A plane electromagnetic wave propagating along x-direction can have the following pairs of mathbfE and mathbfB

NCERT Exemplar · Q29

x

(a)

E_x, B_y

(b)

E_y, B_z

(c)

B_x, E_y

(d)

E_z, B_y

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Concept understanding — Electromagnetic Wave Relation

Electromagnetic Wave Relation: From Intuition to Precision

Imagine you're standing at the beach. You see a wave coming in — it has a certain speed, a certain distance between crests (wavelength), and a certain number of crests passing you per second (frequency). The faster the wave, the more crests pass you in a given time. That's the basic idea: speed = frequency × wavelength.

Now, light is also a wave — an electromagnetic wave. It doesn't need water or air; it travels through empty space at a staggering speed. The relation that governs all waves, including light, is:

v = f \lambda

where $v$ is the wave speed, $f$ is the frequency (in hertz, Hz), and $\lambda$ (lambda) is the wavelength (in metres).

For electromagnetic waves in vacuum, this speed is a universal constant: $c = 3 \times 10^8$ m/s. So the relation becomes:

c = f \lambda

That's it. But let's unpack what this really means.

What is frequency? What is wavelength?

Frequency is how many complete wave cycles pass a fixed point in one second. A radio station broadcasting at 100 MHz means 100 million cycles per second. Higher frequency means more oscillations per second.

Wavelength is the distance between two consecutive crests (or troughs) of the wave. For visible light, wavelengths are tiny — around 400 to 700 nanometres (billionths of a metre).

The product $f \lambda$ always equals the wave speed. So if frequency goes up, wavelength must go down to keep the product constant. This is why:

Gamma rays have extremely high frequency and extremely short wavelength.
Radio waves have low frequency and very long wavelength (metres to kilometres).

Both travel at the same speed $c$ in vacuum.

Why does this matter for exams?

You'll use this relation in three main ways:

Given frequency, find wavelength (or vice versa) — just rearrange: $\lambda = \frac{c}{f}$ or $f = \frac{c}{\lambda}$ .
Compare different regions of the electromagnetic spectrum — know that as frequency increases, wavelength decreases proportionally.
Solve problems involving energy — because photon energy $E = hf$ (where $h$ is Planck's constant), the wave relation links energy to wavelength: $E = \frac{hc}{\lambda}$ .

Watch out

A common mistake: using $c = f \lambda$ for waves in a medium (like glass or water). In a medium, the speed is less than $c$ , so the wavelength changes but frequency stays the same. The relation $v = f \lambda$ still holds, but $v$ is now the speed in that medium.

A concrete example

A microwave oven operates at 2.45 GHz. What is its wavelength in vacuum?

$f = 2.45 \times 10^9$ Hz, $c = 3 \times 10^8$ m/s. …

Why this formula?

Electromagnetic Wave Relation: Why $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Let's build this from first principles — not just memorising the formula, but understanding why light and all EM waves travel at this specific speed.

1. The Starting Point: Maxwell's Equations in Vacuum

In empty space (no charges, no currents), Maxwell's equations simplify to:

Gauss's law for electricity: $\nabla \cdot \mathbf{E} = 0$
Gauss's law for magnetism: $\nabla \cdot \mathbf{B} = 0$
Faraday's law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$
Ampère-Maxwell law: $\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

The key insight: a changing electric field creates a magnetic field, and a changing magnetic field creates an electric field. This mutual induction is what sustains the wave.

2. Deriving the Wave Equation for $\mathbf{E}$

Take the curl of Faraday's law:

\nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left(-\frac{\partial \mathbf{B}}{\partial t}\right) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Now use the vector identity: $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$

Since $\nabla \cdot \mathbf{E} = 0$ in vacuum, this becomes:

-\nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Substitute $\nabla \times \mathbf{B}$ from Ampère-Maxwell:

-\nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} \left( \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right)

Result: The electric field satisfies the wave equation:

\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}

3. Identifying the Wave Speed

Compare with the standard wave equation for any wave travelling at speed $v$ :

\nabla^2 \psi = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}

Matching terms:

\frac{1}{v^2} = \mu_0 \varepsilon_0 \quad \Rightarrow \quad v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

This $v$ is the speed of electromagnetic waves in vacuum — denoted $c$ .

Why this is profound: The constants $\mu_0$ (permeability of free space) and $\varepsilon_0$ (permittivity of free space) come from static electricity and magnetism. Yet their combination gives the speed of light — showing light is an electromagnetic wave.

4. The Magnetic Field Follows Suit

Exactly the same derivation starting from Ampère-Maxwell law gives:

\nabla^2 \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2}

So both $\mathbf{E}$ and $\mathbf{B}$ propagate at the same speed $c$ .

5. The Crucial Relationship Between $\mathbf{E}$ and $\mathbf{B}$

For a plane wave travelling in the $x$ -direction:

$\mathbf{E}$ oscillates along $y$ : $E_y = E_0 \sin(kx - \omega t)$
$\mathbf{B}$ oscillates along $z$ : $B_z = B_0 \sin(kx - \omega t)$

From Faraday's law: $\frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t}$

Differentiating the wave forms:

k E_0 \cos(kx - \omega t) = \omega B_0 \cos(kx - \omega t)

Since $\omega = ck$ , we get:

$\frac{E_0}{B_0} = \frac{\omega}{k} = c$ …

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For a plane EM wave travelling along $+x$ , both $\mathbf E$ and $\mathbf B$ must be transverse (perpendicular to $x$ ), and their cross product $\mathbf E\times\mathbf B$ must point along $+\hat\imath$ . Checking all four listed pairs, only (b) $E_y,B_z$ satisfies both requirements.

The two requirements

A plane electromagnetic wave in vacuum, propagating along $\hat k$ , has:

Transversality: neither $\mathbf E$ nor $\mathbf B$ has a component along $\hat k$ .
Right-handedness: $\mathbf E\times\mathbf B$ points along $\hat k$ (the direction of energy flow, given by the Poynting vector $\mathbf S=\frac{1}{\mu_0}\mathbf E\times\mathbf B$ ).

Here $\hat k=\hat\imath$ ( $+x$ direction).

Checking each pair

(a) $E_x,B_y$ . The electric field has an $x$ -component — i.e. $\mathbf E$ points (at least partly) along the direction of propagation itself. This violates transversality outright. Invalid.

(b) $E_y,B_z$ . Both fields are transverse (one along $y$ , one along $z$ , neither along $x$ ). Check the cross product direction using $\hat\jmath\times\hat{z}=\hat\imath$ (writing $\hat{z}$ for the unit vector along $z$ to avoid clashing with the propagation-direction symbol $\hat k$ ): $\hat\jmath\times\hat{z}=\hat\imath$ , so $\mathbf E\times\mathbf B$ points along $+\hat\imath$ — exactly the stated propagation direction. Valid.

(c) $B_x,E_y$ . The magnetic field has an $x$ -component, again violating transversality (this time for $\mathbf B$ ). Invalid. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$3 \times 10^8$
›Reveal solutionSolution
EM waves propagate through vacuum, carrying mutually perpendicular oscillating electric and magnetic fields at speed $c$ .

Examining the options:

(A) True — a changing electric field creates a magnetic field and vice versa, so the wave is self-sustaining and needs no material medium (sunlight reaching Earth through space proves it).

(B) False — it has both fields, not just electric. …
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KEAM 2026Set eng-2026-04184 marksMCQ
$MHz$
›Reveal solutionSolution

Using $\lambda=c/f$ , the 7.5 MHz end maps to 40 m and the 12 MHz end to 25 m, giving a band of $40\,m$ – $25\,m$ .

$\lambda=\frac{c}{f},\qquad c=3\times10^{8}\,ms^{-1}$
For $f=7.5\,MHz=7.5\times10^{6}\,Hz$ :
$\lambda=\frac{3\times10^{8}}{7.5\times10^{6}}=40\,m$
For $f=12\,MHz=12\times10^{6}\,Hz$ : …
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KEAM 2026Set eng-2026-04204 marksMCQ
$3.6\ J$
›Reveal solutionSolution

For a completely absorbing surface the momentum delivered per unit time is $\dfrac{U/t}{c}$ .

For an electromagnetic wave completely absorbed by a surface, the force (rate of momentum transfer) equals the power divided by the speed of light. With energy delivered per unit time $\dfrac{U}{t}=3.6\ \text{J s}^{-1}$ : …
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KEAM 2026Set eng-2026-04214 marksMCQ
$Q. Microwaves are (A) used in radio and television communications (B) having frequency range from 54 MHz to 890 MHz (C) short wavelength radio waves (D) produced by hot bodies and molecules (E) absorbed by ordinary glass$
›Reveal solutionSolution

Microwaves = short-wavelength radio waves.

Microwaves occupy wavelengths of about $1\\,mm$ to $0.3\\,m$ (frequencies $\sim10^{9}$ – $10^{11}\\,Hz$ ), i.e. the short-wavelength end of the radio band, generated by devices such as klystrons and magnetrons. Options describing radio/TV frequency ran …
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KEAM 2026Set eng-2026-04224 marksMCQ
$^{-1}$
›Reveal solutionSolution

In an EM wave the field amplitudes satisfy $B_0=E_0/c$ . …

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KEAM 2026Set pha-2026-0418F4 marksMCQ
$Q. Identify the two electromagnetic waves A and B having respective wavelengths 2 cm and 580 nm (A) A is microwave and B is visible light (B) A is infrared and B is ultraviolet ray (C) A is radio wave and B is visible light (D) A is infrared and B is visible light (E) A is radio wave and B ultraviolet ray$
›Reveal solutionSolution

$\lambda = 2\,cm$ is microwave, $\lambda = 580\,nm$ is visible light. …

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KEAM 2026Set pha-2026-0420F4 marksMCQ
$\varepsilon_r = 4$
›Reveal solutionSolution

$v=c/\sqrt{\varepsilon_r\mu_r}=c/\sqrt{4\cdot1}=c/2$ .

The speed of an electromagnetic wave in a medium is
$v = \frac{c}{\sqrt{\varepsilon_r\mu_r}}.$
With $\varepsilon_r = 4$ and $\mu_r = 1$ : …
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KEAM 2025Set eng-2025-04234 marksMCQ
$Q. When a ray of light moves from one medium to another medium, (A) its frequency remains unchanged (B) its frequency alone changes (C) its wavelength remains unchanged (D) both its frequency and wavelength change (E) its velocity remains constant$
›Reveal solutionSolution
When light passes into another medium its frequency stays unchanged.

Concept and Intuition

Frequency is fixed by the source and is conserved across a boundary (the fields must oscillate continuously at the interface). Speed and wavelength change with the medium's refractive index, but frequency does not.

Step-by-Step Solution

v = f*lambda; on entering a new medium v changes.

f is set by the source and is conserved at the boundary. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$E = 15 \sin [1.57y + 5.4t]\hat{j}$
›Reveal solutionSolution

The propagation constant is $k = 1.57\ \text{m}^{-1}$ , giving wavelength $\lambda = 2\pi/k = 4.0$ m.

The wave is $E = 15\sin(1.57y + 5.4t)\hat{j}$ , of the form $E = E_0\sin(ky + \omega t)$ .

The angular wavenumber is $k = 1.57\ \text{m}^{-1} = \dfrac{2\pi}{\lambda}$ . …
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KEAM 2025Set eng-2025-04274 marksMCQ
$Q. The speed of electromagnetic waves in a medium depends on the (A) intensity of the wave (B) initial phase of the wave (C) permittivity and permeability of the medium (D) energy it carries (E) reflectivity of the medium$
›Reveal solutionSolution

Electromagnetic wave speed is fixed by the medium's electric and magnetic properties: $v = \dfrac{1}{\sqrt{\mu\varepsilon}}$ .

From Maxwell's equations, the speed of an electromagnetic wave in a medium is determined solely by its permittivity $\varepsilon$ and permeability $\mu$ :

$v = \frac{1}{\sqrt{\mu\varepsilon}}.$ …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$B = 400 \times 10^{-6} \sin\left[(4.0 \times 10^{-4})(t - x/c)\right]$
›Reveal solutionSolution

In an EM wave $E_0=cB_0=3\times10^{8}\times400\times10^{-6}=12\times10^{4}$ V m $^{-1}$ .

In a plane electromagnetic wave the peak electric and magnetic fields are related by
$E_0=cB_0.$
Here the amplitude of the magnetic field is $B_0=400\times10^{-6}\,\text{T}$ , so …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$2.5 \times 10^{-7} s$
›Reveal solutionSolution

The oscillation period equals the reciprocal of the frequency: $T = 1/(2\times10^6\,\mathrm{Hz}) = 5\times10^{-7}\,\mathrm{s}$ .

The accelerated charge oscillates at the same frequency as the electromagnetic wave it radiates. With $f = 2\,\mathrm{MHz} = 2\times10^6\,\mathrm{Hz}$ , …
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Q.A plane electromagnetic wave propagating along $x$ -direction can have the following pairs of $\mathbf{E}$ and $\mathbf{B}$

Concept understanding — Electromagnetic Wave Relation

Electromagnetic Wave Relation: From Intuition to Precision

What is frequency? What is wavelength?

Why does this matter for exams?

A concrete example

Why this formula?

Electromagnetic Wave Relation: Why $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

1. The Starting Point: Maxwell's Equations in Vacuum

2. Deriving the Wave Equation for $\mathbf{E}$

3. Identifying the Wave Speed

4. The Magnetic Field Follows Suit

5. The Crucial Relationship Between $\mathbf{E}$ and $\mathbf{B}$

The two requirements

Checking each pair

Method: Testing Whether a Given (E, B) Component Pair Is Valid for a Stated Propagation Direction

Steps

Concept and Intuition

Step-by-Step Solution

Q.A plane electromagnetic wave propagating along xxx-direction can have the following pairs of E\mathbf{E}E and B\mathbf{B}B

The two requirements

Checking each pair

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Q.A plane electromagnetic wave propagating along $x$ -direction can have the following pairs of $\mathbf{E}$ and $\mathbf{B}$

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