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

GUJCET 2026Set x1 markMCQ
$8 \times 10^9$
›Reveal solutionSolution
[!TLDR]

Using the NCERT/CBSE list of insect-resistant crop varieties: $X$ = flat bean, $Y$ = Pusa A-4, $Z$ = Aphids.

Concept

Under 'Strategies for Enhancement in Food Production', plant breeders release crop varieties resistant to specific insect pests. A standard table pairs each crop with its released variety and the pest it resists. This GSEB/NCERT-aligned question tests recall of three of those pairings.

Solution

Row (i): Rape-seed mustard, variety Pusa Gaurav. This mustard variety is bred for resistance to the mustard aphid, so $Z$ = Aphids.

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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
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(b) 153 V/m
(c) 135 V/m
(d) 170 V/m
›Reveal solutionSolution

In an electromagnetic wave, the electric and magnetic field amplitudes are related by E_0 = c B_0, where c is the speed of light.

Given B_0 = 510 nT = 510 x 10^-9 T, c = 3 x 10^8 m/s.

…

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GUJCET 2023Set 091 markMCQ
$\vec{E}$
›Reveal solutionSolution

The Poynting direction $\vec E\times\vec B$ gives the wave's propagation direction.

Concept: In an electromagnetic wave, $\vec E$ , $\vec B$ , and the propagation direction form a right-handed triad, with p …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. For a given electromagnetic waves the magnitude of electric field is 6.6 V/m at a point in space. The magnitude of magnetic field at this point is ___ T.$
(a) 2.1 x 10^-8
(b) 6.6 x 10^-8
(c) 19.8 x 10^-8
(d) 2.2 x 10^-8
›Reveal solutionSolution

In an EM wave the field magnitudes obey E = cB, so B = E/c = 6.6/(3x10^8) = 2.2 x 10^-8 T.

…

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GUJCET 2022Set 171 markMCQ
$c = 3 \times 10^8$
›Reveal solutionSolution
$\lambda = c/f$ ; higher frequency gives shorter wavelength.

Steps.

At $f = 6$ MHz: $\lambda = \dfrac{3\times10^8}{6\times10^6} = 50$ m. …
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GUJCET 2022Set 171 markMCQ
$10^9$
›Reveal solutionSolution

An oscillating charge radiates EM waves at exactly its own oscillation frequency.

Concept. An accelerating/oscillating charge produces electromagnetic waves whose frequency equals the frequency of oscillation of the charge. …
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GUJCET 2021Set 151 markMCQ
$\vec{B} = 2.1 \times 10^{-8}\,\hat{k}$
›Reveal solutionSolution

For an EM wave $E=cB$ , with $\vec E,\vec B$ and the propagation direction mutually perpendicular ( $\vec E\times\vec B$ points along propagation).

Concept. $E=cB$ and $\hat E\times\hat B=\hat{\text{propagation}}$ . …
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GUJCET 2019Set 131 markMCQ
$E$
›Reveal solutionSolution

Far from the source, radiated $E$ and $B$ are in phase and decrease as $\dfrac{1}{r}$ .

Concept: The radiation (far) field of an accelerating charge dominates at large distances because it decays only as $1/r$ , unlike the static ( $1/r^2$ ) or induction ( $1/r^3$ ) terms. This slow decay is why radiated energy reaches far away.

Steps: …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. The maximum value of E in an electromagnetic wave is equal to 1.8 Vm^-1. Thus the maximum value of B is ___.$
(a) 6 x 10^-8 T
(b) 3 x 10^-6 T
(c) 6 x 10^-9 T
(d) 2 x 10^-10 T
›Reveal solutionSolution

In an electromagnetic wave, the peak electric and magnetic fields are related by $B_0=E_0/c$ .

Given $E_0=1.8$ V/m, $c=3\times10^8$ m/s.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. For a radiation of 6 GHz passing through air, the wave number (number of waves) per 1 m length is ___ (1 GHz = 10^9 Hz).$
(a) 5
(b) 3
(c) 20
(d) 30
›Reveal solutionSolution

The number of complete waves per unit length (wave number in this sense) equals $f/c$ , the reciprocal of wavelength.

Given $f=6$ GHz $=6\times10^9$ Hz, $c=3\times10^8$ m/s.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. In the region closer to the oscillating charges, the phase difference between E (vector) and B (vector) fields is ___ and their magnitude quickly decreases as ___ with distance r from the source.$
(a) 0, r^-1
(b) pi/2, r^-3
(c) pi/2, r^-1
(d) 0, r^-3
›Reveal solutionSolution
Close to the oscillating charges (the near field), E and B are pi/2 out of phase and their amplitudes decrease steeply, as r^-3.

Near an oscillating charge (the induction/near-field zone, distances small compared with the wavelength):

The electric and magnetic fields are out of phase by pi/2 (90 degree). …
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GUJCET 2015Set C1 markMCQ
$Q. To transmit a signal of 3 KHz frequency, the minimum length of antenna is _____ km (A) 25 (B) 20 (C) 50 (D) 75$
›Reveal solutionSolution
[!TLDR]

$\lambda = c/f = 100$ km; minimum antenna length $= \lambda/4 = 25$ km. Answer: (A).

Concept

To radiate a signal efficiently, an antenna should have a length of at least about a quarter of the signal wavelength, $L_{min} = \lambda/4$ , where $\lambda = c/f$ (NCERT/CBSE communication systems).

Solution

Wavelength of the 3 kHz signal: …
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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

Concept

Solution

Concept

Solution

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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