(a) Depict the variation of electric field ($\vec{E}$) and magnetic field ($\vec{B}$) with respect to the direction of propagation of an electromagnetic wave. Write their two important characteristics. (b) Show that $\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$ gives the velocity of an electromagnetic wave in free space.

(a) Depict the variation of electric field (vecE) and magnetic field (vecB) with respect to the direction of p

Question

\vec{E}

(b) Show that

\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}

gives the velocity of an electromagnetic wave in free space.

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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In an electromagnetic wave, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the direction of propagation, forming a transverse wave. The speed of this wave in free space is $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$ , which is derived from Maxwell's equations.

(a) Variation of $\vec{E}$ and $\vec{B}$ with direction of propagation

An electromagnetic wave is a transverse wave. This means the oscillations of the electric and magnetic fields are perpendicular to the direction in which the wave travels.

If the wave propagates along the $x$ -axis, the electric field $\vec{E}$ oscillates along the $y$ -axis and the magnetic field $\vec{B}$ oscillates along the $z$ -axis. At any instant, the fields vary sinusoidally with position $x$ and time $t$ :

\vec{E} = E_0 \sin(kx - \omega t)\,\hat{j}, \quad \vec{B} = B_0 \sin(kx - \omega t)\,\hat{k}

Here, $k = 2\pi/\lambda$ is the wave number and $\omega = 2\pi f$ is the angular frequency.

Note

The three vectors $\vec{E}$ , $\vec{B}$ , and the direction of propagation $\vec{k}$ form a right-handed orthogonal triad. If you curl the fingers of your right hand from $\vec{E}$ to $\vec{B}$ , your thumb points in the direction of propagation.

Two important characteristics of electromagnetic waves:

Transverse nature: Both $\vec{E}$ and $\vec{B}$ are perpendicular to the direction of wave propagation. Neither field has a component along the direction of travel.
Mutual perpendicularity: $\vec{E}$ is perpendicular to $\vec{B}$ . The two fields oscillate in phase — they reach their maximum and minimum values at the same points in space and time.

Watch out

A common mistake is to think $\vec{E}$ and $\vec{B}$ are perpendicular to each other but one of them is along the direction of propagation. That is incorrect — both are transverse, so neither points along the propagation direction.

(b) Showing that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ gives the speed of EM waves in free space

We start from Maxwell's equations in free space (no charges, no currents):

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

The key idea is to derive a wave equation from these. A wave equation has the form $\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}$ , where $v$ is the wave speed.

Step 1: Take the curl of Faraday's law

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

Step 2: Use the Ampere-Maxwell law to replace $\nabla \times \vec{B}$ :

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

Step 3: Apply the vector identity $\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E}$

From Gauss's law, $\nabla \cdot \vec{E} = 0$ , so the first term vanishes. We get:

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

Step 4: Rearrange into the standard wave equation …

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

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$x$
›Reveal solutionSolution
In a propagating electromagnetic wave, the electric and magnetic fields oscillate in phase with each other, so the phase difference is zero. The correct option is (A).

Concept and Intuition

The question tests a fundamental property of electromagnetic waves in free space. Many students carry a vague memory that $\vec{E}$ and $\vec{B}$ are "perpendicular" and mistakenly think that means a $\pi/2$ phase difference. But perpendicularity here refers to direction in space, not a time delay.

In a plane electromagnetic wave, both fields vary sinusoidally with position and time. Maxwell's equations demand that the time-varying electric field produces the magnetic field, and vice versa — they are coupled in such a way that their peaks and zeros occur at the same instant. There is no lag between them.

Watch out

A common mistake is to confuse spatial orthogonality (the fields are perpendicular to each other and to the direction of propagation) with a phase difference. They are perpendicular in space, but they oscillate in time together — zero phase difference.

Step-by-Step Reasoning

Recall the wave equations from Maxwell's laws In free space (no charges or currents), Maxwell's equations yield wave equations for both $\vec{E}$ and $\vec{B}$ . For a wave traveling along the $x$ -axis, the solutions are:

$E_y = E_0 \sin(\omega t - kx + \phi)$ $B_z = B_0 \sin(\omega t - kx + \phi)$
Notice the same argument $(\omega t - kx + \phi)$ appears in both. This is not an assumption — it follows directly from Faraday's law and Ampère's law.

Why the phase must be identical Consider Faraday's law for a plane wave:

$\frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t}$
If $E_y = E_0 \sin(\omega t - kx + \phi_E)$ and $B_z = B_0 \sin(\omega t - kx + \phi_B)$ , then:

Left side: $\frac{\partial E_y}{\partial x} = -k E_0 \cos(\omega t - kx + \phi_E)$

Right side: $-\frac{\partial B_z}{\partial t} = -\omega B_0 \cos(\omega t - kx + \phi_B)$ …
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Figure — This is a 1-mark 'Draw propagation diagram of a linearly polarised EM wave' item where the drawing IS the answ

In a linearly polarised electromagnetic wave, the electric and magnetic fields oscillate along two fixed, mutually perpendicular directions, both perpendicular to the direction the wave travels.

Consider a wave travelling along the x-direction. In a linearly (plane) polarised wave, the electric field vector E oscillates only along one fixed direction, say the y-axis: Ey = E0 sin(kx - omega t). The magnetic field vector B oscillates only along the z-axis, in phase with E: Bz = B0 sin(kx - omega t), with E0/B0 = c (speed of light). …

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In free space E = cB in magnitude, and E, B, and the propagation direction form a right-handed set. Magnitude = 7·2 V/m; direction = ĵ. Option (a).

Step 1 — magnitude: For an electromagnetic wave in vacuum, E = cB = (3×10⁸ m/s)(2·4×10⁻⁸ T) = 7·2 V/m.

…

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$0^\circ$
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The propagation direction lies inside the plane of polarization, so the angle between the plane of polarization and the direction of propagation is $0^\circ$ — option (A).

Concept. An electromagnetic wave is transverse: the electric field $\vec E$ and magnetic field $\vec B$ oscillate perpendicular to the direction of propagation. Two planes are defined:

Plane of vibration — contains the electric-field vibration and the direction of propagation.

Plane of polarization — passes through the direction of propagation and is perpendicular to the plane of vibration (it carries no vibrations). …
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In an electromagnetic wave, the electric field vector E and the magnetic field vector B oscillate sinusoidally, perpendicular to each other, and both are perpendicular to the direction in which the wave travels. Since the oscillating quantities (E and B) vary in a direction perpendicular to the direction of propagation, electromagnetic waves are classif …

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›Reveal solutionSolution
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Key features of the nature of electromagnetic waves:

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E and B are mutually perpendicular to each other as well, and both oscillate in phase.

They require no material medium and can travel through vacuum, unlike mechanical waves (e.g. sound).

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For a plane e.m. wave travelling along, say, the $x$ -axis, with $\vec E$ along $y$ and $\vec B$ along $z$ :

$\hat{y}\times\hat{z} = \hat{x}$ …

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EM waves are transverse.

In an electromagnetic wave the electric field $\vec{E}$ and magnetic field $\vec{B}$ oscillate perpendicular to each other and both are perpendicular to the direction of propagation $\vec{k}$ :

$\vec{E} \perp \vec{B} \perp \vec{k}.$

…

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In an electromagnetic wave the electric field E, magnetic field B, and the direction of propagation are mutually perpendicular, forming a right-handed set. The direction of energy flow (Poynting vector S = (1/μ₀) E × B) gives the direction of propagation:

$\hat{k} \parallel (\vec{E} \times \vec{B})$ …

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Q.(a) Depict the variation of electric field ( $\vec{E}$ ) and magnetic field ( $\vec{B}$ ) with respect to the direction of propagation of an electromagnetic wave. Write their two important characteristics.

Concept understanding — Transverse Nature of Electromagnetic Waves

(a) Variation of $\vec{E}$ and $\vec{B}$ with direction of propagation

(b) Showing that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ gives the speed of EM waves in free space

Concept and Intuition

Step-by-Step Reasoning

Q.(a) Depict the variation of electric field (E⃗\vec{E}E) and magnetic field (B⃗\vec{B}B) with respect to the direction of propagation of an electromagnetic wave. Write their two important characteristics.

(a) Variation of E⃗\vec{E}E and B⃗\vec{B}B with direction of propagation

(b) Showing that 1μ0ε0\frac{1}{\sqrt{\mu_0\varepsilon_0}}μ0​ε0​​1​ gives the speed of EM waves in free space

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Q.(a) Depict the variation of electric field ( $\vec{E}$ ) and magnetic field ( $\vec{B}$ ) with respect to the direction of propagation of an electromagnetic wave. Write their two important characteristics.

(a) Variation of $\vec{E}$ and $\vec{B}$ with direction of propagation

(b) Showing that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ gives the speed of EM waves in free space

Unlock everything free for 14 days