A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}$, where $x$ is in metres and $t$ is in seconds. The velocity of the wave is : (A) $2.0\times10^{8}\ \text{ms}^{-1}$ (B) $4.5\times10^{7}\ \text{ms}^{-1}$ (C) $3.5\times10^{7}\ \text{ms}^{-1}$ (D) $2.5\times10^{8}\ \text{ms}^{-1}$

> [!TLDR] > The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is **(A)**. The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$, where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$. That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves. 1. Identify $\omega$ and $k$ from the equation. The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s. The term multiplying $x$ is $k = 150$ rad/m. 2. Apply the wave velocity formula: $$v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}$$ 3. Simplify: $$v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}$$ > [!WARNING] > A common mistake is to confuse $k$ with wavelength $\lambda$ or to use $v = f\lambda$ without first finding $f$ and $\lambda$ correctly. Here $f = \omega/(2\pi)$ and $\lambda = 2\pi/k$, so $v = f\lambda = (\omega/2\pi)(2\pi/k) = \omega/k$ — same result. But the direct ratio is faster and less error-prone. > [!TIP] > In an exam, whenever you see a wave written as $\sin(\omega t - kx)$ or $\cos(\omega t - kx)$, immediately read off $\omega$ and $k$ and compute $v = \omega/k$. This works for any sinusoidal wave — mechanical or electromagnetic. > [!ANSWER] > The velocity of the wave is $2.0\times10^{8}$ m/s, which corresponds to option **(A)**.

A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by B

Question

B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

✓ Free question

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.

\lambda = \frac{c}{f} = \frac{3 \times 10^8}{2.45 \times 10^9} = 0.122 \text{ m} = 12.2 \text{ cm}

That's why the mesh on a microwave door has holes about 1–2 mm across — much smaller than 12 cm — so microwaves can't escape, but visible light (wavelength ~500 nm) passes through easily.

The big picture

The electromagnetic wave relation $c = f \lambda$ is not a deep law of nature — it's a definitional consequence of what frequency and wavelength mean. But it's the single most useful tool for navigating the electromagnetic spectrum. Memorise it, understand it, and you'll be able to connect wave properties to energy, to colour, to radiation types, and to countless exam problems.

$c = f \lambda$ — that's the relation. Everything else is just applying it.

The relation c = fλ connecting frequency and wavelength across the electromagnetic spectrum is introduced in the NCERT Class 12 Physics chapter on electromagnetic waves, tested in CBSE boards, JEE Main and NEET. Anyone searching "electromagnetic spectrum frequency wavelength relation class 12 physics" will find this wave-speed reasoning, including the medium-versus-vacuum distinction, matches the NCERT treatment.

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

Key result: In vacuum, the magnitudes are related by:

\boxed{E = cB}

This means:

$\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other and to the direction of propagation
They are in phase (peaks and zeros occur together)
The electric field is $c$ times stronger than the magnetic field in SI units

6. Physical Intuition: Why This Speed?

Think of it this way:

$\mu_0$ measures how strongly a current creates a magnetic field
$\varepsilon_0$ measures how strongly a charge creates an electric field
Their product in the denominator means: the more "reluctant" space is to create fields, the slower the wave

If space were more "magnetic" (larger $\mu_0$ ) or more "electric" (larger $\varepsilon_0$ ), EM waves would travel slower. The actual value $c \approx 3 \times 10^8$ m/s emerges from the measured values of these constants.

Summary: The Core Relations

Quantity	Formula	Why
Wave speed	$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$	From wave equation derived from Maxwell's equations
Field ratio	$E = cB$	From Faraday's law applied to plane waves
Direction	$\mathbf{E} \perp \mathbf{B} \perp$ propagation	From cross-product structure of Maxwell's equations

Exam tip: Never just quote $c = 1/\sqrt{\mu_0\varepsilon_0}$ — be ready to show it comes from taking curls of Maxwell's equations and identifying the wave equation form.

The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is (A).

The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$ , where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$ . That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves.

Identify $\omega$ and $k$ from the equation.

The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s.

The term multiplying $x$ is $k = 150$ rad/m.
Apply the wave velocity formula:

v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}

Simplify:

v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}

Watch out

A common mistake is to confuse $k$ with wavelength $\lambda$ or to use $v = f\lambda$ without first finding $f$ and $\lambda$ correctly. Here $f = \omega/(2\pi)$ and $\lambda = 2\pi/k$ , so $v = f\lambda = (\omega/2\pi)(2\pi/k) = \omega/k$ — same result. But the direct ratio is faster and less error-prone.

Tip

In an exam, whenever you see a wave written as $\sin(\omega t - kx)$ or $\cos(\omega t - kx)$ , immediately read off $\omega$ and $k$ and compute $v = \omega/k$ . This works for any sinusoidal wave — mechanical or electromagnetic.

✓Final answer

2.0\times10^{8}

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

Showing the 12 most recent of 32 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}$
›Reveal solutionSolution
The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is (A).

The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$ , where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$ . That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves.

Identify $\omega$ and $k$ from the equation.

The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s.

The term multiplying $x$ is $k = 150$ rad/m.

Apply the wave velocity formula:

$v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}$

Simplify:

$v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}$
Watch out

A common mistake is to confuse $k$ with wavelength $\lambda$ or to use $v = f\lambda$ without first finding $f$ and $\lambda$ correctly. Here $f = \omega/(2\pi)$ and $\lambda = 2\pi/k$ , so $v = f\lambda = (\omega/2\pi)(2\pi/k) = \omega/k$ — same result. But the direct ratio is faster and less error-prone.

Tip

In an exam, whenever you see a wave written as $\sin(\omega t - kx)$ or $\cos(\omega t - kx)$ , immediately read off $\omega$ and $k$ and compute $v = \omega/k$ . This works for any sinusoidal wave — mechanical or electromagnetic.

✓Final answer
$2.0\times10^{8}$
CBSE 2026Set A1 markMCQ
$Q. The dimensions of B0^2/μ0 will be the same as that of (A) energy density (B) work (C) momentum (D) electric flux$
›Reveal solutionSolution

Magnetic energy density = B²/2μ₀, so B²/μ₀ carries the dimensions of energy density.

The energy stored per unit volume in a magnetic field is $u_B = \dfrac{B^2}{2\mu_0}$ .

Apart from the numerical factor ½, the quantity $\dfrac{B^2}{\mu_0}$ therefore has the dimensions of energy density (J/m³).

Checking: $[B] =$ T $=$ kg·s⁻²·A⁻¹, $[\mu_0] =$ kg·m·s⁻²·A⁻². Then $\dfrac{B^2}{\mu_0}$ = kg·m⁻¹·s⁻² = J/m³ = energy density. ✓

✓Final answer
$(A) energy density.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The velocity of electromagnetic waves in vacuum is:$
(a) c = 1/√(μ₀ε₀)
(b) c = √(μ₀ε₀)
(c) c = √(μ₀/ε₀)
(d) c = √(ε₀/μ₀)
›Reveal solutionSolution

Maxwell's equations predict electromagnetic waves travelling in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ , which numerically matches the measured speed of light.

Starting from Maxwell's equations in free space (no charges or currents), the wave equations for $\vec E$ and $\vec B$ both take the standard wave-equation form with wave speed $v=1/\sqrt{\mu_0\varepsilon_0}$ . Substituting $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$ and $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ gives $v\approx3\times10^{8}\ \text{m/s}$ , exactly the speed of light — this agreement is what led Maxwell to conclude that light itself is an electromagnetic wave. The other options mix up the reciprocal/square-root arrangement and do not have the right dimensions of speed.

✓Final answer
$c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$
CBSE 2026Set ANNUAL1 markMCQ
$2.5\times10^{10}Hz$
(a) $1.52\times10^{-8}\ weber/m^2$
(b) $1.52\times10^{-7}\ weber/m^2$
(c) $1.6\times10^{-6}\ weber/m^2$
(d) $1.6\times10^{-7}\ weber/m^2$
›Reveal solutionSolution

$B_0 = E_0/c$ .

In a plane electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $B_0 = \dfrac{E_0}{c}$ :

$B_0 = \frac{480}{3\times10^8} = 1.6\times10^{-6}\ \text{T} = 1.6\times10^{-6}\ \text{weber/m}^2$

✓Final answer
$1.6\times10^{-6}$
CBSE 2025Set 55/4/11 markMCQ
$1000$
›Reveal solutionSolution
In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light: $E_0 = c B_0$ . With $E_0 = 1000$ V/m, we find $B_0 = 3.33 \times 10^{-6}$ T.

Why the fields are linked by $c$

An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that propagate together through space. Maxwell's equations demand a precise relationship between these two fields: at every instant and every point in the wave, the ratio of the electric field amplitude to the magnetic field amplitude equals the speed of light in that medium.

In free space, this relationship is beautifully simple:
$E_0 = c B_0$
where $E_0$ is the amplitude of the electric field, $B_0$ is the amplitude of the magnetic field, and $c = 3 \times 10^8$ m/s is the speed of light in vacuum. This isn't arbitrary—it emerges directly from the wave equations derived from Maxwell's laws. The electric and magnetic fields are perpendicular to each other and to the direction of propagation, oscillating in phase, with their amplitudes locked in this ratio.

Finding the magnetic field amplitude

We're given the electric field amplitude and need to find the magnetic field amplitude.

Write down the fundamental relationship:

$E_0 = c B_0$

Rearrange to solve for $B_0$ :

$B_0 = \frac{E_0}{c}$

Substitute the given values:

$E_0 = 1000$ V/m

$c = 3 \times 10^8$ m/s

$B_0 = \frac{1000}{3 \times 10^8}$

Simplify the fraction:

$B_0 = \frac{1000}{3 \times 10^8} = \frac{10^3}{3 \times 10^8} = \frac{1}{3} \times 10^{-5}$ $B_0 = 0.333... \times 10^{-5} = 3.33 \times 10^{-6} \text{ T}$
Tip

A quick sanity check: magnetic field amplitudes in EM waves are always much smaller than electric field amplitudes (numerically) because $c$ is such a large number. If you ever get $B_0 > E_0$ in magnitude, something has gone wrong.

✓Final answer
$3.33 \times 10^{-6}$
CBSE 2025Set ANNUAL1 markMCQ
$\vec{E} = 9.0\hat{j}$
(a) $3\times10^{-8}$ T
(b) $9\times10^{-8}$ T
(c) $27\times10^{9}$ T
(d) $2.1\times10^{-8}$ T
›Reveal solutionSolution

In a plane EM wave in free space, $B = E/c$ .

For a plane electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic fields are related by

$B = \frac{E}{c}$

Here $E = 9.0$ V/m and $c = 3\times10^{8}$ m/s, so

$B = \frac{9.0}{3\times10^{8}} = 3\times10^{-8}\ \text{T}$

✓Final answer
$3\times10^{-8}$
CBSE 2025Set D1 markMCQ
$Q. Unit of \sqrt(μ0/ε0) is (A) newton/coulomb (B) ohm (C) henry (D) farad$
›Reveal solutionSolution

√(μ₀/ε₀) has the dimensions of resistance; it is the intrinsic impedance of free space, about 377 ohm.

The ratio of the electric to magnetic field amplitude of an electromagnetic wave in vacuum is c = E/B, and the quantity

$Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$

is the impedance of free space. Numerically

$Z_0 = \sqrt{\frac{4\pi\times10^{-7}}{8.85\times10^{-12}}} \approx 377\,\Omega$

Since impedance is measured in ohms, the unit of √(μ₀/ε₀) is the ohm.

✓Final answer
$(B) ohm.$
CBSE 2025Set ANNUAL1 mark
$B_y = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\ T$
›Reveal solutionSolution

$\lambda = 2\pi/k$ , with $k=0.5\times10^3$ rad/m read off the wave equation.

Comparing $B_y = 2\times10^{-7}\sin(0.5\times10^3 x + 1.5\times10^{11}t)$ with the standard form $B_y = B_0\sin(kx+\omega t)$ , the wave number is $k = 0.5\times10^3\ \text{rad/m}$ . The wavelength is

$\lambda = \frac{2\pi}{k} = \frac{2\pi}{500} \approx 1.257\times10^{-2}\ \text{m} \approx 1.26\ \text{cm}$

✓Final answer
$\lambda \approx 1.26\times10^{-2}$
CBSE 2024Set A1 markMCQ
$Q. The value of (μ₀ε₀)^-1/2 is (A) 3 \times 10^8 cm/second (B) 3 \times 10^10 cm/second (C) 3 \times 10^9 cm/second (D) 3 \times 10^8 km/second$
›Reveal solutionSolution

1/√(μ₀ε₀) is the speed of light c = 3×10⁸ m/s = 3×10¹⁰ cm/s.

Maxwell showed the speed of electromagnetic waves in vacuum is

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3\times10^8\ \text{m/s}$

Converting to CGS: 3×10⁸ m/s × 100 cm/m = 3×10¹⁰ cm/s.

✓Final answer
$(B) 3 \times 10¹⁰ cm/second.$
CBSE 2024Set ANNUAL1 markMCQ
$3 \times 10^{-6}$
(a) 600 Vm $^{-1}$
(b) 100 Vm $^{-1}$
(c) 900 Vm $^{-1}$
(d) 300 Vm $^{-1}$
›Reveal solutionSolution

Using $E_0 = cB_0$ with $c = 3\times10^8\ \text{ms}^{-1}$ and $B_0 = 3\times10^{-6}$ T gives $E_0 = 900\ \text{Vm}^{-1}$ .

Working

For an electromagnetic wave travelling in free space, the peak electric and magnetic field amplitudes are related by

$E_0 = c B_0$

Given $B_0 = 3\times10^{-6}$ T:

$E_0 = (3\times10^{8})\times(3\times10^{-6}) = 9\times10^{2} = 900\ \text{Vm}^{-1}$

✓Final answer
$E_0 = 900\ \text{Vm}^{-1}$
CBSE 2023Set TERM21 markMCQ
$Q. Amplitude of the magnetic field part of a harmonic Electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave ?$
(a) 150 NC-1
(b) 160 NC-1
(c) 153 NC-1
(d) 163 NC-1
›Reveal solutionSolution

In vacuum the ratio of the electric-field amplitude to the magnetic-field amplitude of an EM wave equals the speed of light: $E_0 = cB_0$ .

For a plane electromagnetic wave travelling in vacuum, Maxwell's equations require the electric and magnetic field amplitudes to be related by

$E_0 = c\,B_0$

where $c = 3\times10^{8}\ \text{m/s}$ is the speed of light in vacuum.

Given $B_0 = 510\ \text{nT} = 510\times10^{-9}\ \text{T}$ :

$E_0 = (3\times10^{8})(510\times10^{-9}) = 153\ \text{NC}^{-1}$

✓Final answer
$153\ \text{NC}^{-1}$
CBSE 2023Set ANNUAL1 markMCQ
$3 \times 10^{-6}$
(a) 600 Vm $^{-1}$
(b) 100 Vm $^{-1}$
(c) 900 Vm $^{-1}$
(d) 300 Vm $^{-1}$
›Reveal solutionSolution

Using $E=cB$ for an electromagnetic wave with $c=3\times10^8\ \text{m/s}$ and $B=3\times10^{-6}$ T gives $E=900\ \text{Vm}^{-1}$ .

Working

For an electromagnetic wave travelling in free space (or air), the magnitudes of the electric and magnetic field amplitudes are related by

$E = cB$

where $c = 3\times10^8\ \text{ms}^{-1}$ is the speed of light.

Given $B = 3\times10^{-6}$ T:

$E = (3\times10^{8})\times(3\times10^{-6}) = 9\times10^{2} = 900\ \text{Vm}^{-1}$

✓Final answer
$E = 900\ \text{Vm}^{-1}$

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