How much of JEE Main comes from Electromagnetic Waves?

JEEGenie estimates about 1% of JEE Main marks come from Electromagnetic Waves, based on recent papers. The share varies from year to year.

Which formulas matter most in Electromagnetic Waves?

The Electromagnetic Waves sheet on this page has 16 formula entries, starting with Maxwell's Equations (Integral Form), Displacement Current, Speed of EM Waves in Vacuum, Relation Between E and B in EM Wave, Energy Density of EM Wave. Each one comes with when to use it and the trap to avoid.

What topics does Electromagnetic Waves cover for JEE?

Electromagnetic Waves covers Displacement Current, Properties of EM Waves, Electromagnetic Spectrum, Energy in EM Waves.

How should I revise Electromagnetic Waves before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Electromagnetic Waves questions on JEEGenie.

How many previous-year questions are there on Electromagnetic Waves?

JEEGenie has 178 previous-year questions on Electromagnetic Waves, and this page shows a few with worked explanations.

Is practice on Electromagnetic Waves free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Electromagnetic Waves for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 1% of JEE Main marks come from Electromagnetic Waves. This page covers 16 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing EM waves with mechanical waves; EM waves do not require a medium.

Key points to know

Electromagnetic Waves (EM Waves). EM waves are transverse waves consisting of oscillating electric field (E) and magnetic field (B) perpendicular to each other and to the direction of propagation.
Maxwell's Equations (Integral Form). Four equations governing electromagnetism: Gauss's law for electricity, Gauss's law for magnetism, Faraday's law, and Ampere-Maxwell law.
Displacement Current. A current-like term $ϵ_{0} \frac{d Φ_{E}}{d t}$ that appears in Ampere-Maxwell law, allowing EM wave propagation even in vacuum.
Electromagnetic Spectrum. The continuous range of EM waves ordered by frequency/wavelength, from radio waves to gamma rays.
Transverse Nature of EM Waves. In EM waves, E and B are perpendicular to each other and to the direction of propagation, making them transverse.
Polarization of EM Waves. Polarization describes the orientation of the electric field oscillation; only transverse waves can be polarized.
Applications of EM Waves. Different EM wave bands have specific applications: radio for communication, microwaves for radar/cooking, IR for heating, visible for sight, UV for sterilization, X-rays for medical imaging, gamma for cancer treatment.
Properties of EM Waves. EM waves travel at speed c in vacuum, carry energy and momentum, can be polarized, and obey superposition and reflection/refraction.
Hertz's Experiment. Heinrich Hertz first generated and detected EM waves (radio waves) using a spark gap transmitter and receiver, confirming Maxwell's theory.
Doppler Effect in EM Waves. For EM waves, Doppler shift depends on relative velocity; used in radar and astronomy.

Formula sheet

Name	Formula	When to use
Maxwell's Equations (Integral Form)	$\oint \vec{E} \cdot d \vec{A} = \frac{Q_{enc}}{ϵ_{0}}$ $\oint \vec{B} \cdot d \vec{A} = 0$ $\oint \vec{E} \cdot d \vec{l} = - \frac{d Φ_{B}}{d t}$ $\oint \vec{B} \cdot d \vec{l} = μ_{0} I_{enc} + μ_{0} ϵ_{0} \frac{d Φ_{E}}{d t}$	When deriving wave equations or understanding the origin of EM waves.
Displacement Current	$I_{d} = ϵ_{0} \frac{d Φ_{E}}{d t}$	When analyzing circuits with capacitors or EM wave generation.
Speed of EM Waves in Vacuum	$c = \frac{1}{\sqrt{μ_{0} ϵ_{0}}} = 3 \times 10^{8} m/s$	When calculating wavelength, frequency, or time of flight for EM waves.
Relation Between E and B in EM Wave	$E_{0} = c B_{0}$ or $E = c B$ (instantaneous)	When given one field amplitude to find the other.
Energy Density of EM Wave	$u = \frac{1}{2} ϵ_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}} = ϵ_{0} E^{2} = \frac{B^{2}}{μ_{0}}$ (in vacuum)	When calculating energy stored in an EM wave or comparing fields.
Intensity of EM Wave	$I = \frac{P}{A} = \frac{1}{2} ϵ_{0} c E_{0}^{2} = \frac{1}{2} \frac{c}{μ_{0}} B_{0}^{2} = \frac{E_{0}^{2}}{2 μ_{0} c}$	When calculating power radiated by a source or received by a detector.
Poynting Vector	$\vec{S} = \frac{1}{μ_{0}} \vec{E} \times \vec{B}$ Magnitude: $S = \frac{E B}{μ_{0}}$	When finding direction of energy propagation or instantaneous power flow.
Electromagnetic Spectrum	$c = f λ$	When identifying wave type based on frequency or wavelength.
Solving EM Wave Problems Using Wave Equation	$k = \frac{2 π}{λ}$ , $ω = 2 π f$ , $c = \frac{ω}{k}$	When given a sinusoidal EM wave and asked to find frequency, wavelength, or field values.
Calculating Displacement Current in Capacitor	$I_{d} = ϵ_{0} \frac{d Φ_{E}}{d t} = ϵ_{0} A \frac{d E}{d t}$ (for parallel plate capacitor)	When analyzing current in capacitor circuit or EM wave generation.
Finding Radiation Pressure	For perfect absorption: $P = \frac{I}{c}$ For perfect reflection: $P = \frac{2 I}{c}$	When calculating force on a surface due to EM wave.
Speed of EM Wave in a Medium	$v = \frac{1}{\sqrt{μ ϵ}} = \frac{c}{n}$ where $n = \sqrt{μ_{r} ϵ_{r}}$ is refractive index	When calculating speed, wavelength, or frequency in a medium.
Calculating Energy and Momentum of EM Wave	Momentum density: $\frac{u}{c}$ Momentum per unit area per time: $\frac{I}{c}$ (absorption)	When calculating force or momentum transfer from EM waves.
Average Intensity from RMS Fields	$I = ϵ_{0} c E_{rms}^{2} = \frac{c}{μ_{0}} B_{rms}^{2}$ where $E_{rms} = \frac{E_{0}}{\sqrt{2}}$ , $B_{rms} = \frac{B_{0}}{\sqrt{2}}$	When given RMS values of fields to find intensity.
Solving Problems with EM Wave in Dielectric	$v = \frac{c}{n}$ , $λ_{medium} = \frac{λ_{0}}{n}$ , $f$ unchanged	When EM wave enters a medium from vacuum.
Doppler Effect in EM Waves	$f^{'} = f \sqrt{\frac{1 \pm β}{1 \mp β}}$ where $β = v / c$ (relativistic formula)	When calculating frequency shift due to relative motion.

Open the Electromagnetic Waves formula page

What JEE asks

Electromagnetic Waves (EM Waves). JEE often asks to identify the correct orientation of E, B, and propagation direction using right-hand rule.
Maxwell's Equations (Integral Form). JEE may ask to identify which equation predicts EM waves or the significance of the displacement current.
Displacement Current. JEE often asks to calculate displacement current in a charging capacitor or its role in EM waves.
Speed of EM Waves in Vacuum. JEE may ask to derive $c$ from Maxwell's equations or use it in relation $c = f λ$ .
Relation Between E and B in EM Wave. JEE often gives E-field amplitude and asks for B-field amplitude or vice versa.
Energy Density of EM Wave. JEE may ask to find energy density given E or B, or to show equality of densities.
Intensity of EM Wave. JEE often asks to find intensity from field amplitudes or vice versa.
Poynting Vector. JEE may ask to determine direction of S given E and B, or to calculate average S.
Electromagnetic Spectrum. JEE often asks to arrange waves in order of frequency/wavelength or identify applications.
Solving EM Wave Problems Using Wave Equation. JEE may give equation of E or B and ask for amplitude, frequency, or direction.
Calculating Displacement Current in Capacitor. JEE often asks to find displacement current between plates given charging rate.
Finding Radiation Pressure. JEE may ask to find pressure on a solar sail or mirror.
Transverse Nature of EM Waves. JEE may ask to identify which vectors are perpendicular in an EM wave.
Polarization of EM Waves. JEE may ask to identify polarization type or its effect on intensity.
Speed of EM Wave in a Medium. JEE may ask to find refractive index from given $ϵ_{r}$ and $μ_{r}$ .
Using Right-Hand Rule for EM Wave Propagation. JEE often asks to determine propagation direction from given field orientations.
Applications of EM Waves. JEE may ask to match wave type with application or frequency range.
Properties of EM Waves. JEE may ask to identify which property is unique to EM waves (e.g., polarization).
Calculating Energy and Momentum of EM Wave. JEE may ask to find momentum transferred to a surface over time.
Hertz's Experiment. JEE may ask about the setup or significance of Hertz's experiment.
Average Intensity from RMS Fields. JEE may give $E_{rms}$ and ask for intensity or vice versa.
Solving Problems with EM Wave in Dielectric. JEE may ask to find new wavelength or speed given refractive index.
Doppler Effect in EM Waves. JEE may ask to find velocity from redshift or blueshift.

Common traps

Confusing EM waves with mechanical waves; EM waves do not require a medium. Electromagnetic Waves (EM Waves)
Forgetting the displacement current term in Ampere-Maxwell law. Maxwell's Equations (Integral Form)
Thinking displacement current is a real current; it is a concept to maintain continuity. Displacement Current
Using $c$ as speed in a medium; it is only in vacuum. Speed of EM Waves in Vacuum
Using $E = c B$ in a medium; it holds only in vacuum. Relation Between E and B in EM Wave
Forgetting that in vacuum, electric and magnetic energy densities are equal. Energy Density of EM Wave
Using RMS values instead of peak values in the formula; $I = \frac{1}{2} ϵ_{0} c E_{0}^{2}$ uses peak E. Intensity of EM Wave
Forgetting the cross product order; it's $\vec{E} \times \vec{B}$ , not $\vec{B} \times \vec{E}$ . Poynting Vector
Confusing the order: increasing frequency: radio, microwave, IR, visible, UV, X-ray, gamma. Electromagnetic Spectrum
Forgetting that E and B are in phase in vacuum. Solving EM Wave Problems Using Wave Equation
Forgetting that $Φ_{E} = E A$ only if E is uniform. Calculating Displacement Current in Capacitor
Using $P = I / c$ for reflection; it's double. Finding Radiation Pressure
Thinking EM waves can be longitudinal; they are always transverse in vacuum. Transverse Nature of EM Waves
Thinking only light can be polarized; all EM waves can. Polarization of EM Waves

Topics in Electromagnetic Waves

Displacement Current
Properties of EM Waves
Electromagnetic Spectrum
Energy in EM Waves

Previous-year questions on Electromagnetic Waves

JEEGenie has 178 previous-year questions on Electromagnetic Waves. A few worked examples follow.

Year	Questions
2026	17
2025	11
2024	15
2023	24
2022	25
2021	25
2020	15
2019	16
2018	4
2017	3
2016	4
2014	6
2013	5
2012	3
2006	1
2004	2
2002	2

Question 1 easy

An electromagnetic wave travelling in x-direction is described by field equation

E_{y} = 300 \sin ω (t - \frac{x}{c})

. If the electron is restricted to move in y-direction only with speed of

1.5 \times 10^{6}

m/s then ratio of maximum electric and magnetic forces acting on the electron is ______.

Answer: A

Explanation. Given

E_{y} = 300 \sin ω (t - \frac{x}{c}),

the maximum electric field is

E_{0} = 300 V/m .

The maximum electric force on the electron is

F_{e, \max} = e E_{0} .

For an electromagnetic wave in vacuum,

E_{0} = c B_{0} \Rightarrow B_{0} = \frac{E_{0}}{c} .

The magnetic force on a charge moving with velocity

v

F_{m} = e v B .

Since the electron moves in the

y

-direction and the wave travels in the

x

-direction, the magnetic field is along

z

, so

v ⊥ B

, giving

F_{m, \max} = e v B_{0} = e v \frac{E_{0}}{c} .

Thus the ratio is

\frac{F_{e, \max}}{F_{m, \max}} = \frac{e E_{0}}{e v E_{0} / c} = \frac{c}{v} .

Substitute

c = 3 \times 10^{8}

m/s and

v = 1.5 \times 10^{6}

m/s:

\frac{F_{e, \max}}{F_{m, \max}} = \frac{3 \times 10^{8}}{1.5 \times 10^{6}} = 200.

200

**Answer: A) 200**

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 medium

The electric field of an electromagnetic wave in free space is

\vec{E} = 57 \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (4 \hat{i} - 3 \hat{j}) N / C

. The associated magnetic field in Tesla is

$\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$
$\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$
$\vec{B} = - \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$
$\vec{B} = - \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

Answer: D

Explanation. For a plane electromagnetic wave in free space, the magnetic field is related to the electric field by

\vec{B} = \frac{1}{c} \hat{n} \times \vec{E}

where

\hat{n}

is the unit vector along propagation. The phase is

7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)

so the propagation vector is along

3 \hat{i} + 4 \hat{j}

whose unit vector is

\hat{n} = \frac{3 \hat{i} + 4 \hat{j}}{5} .

Given

\vec{E} = 57 \cos ϕ (4 \hat{i} - 3 \hat{j}),

\hat{n} \times \vec{E} = \frac{1}{5} (3 \hat{i} + 4 \hat{j}) \times 57 (4 \hat{i} - 3 \hat{j}) \cos ϕ .

Now,

(3 \hat{i} + 4 \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = - 9 \hat{k} - 16 \hat{k} = - 25 \hat{k} .

Thus

\hat{n} \times \vec{E} = \frac{57}{5} (- 25 \hat{k}) \cos ϕ = - 285 \hat{k} \cos ϕ .

\vec{B} = \frac{1}{3 \times 10^{8}} (- 285 \hat{k}) \cos ϕ = - \frac{57}{3 \times 10^{8}} (5 \hat{k}) \cos ϕ .

Hence

\vec{B} = - \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k}) .

Answer: (D)

From NTA/IIT papers: JEE Main 2025 · 23 Jan Shift 1

Question 3 easy

The electric field in an electromagnetic wave is given by

\vec{E} = \hat{i} 40 \cos ω (t - z / c) {N C}^{- 1}

. The magnetic field induction of this wave is (in SI unit) :

$\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - z / c)$
$\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - z / c)$
$\vec{B} = \hat{j} 40 \cos ω (t - z / c)$
$\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - z / c)$

Answer: A

Explanation. For a plane electromagnetic wave, the electric field

\vec{E}

, magnetic field

\vec{B}

, and propagation direction

\hat{n}

are mutually perpendicular, with

\vec{B} = \frac{\hat{n} \times \vec{E}}{c} .

Here

\vec{E} = \hat{i} 40 \cos ω (t - z / c) .

The wave travels along

+ z

, so

\hat{n} = \hat{k}

. Then

\vec{B} = \frac{\hat{k} \times (\hat{i} 40 \cos ω (t - z / c))}{c} .

Using

\hat{k} \times \hat{i} = \hat{j}

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - z / c) .

The magnetic field must be perpendicular to both

\vec{E}

and the propagation direction. Since

\vec{E}

is along

\hat{i}

and propagation is along

\hat{k}

\vec{B}

must be along

\hat{j}

. Options (B), (C), and (D) fail because they give

\vec{B}

along

\hat{i}

, omit the

1 / c

factor, or give

\hat{k}

, respectively. Answer: (A)

From NTA/IIT papers: JEE Main 2024 · 4 Apr Shift 1

Question 4 medium

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of

24 W

. The radius of curvature of hemisphere is

10 c m

and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________

\times 10^{- 8} N

Answer: 4

Explanation. **Principle:** The source is at the centre of curvature, so every ray meets the inner surface normally and is reflected straight back. Each reflected photon transfers momentum

2 p

along its ray. Intensity on the surface (

R = 0.1

m):

I = \frac{P}{4 π R^{2}}

A surface element

d A

at angle

θ

from the symmetry axis receives force

d F = \frac{2 I}{c} d A

along its radius. By symmetry only the axial components

d F \cos θ

add up:

F = \frac{2 I}{c} \int \cos θ d A

\int \cos θ d A

over the hemisphere is its projection on the base plane,

π R^{2}

. Hence

F = \frac{2}{c} \cdot \frac{P}{4 π R^{2}} \cdot π R^{2} = \frac{P}{2 c}

F = \frac{24}{2 \times 3 \times 10^{8}} = 4 \times 10^{- 8} N

The radius cancels, so the result does not depend on the 10 cm value. Answer: 4

From NTA/IIT papers: JEE Main 2023 · 30 Jan Shift 1

Question 5 hard

Seawater at a frequency f = 9

\times

102 Hz, has permittivity

ε

= 80

ε

0 and resistivity

ρ

= 0.25

Ω

m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0

\sin

π

ft). Then the conduction current density becomes 10x times the displacement current density after time t =

\frac{1}{800}

s. The value of x is _____________. (Given :

\frac{1}{4 π ε_{0}} = 9 \times 10^{9}

Nm2C

-

Answer: 6

Explanation. **Principle:** In a conducting dielectric, total current is the sum of conduction current density

J_{c} = σ E

and displacement current density

J_{d} = ε \frac{d E}{d t}

. Given

ρ = 0.25 Ω m

, so

σ = 1 / ρ = 4 S/m

ε = 80 ε_{0}

, and

f = 9 \times 10^{2} Hz

ω = 2 π f

. For

V (t) = V_{0} \sin ω t

, the field is

E = E_{0} \sin ω t

. Then

J_{c} = σ E_{0} \sin ω t, J_{d} = ε ω E_{0} \cos ω t .

The ratio is

\frac{J_{c}}{J_{d}} = \frac{σ}{ε ω} \tan ω t .

t = \frac{1}{800} s

ω t = 2 π (900) (\frac{1}{800}) = \frac{9 π}{4} .

Thus

\tan \frac{9 π}{4} = 1.

\frac{J_{c}}{J_{d}} = \frac{σ}{ε ω} = \frac{4}{80 ε_{0} (2 π) (900)} .

Using

\frac{1}{4 π ε_{0}} = 9 \times 10^{9}

\frac{1}{ε_{0}} = 36 π \times 10^{9} .

Therefore

\frac{J_{c}}{J_{d}} = \frac{4}{80 (2 π) (900)} \cdot 36 π \times 10^{9} = 10^{6} .

Hence

J_{c} = 10^{6} J_{d}

, so

x = 6

. Answer: 6

From NTA/IIT papers: JEE Main 2021 · 17 Mar Shift 2

Frequently asked questions

How much of JEE Main comes from Electromagnetic Waves?: JEEGenie estimates about 1% of JEE Main marks come from Electromagnetic Waves, based on recent papers. The share varies from year to year.
Which formulas matter most in Electromagnetic Waves?: The Electromagnetic Waves sheet on this page has 16 formula entries, starting with Maxwell's Equations (Integral Form), Displacement Current, Speed of EM Waves in Vacuum, Relation Between E and B in EM Wave, Energy Density of EM Wave. Each one comes with when to use it and the trap to avoid.
What topics does Electromagnetic Waves cover for JEE?: Electromagnetic Waves covers Displacement Current, Properties of EM Waves, Electromagnetic Spectrum, Energy in EM Waves.
How should I revise Electromagnetic Waves before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Electromagnetic Waves questions on JEEGenie.
How many previous-year questions are there on Electromagnetic Waves?: JEEGenie has 178 previous-year questions on Electromagnetic Waves, and this page shows a few with worked explanations.
Is practice on Electromagnetic Waves free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.