The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM

NCERT Exemplar · Q6

Q. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

(a)

c : 1

(b)

c^2 : 1

(c)

1 : 1

(d)

\sqrt{c} : 1

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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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In an electromagnetic wave, the electric and magnetic fields contribute equally to the intensity. The ratio of their contributions is 1 : 1.

The key to this question lies in understanding how energy is stored in an electromagnetic wave. An EM wave carries energy in both its electric field and its magnetic field. The intensity of the wave — the power per unit area — is the rate at which this energy flows.

The energy per unit volume (energy density) associated with the electric field is given by $u_E = \frac{1}{2} \varepsilon_0 E^2$ , and for the magnetic field it is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$ .

For an electromagnetic wave in vacuum, the magnitudes of $E$ and $B$ are not independent. They are linked by the fundamental relation $E = cB$ , where $c$ is the speed of light. This relation is the direct consequence of Maxwell's equations and is the physical reason the two contributions balance perfectly.

Let's see what happens when we substitute this relation into the expression for magnetic energy density.

Write the magnetic energy density. We start with $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$ .
Use the EM wave relation. We know that for an EM wave, $E = cB$ , which means $B = E/c$ . Substitute this into the expression for $u_B$ :

u_B = \frac{1}{2} \frac{(E/c)^2}{\mu_0} = \frac{1}{2} \frac{E^2}{\mu_0 c^2}

Recall the speed of light. The speed of light in vacuum is defined by the constants of electromagnetism: $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ . Squaring this gives $c^2 = \frac{1}{\mu_0 \varepsilon_0}$ .
Substitute $c^2$ into $u_B$ . Replacing $c^2$ in the denominator of $u_B$ :

u_B = \frac{1}{2} \frac{E^2}{\mu_0 \cdot \frac{1}{\mu_0 \varepsilon_0}} = \frac{1}{2} \frac{E^2}{\frac{1}{\varepsilon_0}} = \frac{1}{2} \varepsilon_0 E^2

Compare the two energy densities. The result $u_B = \frac{1}{2} \varepsilon_0 E^2$ is exactly the expression for the electric energy density $u_E$ . …

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Q.The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

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 $\mathbf{B}$

Method: Comparing the Electric- and Magnetic-Field Contributions to EM Wave Intensity

Steps

Q.The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

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