here: forgetting to double the electric energy density to get total energy density. Since $u_E = u_B$ at all times (not just on average), total = $2 \times \langle u_E\rangle$ .

Difficulty Distribution

For JEE Main, EM Waves questions fall as:

Difficulty	% of Questions	What They Test
Easy	50%	Spectrum order, $E = cB$ , identifying EM wave properties
Medium	40%	Displacement current calculation, energy density, intensity
Hard	10%	Radiation pressure, Poynting vector direction, multi-concept

The hard questions in this chapter usually combine EM waves with optics (reflection from surfaces → radiation pressure) or modern physics (photon momentum). Pure EM Waves questions rarely go above medium difficulty.

Expert Strategy

Step 1 — Clear the spectrum table first. Spend 30 minutes memorising the EM spectrum table once. Every PYQ set has at least one spectrum-order question. It’s free marks if you know the table; lost marks if you don’t.

Step 2 — Understand displacement current physically. Don’t just memorise the formula. Visualise: current flows into a charging capacitor, the electric field between the plates increases, and this changing $\Phi_E$ acts exactly like a current for magnetic field purposes. This physical picture solves 80% of conceptual questions.

Step 3 — The four Maxwell equations — know them by name and physical meaning, not by solving. JEE asks “which equation represents Faraday’s law” or “Gauss’s law for magnetism says magnetic monopoles don’t exist.” One-liners, high yield.

Maxwell’s equations summary for JEE:

Gauss (Electric): Electric flux out of closed surface = charge enclosed / $\varepsilon_0$
Gauss (Magnetic): Magnetic flux through any closed surface = 0 (no monopoles)
Faraday: Changing $\vec{B}$ induces $\vec{E}$ (EMF = $-d\Phi_B/dt$ )
Ampere-Maxwell: Changing $\vec{E}$ + real current both create $\vec{B}$

Step 4 — Practice the $E_0/B_0 = c$ calculation until it’s reflex. It appears in at least one question almost every year, and students who hesitate here lose 30 seconds they can’t afford.

Step 5 — Allocate time correctly. This chapter should take you 4–5 days to prepare thoroughly: 2 days for theory + formulas, 2 days for PYQs from the last 10 years (Kiran’s or Arihant collections), 1 day for revision. Any more time spent here is overkill — redirect to Electrostatics or Magnetism.

Common Traps

Trap 1: Confusing $E = cB$ with $E_{\text{rms}} = cB_{\text{rms}}$

The relation $E = cB$ holds at every instant, not just for amplitudes or rms values. So $E_0 = cB_0$ , $E_{\text{rms}} = cB_{\text{rms}}$ , and $E(t) = cB(t)$ all hold. But examiners sometimes give you rms values — the formula still applies directly.

Trap 2: Direction of propagation

EM wave travels in the direction of $\vec{E} \times \vec{B}$ . If a question says the wave travels in the $+z$ direction and $\vec{E}$ is along $+x$ , then $\vec{B}$ must be along $+y$ (since $\hat{x} \times \hat{y} = \hat{z}$ ). Students often get the sign wrong here. Draw the right-hand system explicitly.

Trap 3: Gamma rays vs X-rays wavelength overlap

The ranges of X-rays and gamma rays overlap in wavelength. They’re distinguished by source, not wavelength: nuclear decay → gamma; electron deceleration → X-ray. Examiners exploit this — a question saying “wavelength = $10^{-11}$ m, source = nucleus” is asking about gamma, not X-rays.

Trap 4: Ozone layer absorbs UV, not IR

The ozone layer blocks UV radiation. The greenhouse effect involves IR radiation being trapped. These two facts get swapped surprisingly often under exam pressure. Ozone = UV blocker. CO₂/water vapour = IR absorbers (greenhouse gases).

Trap 5: Displacement current depends on $d\Phi_E/dt$ , not $\Phi_E$

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ . If the electric field is constant (DC steady state after capacitor is fully charged), $I_d = 0$ even though $\Phi_E$ is non-zero. The current exists only while the field is changing.