A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\ \text{Hz}$. The electromagnetic waves produced: (a) will have frequency of $10^9$ Hz. (b) will have frequency of $2 \times 10^9$ Hz. (c) will have a wavelength of 0.3 m. (d) fall in the region of radiowaves.

A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹Hz. The electromagnet

NCERT Exemplar · Q30

10^9\ \text{Hz}

(a) will have frequency of

10^9

Hz.

(b) will have frequency of

2 \times 10^9

Hz.

(d) fall in the region of radiowaves.

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Concept understanding — Speed Of Light

Speed of Light

What It Is

The speed of light is the speed at which light — and every other electromagnetic wave — travels through empty space (vacuum). It is one of the most important constants in physics, denoted $c$ :

c \approx 3 \times 10^8 \ \text{m/s} = 3\times10^{5}\ \text{km/s}

More precisely $c = 2.998 \times 10^8\ \text{m/s}$ . In one second light travels about 300,000 km — roughly seven and a half times around the Earth.

Where the Value Comes From (Maxwell)

The speed of light is not just measured; it is predicted by Maxwell's equations. When Maxwell combined his laws of electricity and magnetism, he found that electromagnetic waves must travel through vacuum at a speed fixed entirely by two constants of free space:

c = \frac{1}{\sqrt{\mu_0\,\varepsilon_0}}

where

$\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, and
$\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space.

Plugging in these numbers gives $c \approx 3\times10^8$ m/s — matching the measured speed of light. This agreement was the decisive clue that light itself is an electromagnetic wave.

Key Properties

Same for all electromagnetic waves. Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays all travel at $c$ in vacuum, regardless of their frequency or wavelength.
Universal constant. In vacuum, $c$ is the same for every observer and does not depend on the motion of the source — the starting postulate of Einstein's special relativity.
The cosmic speed limit. No material object or signal carrying information can travel faster than $c$ .
Links wavelength and frequency. For any EM wave in vacuum,

c = f\lambda

so a high-frequency wave has a short wavelength and vice versa.

Speed of Light in a Medium

Inside a transparent material (glass, water, etc.) light slows down. Its speed becomes

v = \frac{c}{n} = \frac{1}{\sqrt{\mu\,\varepsilon}}

where $n = \sqrt{\mu_r\varepsilon_r}$ is the refractive index of the medium and is always greater than 1. For example, in water $n \approx 1.33$ , so light travels at about $2.25\times10^8$ m/s. The frequency stays the same, but the wavelength shortens because $v = f\lambda$ . …

Why this formula?

Speed of Light: Why the Formula Holds

The speed of light ( $c$ ) is not just a number — it emerges from the fundamental laws of electricity and magnetism. Let's understand why its value is fixed and where the formula comes from.

1. The Core Formula

The speed of light in vacuum is given by:

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

Where:

$\mu_0$ = permeability of free space (how easily a magnetic field forms)
$\varepsilon_0$ = permittivity of free space (how easily an electric field forms)

2. Why This Formula? — The Derivation

Step 1: Maxwell's Equations

James Clerk Maxwell unified electricity and magnetism into four equations. Two key ones for light:

Faraday's Law: A changing magnetic field creates an electric field

\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}

Ampère's Law (with Maxwell's correction): A changing electric field creates a magnetic field

\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}

Step 2: The Wave Emerges

Take the curl of Faraday's Law:

\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Using the vector identity $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$ and noting that in vacuum $\nabla \cdot \mathbf{E} = 0$ , we get:

-\nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Now substitute Ampère's Law for $\nabla \times \mathbf{B}$ :

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

Step 3: The Wave Equation

This simplifies to:

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

This is the wave equation. For any wave, the general form is:

\nabla^2 \mathbf{E} = \frac{1}{v^2} \frac{\partial^2 \mathbf{E}}{\partial t^2}

Comparing the two, the wave speed $v$ must satisfy:

\frac{1}{v^2} = \mu_0 \varepsilon_0

Hence:

v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

This $v$ is the speed of light — denoted $c$ .

3. Why Is It Constant?

$\mu_0$ and $\varepsilon_0$ are fundamental constants of nature — they don't depend on the observer or the source.
Therefore, $c$ is also a universal constant. …

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An oscillating charge radiates EM waves at exactly its own oscillation frequency, so with $f=10^9$ Hz: (a) is true, (b) [doubled frequency] is false, (c) [ $\lambda=0.3$ m] is true, and (d) [radio-wave region] is true. The correct options are (a), (c), and (d).

The governing principle

An oscillating charged particle is an accelerating charge, and Maxwell's theory shows that an accelerating charge radiates electromagnetic waves at exactly the same frequency as its own oscillation -- the emitted wave's electric and magnetic fields vary in step with the source's motion. There is no classical mechanism by which a simply-oscillating dipole charge would radiate at double (or any other multiple of) its own frequency.

Checking each option

(a) Frequency $=10^9$ Hz. Since the emitted wave's frequency equals the source's oscillation frequency, $f=10^9$ Hz. TRUE.

(b) Frequency $=2\times10^9$ Hz. This would require a frequency-doubling mechanism (e.g. a nonlinear process), which doesn't apply to a simple oscillating charge. FALSE.

(c) Wavelength $=0.3$ m. In vacuum, $c=f\lambda$ , so

\lambda=\frac{c}{f}=\frac{3\times10^8}{10^9}=0.3\ \text{m}

TRUE. …

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

COMEDK 2026Set 2026-A1 markMCQ
$\mu_0$
›Reveal solutionSolution
The quantity $(\mu_0 \varepsilon_0)^{1/2}$ has the dimension of inverse velocity, specifically $[L^{-1}T]$ , which corresponds to option (C).

The key insight is that $\mu_0$ and $\varepsilon_0$ are not arbitrary constants — they are deeply linked to the speed of light in vacuum. In electromagnetism, the wave equation for light emerges from Maxwell’s equations, and the speed of light $c$ is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ . Therefore, $\sqrt{\mu_0 \varepsilon_0}$ is simply $1/c$ . Since velocity has dimension $[L T^{-1}]$ , its reciprocal has dimension $[L^{-1} T]$ . No need to derive dimensions from scratch — just remember this fundamental relation.

Let’s walk through it carefully:

Recall the relation between $\mu_0$ , $\varepsilon_0$ , and the speed of light From Maxwell’s equations in vacuum, the speed of electromagnetic waves is

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}.$
This is a standard result — it tells us that $\sqrt{\mu_0 \varepsilon_0} = \frac{1}{c}$ .

Find the dimension of $c$ Speed is distance per time, so

$[c] = [L T^{-1}].$

Take the reciprocal Since $\sqrt{\mu_0 \varepsilon_0} = 1/c$ , its dimension is the inverse of the dimension of $c$ :

$\left[ \sqrt{\mu_0 \varepsilon_0} \right] = \frac{1}{[L T^{-1}]} = [L^{-1} T].$

Match with the options The dimension $[L^{-1} T]$ is exactly option (C). …
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KCET 2022Set B-31 markMCQ
$R=300\,\Omega$
›Reveal solutionSolution
The impedance of a series LCR circuit is the vector sum of resistance and net reactance. For the given values, the impedance comes out to $500\,\Omega$ , so option (A) is correct.

The key idea is that in a series LCR circuit, the total opposition to current is not just the resistance $R$ — the inductor and capacitor each oppose current in a frequency-dependent way, called reactance. Inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = 1/(\omega C)$ . Because the voltage across $L$ leads the current by $90^\circ$ and across $C$ lags by $90^\circ$ , these two reactances oppose each other. The net reactance is $|X_L - X_C|$ , and the total impedance $Z$ is found by the Pythagorean sum: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ .

Let’s work through the numbers step by step.

Calculate inductive reactance $X_L$

$X_L = \omega L = 1000 \times 0.9 = 900\,\Omega$

Calculate capacitive reactance $X_C$

$C = 2.0\,\mu\text{F} = 2.0 \times 10^{-6}\,\text{F}$

$X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2.0 \times 10^{-6}} = \frac{1}{2.0 \times 10^{-3}} = 500\,\Omega$

Find the net reactance

Since $X_L > X_C$ , the circuit is inductive overall:

$X_L - X_C = 900 - 500 = 400\,\Omega$

Compute the impedance $Z$

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{300^2 + 400^2}$ …
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KCET 2021Set B-21 markMCQ
$E_0 = 120~\text{NC}^{-1}$
›Reveal solutionSolution

Compute $B_0$ , $\omega$ , $\lambda$ and $k$ from $E_0=120\ \mathrm{N\,C^{-1}}$ and $f=50\ \mathrm{MHz}$ — three match the options, but the wave number comes out $1.05\ \mathrm{rad/m}$ , so the option quoting $2.1$ is the wrong one.

Given
$E_0 = 120\ \mathrm{N\,C^{-1}}, \qquad f = 50\ \mathrm{MHz} = 5\times10^{7}\ \mathrm{Hz}, \qquad c = 3\times10^{8}\ \mathrm{m\,s^{-1}}$
Note the question asks which value is INCORRECTLY computed — so we must verify all four and find the odd one out.

1. Check (A) — magnetic field amplitude

In an electromagnetic wave in vacuum, the two field amplitudes are locked together by
$c = \frac{E_0}{B_0} \qquad\Longrightarrow\qquad B_0 = \frac{E_0}{c}$ $B_0 = \frac{120}{3\times10^{8}} = 40\times10^{-8} = 4\times10^{-7}\ \mathrm{T} = 400\times10^{-9}\ \mathrm{T} = \mathbf{400\ nT}$
(A) is CORRECT ✓

2. Check (B) — angular frequency
$\omega = 2\pi f = 2\pi \times 5\times10^{7} = 10\pi \times 10^{7} = \boldsymbol{\pi \times 10^{8}\ \mathrm{rad/s}}$
(B) is CORRECT ✓

3. Check (D) — wavelength
$\lambda = \frac{c}{f} = \frac{3\times10^{8}}{5\times10^{7}} = \mathbf{6\ m}$
(D) is CORRECT ✓

4. Check (C) — propagation constant (angular wave number)
$k = \frac{2\pi}{\lambda} = \frac{\omega}{c}$ $k = \frac{2\pi}{6} = \frac{\pi}{3} = \frac{3.1416}{3} \approx \mathbf{1.05\ rad/m}$
(Cross-check via $\omega/c$ : $k = \dfrac{\pi\times10^{8}}{3\times10^{8}} = \dfrac{\pi}{3} \approx 1.05\ \mathrm{rad/m}$ ✓ — same answer, so the arithmetic is sound.) …
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Q.A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\ \text{Hz}$ . The electromagnetic waves produced:

Concept understanding — Speed Of Light

Speed of Light

What It Is

Where the Value Comes From (Maxwell)

Key Properties

Speed of Light in a Medium

Why this formula?

Speed of Light: Why the Formula Holds

1. The Core Formula

2. Why This Formula? — The Derivation

Step 1: Maxwell's Equations

Step 2: The Wave Emerges

Step 3: The Wave Equation

3. Why Is It Constant?

The governing principle

Checking each option

Method: Relating an Oscillating Source's Frequency to the Emitted EM Wave's Frequency and Wavelength

Steps

Given

1. Check (A) — magnetic field amplitude

2. Check (B) — angular frequency

3. Check (D) — wavelength

4. Check (C) — propagation constant (angular wave number)

Q.A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz10^9\ \text{Hz}109 Hz. The electromagnetic waves produced:

The governing principle

Checking each option

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Q.A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\ \text{Hz}$ . The electromagnetic waves produced:

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