An electromagnetic wave travels in vacuum along z- – Class 12 Physics

NCERT Exemplar · Q27

z

(a) The associated magnetic field is given as

\mathbf{B} = \dfrac{1}{c}(E_1\hat{i} - E_2\hat{j})\cos(kz - \omega t)

(b) The associated magnetic field is given as

\mathbf{B} = \dfrac{1}{c}(E_1\hat{i} + E_2\hat{j})\cos(kz - \omega t)

(d) The given electromagnetic wave is plane polarised.

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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 $\mathbf E=(E_1\hat\imath+E_2\hat\jmath)\cos(kz-\omega t)$ travelling along $+z$ in vacuum, the associated magnetic field is $\mathbf B=\dfrac1c(\hat k\times\mathbf E)=\dfrac1c(E_1\hat\jmath-E_2\hat\imath)\cos(kz-\omega t)$ . Both components oscillate with the same phase, so the wave is plane (linearly) polarised — option (d) is correct, option (c) is false. Checked against the stem's own printed options, neither (a) nor (b) algebraically equals this correct $\mathbf B$ (option (b) would make $\mathbf B\parallel\mathbf E$ , which is physically impossible; option (a) has its $\hat\imath$ / $\hat\jmath$ components swapped relative to the correct result).

Setting up the relation between E and B

For a plane electromagnetic wave in vacuum travelling along $\hat k$ , Maxwell's equations require $\mathbf E$ , $\mathbf B$ and $\hat k$ to form a mutually perpendicular, right-handed set, related by

\mathbf B=\frac{1}{c}\big(\hat k\times \mathbf E\big)

Here the wave travels along $+z$ , so $\hat k=\hat k$ (the unit vector along $z$ ), and $\mathbf E=(E_1\hat\imath+E_2\hat\jmath)\cos(kz-\omega t)$ lies entirely in the $xy$ -plane (no $z$ -component — confirming the wave is transverse).

Step 1 — compute the cross product

Using $\hat k\times\hat\imath=\hat\jmath$ and $\hat k\times\hat\jmath=-\hat\imath$ :

\hat k\times\mathbf E=\hat k\times(E_1\hat\imath+E_2\hat\jmath)\cos(kz-\omega t)=(E_1\hat\jmath-E_2\hat\imath)\cos(kz-\omega t)

Step 2 — write B explicitly

\mathbf B=\frac{1}{c}(E_1\hat\jmath-E_2\hat\imath)\cos(kz-\omega t)=\frac{1}{c}(-E_2\hat\imath+E_1\hat\jmath)\cos(kz-\omega t)

This $\mathbf B$ lies in the $xy$ -plane, perpendicular to $\mathbf E$ (their dot product $E_1(-E_2)+E_2(E_1)=0$ confirms this), with magnitude $|\mathbf B|=\sqrt{E_1^2+E_2^2}/c=|\mathbf E|/c$ , exactly as required.

Step 3 — is the wave plane or circularly polarised?

Both the $\hat\imath$ and $\hat\jmath$ components of $\mathbf E$ share the identical factor $\cos(kz-\omega t)$ — they rise and fall together, in phase, always in the fixed ratio $E_1:E_2$ . The tip of $\mathbf E$ therefore traces a straight line (not a circle or ellipse) in the $xy$ -plane as time passes — this is the defining signature of a plane- (linearly-) polarised wave, not a circularly polarised one (which would need two components of equal amplitude, $90^\circ$ out of phase). So statement (d) is true and (c) is false.

Step 4 — checking the stem's own printed options for B …

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Q.An electromagnetic wave travels in vacuum along $z$ -direction: $\mathbf{E} = (E_1\hat{i} + E_2\hat{j})\cos(kz - \omega t)$ . Choose the correct options from the following:

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 $\mathbf{B}$

Setting up the relation between E and B

Step 1 — compute the cross product

Step 2 — write B explicitly

Step 3 — is the wave plane or circularly polarised?

Step 4 — checking the stem's own printed options for B …

Method: Finding B from a Given E and Testing the Polarization State

Steps

Q.An electromagnetic wave travels in vacuum along zzz-direction: E=(E1i^+E2j^)cos⁡(kz−ωt)\mathbf{E} = (E_1\hat{i} + E_2\hat{j})\cos(kz - \omega t)E=(E1​i^+E2​j^​)cos(kz−ωt). Choose the correct options from the following:

Setting up the relation between E and B

Step 1 — compute the cross product

Step 2 — write B explicitly

Step 3 — is the wave plane or circularly polarised?

Step 4 — checking the stem's own printed options for B …

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Q.An electromagnetic wave travels in vacuum along $z$ -direction: $\mathbf{E} = (E_1\hat{i} + E_2\hat{j})\cos(kz - \omega t)$ . Choose the correct options from the following:

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