A plane EM wave travelling along z-direction is de – Class 12 Physics

NCERT Exemplar · Q26

z

(i) the average energy density of the wave is given by

u_{av} = \dfrac{1}{4}\varepsilon_0 E_0^2 + \dfrac{1}{4}\dfrac{B_0^2}{\mu_0}

(ii) the time averaged intensity of the wave is given by

I_{av} = \dfrac{1}{2}c\varepsilon_0 E_0^2

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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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The average energy density and intensity of a plane EM wave are derived by time-averaging the instantaneous electric and magnetic energy densities over one cycle. The key result is that the electric and magnetic contributions are equal, leading to $u_{av} = \frac{1}{4}\varepsilon_0 E_0^2 + \frac{1}{4}\frac{B_0^2}{\mu_0}$ and $I_{av} = \frac{1}{2}c\varepsilon_0 E_0^2$ .

Why This Approach Works

For an electromagnetic wave, energy is stored in both the electric and magnetic fields. The instantaneous energy density is the sum of the electric energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$ and the magnetic energy density $u_B = \frac{1}{2}\frac{B^2}{\mu_0}$ . Since the wave varies sinusoidally in time, the energy density also oscillates. What we measure or care about in most practical situations is the average over a full cycle — this is what the problem asks for.

The intensity (power per unit area) is the rate at which energy flows through a surface. For a plane wave, this is given by the magnitude of the Poynting vector $\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$ . Again, we time-average this to get the average intensity.

The trick: for sinusoidal functions, the average of $\sin^2$ or $\cos^2$ over a full period is exactly $1/2$ . This single fact does all the heavy lifting.

Step-by-Step Solution

1. Write the instantaneous energy density

The total instantaneous energy density is:

u = u_E + u_B = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2}\frac{B^2}{\mu_0}

Given $\mathbf{E} = E_0 \sin(kz - \omega t)\,\hat{i}$ and $\mathbf{B} = B_0 \sin(kz - \omega t)\,\hat{j}$ , we have:

u = \frac{1}{2}\varepsilon_0 E_0^2 \sin^2(kz - \omega t) + \frac{1}{2}\frac{B_0^2}{\mu_0} \sin^2(kz - \omega t)

2. Time-average over one cycle

The time average of $\sin^2(kz - \omega t)$ over a full period $T = 2\pi/\omega$ is:

\langle \sin^2(kz - \omega t) \rangle = \frac{1}{T} \int_0^T \sin^2(kz - \omega t)\, dt = \frac{1}{2}

Tip

This is a standard result: the average of $\sin^2$ or $\cos^2$ over any integer number of half-cycles is $1/2$ . You can derive it quickly using $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ — the cosine term averages to zero.

Therefore:

u_{av} = \frac{1}{2}\varepsilon_0 E_0^2 \cdot \frac{1}{2} + \frac{1}{2}\frac{B_0^2}{\mu_0} \cdot \frac{1}{2}

u_{av} = \frac{1}{4}\varepsilon_0 E_0^2 + \frac{1}{4}\frac{B_0^2}{\mu_0}

This proves part (i).

Note

For an EM wave in vacuum, $E_0 = c B_0$ and $c = 1/\sqrt{\varepsilon_0 \mu_0}$ , so the two terms are actually equal. You can check: $\frac{1}{4}\frac{B_0^2}{\mu_0} = \frac{1}{4}\frac{E_0^2}{c^2 \mu_0} = \frac{1}{4}\varepsilon_0 E_0^2$ . So $u_{av} = \frac{1}{2}\varepsilon_0 E_0^2$ — but the problem asks you to show the given form, which is more general.

3. Find the instantaneous intensity (Poynting vector)

The Poynting vector gives the power per unit area carried by the wave:

\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}

Here $\mathbf{E} = E_0 \sin(kz - \omega t)\,\hat{i}$ and $\mathbf{B} = B_0 \sin(kz - \omega t)\,\hat{j}$ . Their cross product:

\mathbf{E} \times \mathbf{B} = E_0 B_0 \sin^2(kz - \omega t)\, (\hat{i} \times \hat{j}) = E_0 B_0 \sin^2(kz - \omega t)\,\hat{k}

So:

$\mathbf{S} = \frac{E_0 B_0}{\mu_0} \sin^2(kz - \omega t)\,\hat{k}$ …

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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 32 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}$
›Reveal solutionSolution
The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is (A).

The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$ , where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$ . That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves.

Identify $\omega$ and $k$ from the equation.

The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s.

The term multiplying $x$ is $k = 150$ rad/m.

Apply the wave velocity formula:

$v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}$

Simplify: $v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}$ …
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CBSE 2026Set A1 markMCQ
$Q. The dimensions of B0^2/μ0 will be the same as that of (A) energy density (B) work (C) momentum (D) electric flux$
›Reveal solutionSolution

Magnetic energy density = B²/2μ₀, so B²/μ₀ carries the dimensions of energy density.

The energy stored per unit volume in a magnetic field is $u_B = \dfrac{B^2}{2\mu_0}$ .

Apart from the numerical factor ½, the quantity $\dfrac{B^2}{\mu_0}$ therefore has the dimensions of energy density (J/m³).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The velocity of electromagnetic waves in vacuum is:$
(a) c = 1/√(μ₀ε₀)
(b) c = √(μ₀ε₀)
(c) c = √(μ₀/ε₀)
(d) c = √(ε₀/μ₀)
›Reveal solutionSolution

Maxwell's equations predict electromagnetic waves travelling in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ , which numerically matches the measured speed of light.

Starting from Maxwell's equations in free space (no charges or currents), the wave equations for $\vec E$ and $\vec B$ both take the standard wave-equation form with wave speed $v=1/\sqrt{\mu_0\varepsilon_0}$ . Substituting $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$ and $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ gives $v\approx3\times10^{8}\ \text{m/s}$ , exactly the speed of light — this a …

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CBSE 2026Set ANNUAL1 markMCQ
$2.5\times10^{10}Hz$
(a) $1.52\times10^{-8}\ weber/m^2$
(b) $1.52\times10^{-7}\ weber/m^2$
(c) $1.6\times10^{-6}\ weber/m^2$
(d) $1.6\times10^{-7}\ weber/m^2$
›Reveal solutionSolution

$B_0 = E_0/c$ .

In a plane electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $B_0 = \dfrac{E_0}{c}$ :

…

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CBSE 2025Set 55/4/11 markMCQ
$1000$
›Reveal solutionSolution
In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light: $E_0 = c B_0$ . With $E_0 = 1000$ V/m, we find $B_0 = 3.33 \times 10^{-6}$ T.

Why the fields are linked by $c$

An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that propagate together through space. Maxwell's equations demand a precise relationship between these two fields: at every instant and every point in the wave, the ratio of the electric field amplitude to the magnetic field amplitude equals the speed of light in that medium.

In free space, this relationship is beautifully simple:
$E_0 = c B_0$
where $E_0$ is the amplitude of the electric field, $B_0$ is the amplitude of the magnetic field, and $c = 3 \times 10^8$ m/s is the speed of light in vacuum. This isn't arbitrary—it emerges directly from the wave equations derived from Maxwell's laws. The electric and magnetic fields are perpendicular to each other and to the direction of propagation, oscillating in phase, with their amplitudes locked in this ratio.

Finding the magnetic field amplitude

We're given the electric field amplitude and need to find the magnetic field amplitude.

Write down the fundamental relationship:

$E_0 = c B_0$

Rearrange to solve for $B_0$ :

$B_0 = \frac{E_0}{c}$

Substitute the given values: …
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CBSE 2025Set ANNUAL1 markMCQ
$\vec{E} = 9.0\hat{j}$
(a) $3\times10^{-8}$ T
(b) $9\times10^{-8}$ T
(c) $27\times10^{9}$ T
(d) $2.1\times10^{-8}$ T
›Reveal solutionSolution

In a plane EM wave in free space, $B = E/c$ .

For a plane electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic fields are related by

$B = \frac{E}{c}$ …

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CBSE 2025Set D1 markMCQ
$Q. Unit of \sqrt(μ0/ε0) is (A) newton/coulomb (B) ohm (C) henry (D) farad$
›Reveal solutionSolution

√(μ₀/ε₀) has the dimensions of resistance; it is the intrinsic impedance of free space, about 377 ohm.

The ratio of the electric to magnetic field amplitude of an electromagnetic wave in vacuum is c = E/B, and the quantity

$Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$

is the impedance of free space. Numerically

…

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CBSE 2025Set ANNUAL1 mark
$B_y = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\ T$
›Reveal solutionSolution

$\lambda = 2\pi/k$ , with $k=0.5\times10^3$ rad/m read off the wave equation.

Comparing $B_y = 2\times10^{-7}\sin(0.5\times10^3 x + 1.5\times10^{11}t)$ with the standard form $B_y = B_0\sin(kx+\omega t)$ , the wave number is $k = 0.5\times10^3\ \text{rad/m}$ . The wavelength is

…

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CBSE 2024Set A1 markMCQ
$Q. The value of (μ₀ε₀)^-1/2 is (A) 3 \times 10^8 cm/second (B) 3 \times 10^10 cm/second (C) 3 \times 10^9 cm/second (D) 3 \times 10^8 km/second$
›Reveal solutionSolution

1/√(μ₀ε₀) is the speed of light c = 3×10⁸ m/s = 3×10¹⁰ cm/s.

Maxwell showed the speed of electromagnetic waves in vacuum is

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3\times10^8\ \text{m/s}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$3 \times 10^{-6}$
(a) 600 Vm $^{-1}$
(b) 100 Vm $^{-1}$
(c) 900 Vm $^{-1}$
(d) 300 Vm $^{-1}$
›Reveal solutionSolution

Using $E_0 = cB_0$ with $c = 3\times10^8\ \text{ms}^{-1}$ and $B_0 = 3\times10^{-6}$ T gives $E_0 = 900\ \text{Vm}^{-1}$ .

Working

For an electromagnetic wave travelling in free space, the peak electric and magnetic field amplitudes are related by

$E_0 = c B_0$

Given $B_0 = 3\times10^{-6}$ T: …

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CBSE 2023Set TERM21 markMCQ
$Q. Amplitude of the magnetic field part of a harmonic Electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave ?$
(a) 150 NC-1
(b) 160 NC-1
(c) 153 NC-1
(d) 163 NC-1
›Reveal solutionSolution

In vacuum the ratio of the electric-field amplitude to the magnetic-field amplitude of an EM wave equals the speed of light: $E_0 = cB_0$ .

For a plane electromagnetic wave travelling in vacuum, Maxwell's equations require the electric and magnetic field amplitudes to be related by

$E_0 = c\,B_0$

where $c = 3\times10^{8}\ \text{m/s}$ is the speed of light in vacuum.

…

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CBSE 2023Set ANNUAL1 markMCQ
$3 \times 10^{-6}$
(a) 600 Vm $^{-1}$
(b) 100 Vm $^{-1}$
(c) 900 Vm $^{-1}$
(d) 300 Vm $^{-1}$
›Reveal solutionSolution

Using $E=cB$ for an electromagnetic wave with $c=3\times10^8\ \text{m/s}$ and $B=3\times10^{-6}$ T gives $E=900\ \text{Vm}^{-1}$ .

Working

For an electromagnetic wave travelling in free space (or air), the magnitudes of the electric and magnetic field amplitudes are related by

$E = cB$

where $c = 3\times10^8\ \text{ms}^{-1}$ is the speed of light.

…

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Q.A plane EM wave travelling along $z$ -direction is described by $\mathbf{E} = E_0\sin(kz - \omega t)\hat{i}$ and $\mathbf{B} = B_0\sin(kz - \omega t)\hat{j}$ . Show that

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 $\mathbf{B}$

Why This Approach Works

Step-by-Step Solution

1. Write the instantaneous energy density

2. Time-average over one cycle

3. Find the instantaneous intensity (Poynting vector)

Method: Time-Averaging the Energy Density and Intensity of a Plane EM Wave

Steps

Why the fields are linked by $c$

Finding the magnetic field amplitude

Q.A plane EM wave travelling along zzz-direction is described by E=E0sin⁡(kz−ωt)i^\mathbf{E} = E_0\sin(kz - \omega t)\hat{i}E=E0​sin(kz−ωt)i^ and B=B0sin⁡(kz−ωt)j^\mathbf{B} = B_0\sin(kz - \omega t)\hat{j}B=B0​sin(kz−ωt)j^​. Show that

Why This Approach Works

Step-by-Step Solution

1. Write the instantaneous energy density

2. Time-average over one cycle

3. Find the instantaneous intensity (Poynting vector)

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Q.A plane EM wave travelling along $z$ -direction is described by $\mathbf{E} = E_0\sin(kz - \omega t)\hat{i}$ and $\mathbf{B} = B_0\sin(kz - \omega t)\hat{j}$ . Show that

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