An EM wave radiates outwards from a dipole antenna, with $E_0$ as the amplitude of its electric field vector. The electric field $E_0$ which transports significant energy from the source falls off as (a) $\dfrac{1}{r^3}$ (b) $\dfrac{1}{r^2}$ (c) $\dfrac{1}{r}$ (d) remains constant.

An EM wave radiates outwards from a dipole antenna, with E₀ as the amplitude of its electric field vector. The

NCERT Exemplar · Q7

E_0

(a)

\dfrac{1}{r^3}

(b)

\dfrac{1}{r^2}

(c)

\dfrac{1}{r}

(d) remains constant.

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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 electric field amplitude $E_0$ of an electromagnetic wave from a dipole antenna falls off as $1/r$ in the far-field (radiation) zone, because the Poynting vector (power per unit area) must conserve energy flux through an expanding spherical surface.

The key insight here is that an antenna radiates energy outward, and that energy spreads over an ever-larger sphere as it travels. For the wave to carry significant energy to large distances, the field cannot drop too quickly — otherwise, the power would vanish before reaching a receiver.

1. Why the far-field matters

Close to the antenna (the "near-field" region), the fields are complicated — they include static-like components that fall off as $1/r^2$ or $1/r^3$ . These store energy locally but don't radiate it away. The part that actually transports energy to distant points is the radiation field, which dominates only when $r$ is much larger than the wavelength $\lambda$ and the antenna size.

Watch out

A common mistake is to think the field from any source always falls as $1/r^2$ (like Coulomb's law). That's true for static fields, but radiated fields behave differently — they must fall more slowly to carry energy far away.

2. Energy flow and the Poynting vector

The power per unit area carried by an EM wave is given by the magnitude of the Poynting vector:

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

For a plane wave in vacuum, $B = E/c$ , so the time-averaged power per unit area is proportional to $E_0^2$ :

\langle S \rangle = \frac{1}{2} \varepsilon_0 c \, E_0^2

This is the intensity $I$ of the wave.

3. Conservation of energy through a sphere

Imagine a dipole antenna at the origin radiating total power $P$ equally in all directions (isotropic approximation for simplicity). At a distance $r$ , this power spreads uniformly over a sphere of surface area $4\pi r^2$ .

The intensity $I(r)$ at distance $r$ must satisfy:

I(r) \times 4\pi r^2 = P

Since $P$ is constant (energy is conserved, ignoring absorption), we get:

I(r) \propto \frac{1}{r^2}

--- …

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Q.An EM wave radiates outwards from a dipole antenna, with $E_0$ as the amplitude of its electric field vector. The electric field $E_0$ which transports significant energy from the source falls off as

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 $\mathbf{B}$

1. Why the far-field matters

2. Energy flow and the Poynting vector

3. Conservation of energy through a sphere

Method: Finding How a Radiated Field Amplitude Falls Off with Distance

Steps

Q.An EM wave radiates outwards from a dipole antenna, with E0E_0E0​ as the amplitude of its electric field vector. The electric field E0E_0E0​ which transports significant energy from the source falls off as

1. Why the far-field matters

2. Energy flow and the Poynting vector

3. Conservation of energy through a sphere

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Q.An EM wave radiates outwards from a dipole antenna, with $E_0$ as the amplitude of its electric field vector. The electric field $E_0$ which transports significant energy from the source falls off as

Unlock everything free for 14 days