Q.Suppose that the electric field amplitude of an electromagnetic wave is E0=120 N/C and that its frequency is ν=50.0 MHz.
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🔒 Start your 14-day free trial to unlock the full solution →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=fλ
where v is the wave speed, f is the frequency (in hertz, Hz), and λ (lambda) is the wavelength (in metres).
For electromagnetic waves in vacuum, this speed is a universal constant: c=3×108 m/s. So the relation becomes:
c=fλ
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 fλ 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: λ=fc or f=λc.
- 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=λhc.
A common mistake: using c=fλ 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=fλ 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=2.45×109 Hz, c=3×108 m/s. …
Why this formula?
Electromagnetic Wave Relation: Why c=μ0ε01
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: ∇⋅E=0
- Gauss's law for magnetism: ∇⋅B=0
- Faraday's law: ∇×E=−∂t∂B
- Ampère-Maxwell law: ∇×B=μ0ε0∂t∂E
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 E
Take the curl of Faraday's law:
∇×(∇×E)=∇×(−∂t∂B)=−∂t∂(∇×B)
Now use the vector identity: ∇×(∇×E)=∇(∇⋅E)−∇2E
Since ∇⋅E=0 in vacuum, this becomes:
−∇2E=−∂t∂(∇×B)
Substitute ∇×B from Ampère-Maxwell:
−∇2E=−∂t∂(μ0ε0∂t∂E)
Result: The electric field satisfies the wave equation:
∇2E=μ0ε0∂t2∂2E
3. Identifying the Wave Speed
Compare with the standard wave equation for any wave travelling at speed v:
∇2ψ=v21∂t2∂2ψ
Matching terms:
v21=μ0ε0⇒v=μ0ε01
This v is the speed of electromagnetic waves in vacuum — denoted c.
Why this is profound: The constants μ0 (permeability of free space) and ε0 (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:
∇2B=μ0ε0∂t2∂2B
So both E and B propagate at the same speed c.
5. The Crucial Relationship Between E and B
For a plane wave travelling in the x-direction:
- E oscillates along y: Ey=E0sin(kx−ωt)
- B oscillates along z: Bz=B0sin(kx−ωt)
From Faraday's law: ∂x∂Ey=−∂t∂Bz
Differentiating the wave forms:
kE0cos(kx−ωt)=ωB0cos(kx−ωt)
Since ω=ck, we get:
B0E0=kω=c …
Concept: Electromagnetic Wave Relation — in a vacuum, E and B are linked by c, and the wave parameters ω, k, λ follow from frequency.
(a)
Speed of light c=3.00×108 m/s.
Magnetic amplitude:
B0=cE0=3.00×108120=4.00×10−7 T
Angular frequency:
ω=2πν=2π(50.0×106)=3.14×108 rad/s
Wavelength:
λ=νc=50.0×1063.00×108=6.00 m
Wave number:
k=λ2π=6.002π=1.05 rad/m
(b)
Assuming propagation along +x with E along y and B along z:
E(x,t)=(120 N/C)sin(1.05x−3.14×108t)j^ …
For an electromagnetic wave in vacuum, the magnetic amplitude is B0=E0/c, the angular frequency is ω=2πν, the wave number is k=ω/c, and the wavelength is λ=c/ν. Here B0=4.00×10−7 T, ω=3.14×108 rad/s, k=1.05 rad/m, λ=6.00 m, and the wave travels along x with E along y and B along z.
The core idea is that in an electromagnetic wave, the electric and magnetic fields are intimately linked through Maxwell’s equations. In free space, the wave speed is c=3.00×108 m/s, and the amplitudes satisfy E0=cB0. The frequency ν tells us how fast the fields oscillate, and from it we get the angular frequency ω=2πν. The wavelength λ follows from λ=c/ν, and the wave number k=2π/λ. Once we have these parameters, we can write the full expressions for E and B as sinusoidal waves, choosing a direction of propagation (say +x) and mutually perpendicular polarization directions.
Let’s work through each part.
- Magnetic amplitude B0 The relation between the peak electric and magnetic fields in vacuum is E0=cB0. So
B0=cE0=3.00×108 m/s120 N/C=4.00×10−7 T.
This is a very small field — typical for radio waves.
- Angular frequency ω ω=2πν. With ν=50.0 MHz=50.0×106 Hz,
ω=2π(50.0×106)=3.14×108 rad/s.
(Using π≈3.14.)
- Wavelength λ For any wave, λ=c/ν.
λ=50.0×1063.00×108=6.00 m.
This is in the radio band — about the length of a car.
- Wave number k k=2π/λ, so
k=6.002π=1.05 rad/m.
Alternatively, k=ω/c gives the same: 3.14×108/3.00×108=1.05 rad/m.
You can always check consistency: kλ=2π and ω=kc must hold. Here 1.05×6.00=6.30≈2π, good enough.
Now for part (b): we need expressions for E and B. We must choose a direction of propagation and orientations for the fields. The standard choice: let the wave travel along the +x axis. Then E and B are perpendicular to each other and to the direction of propagation. A common convention is to take E along the y-axis and B along the z-axis. The wave is sinusoidal, so we write:
E(x,t)=E0sin(kx−ωt)j^,
B(x,t)=B0sin(kx−ωt)k^. …
Method: Standard Wave Parameter Extraction from Field Amplitude and Frequency
This method uses the fundamental relations of electromagnetic waves in vacuum — the wave equation, Maxwell’s speed relation, and the angular frequency/wave number definitions.
Step 1: Identify given data
- E0=120 N/C
- ν=50.0 MHz=50.0×106 Hz
- Speed of light in vacuum: c=3.00×108 m/s
Step 2: Find B0 (magnetic field amplitude)
From Maxwell’s relation for EM waves in vacuum:
E0=cB0
So:
B0=cE0=3.00×108120
B0=4.00×10−7 T
Step 3: Find ω (angular frequency)
Angular frequency is related to ordinary frequency by:
ω=2πν
ω=2π(50.0×106)
ω=3.14×108 rad/s
Step 4: Find k (wave number)
Wave number is related to angular frequency and speed:
k=cω
k=3.00×1083.14×108
k=1.05 rad/m
Step 5: Find λ (wavelength)
Wavelength from frequency:
λ=νc
λ=50.0×1063.00×108 …
Common Mistakes & How to Avoid Them
Mistake 1: Using c=3×108 without checking units
The error: Students plug c=3×108 directly, forgetting that frequency is given in MHz, not Hz.
How to avoid: Always convert to SI units first.
- ν=50.0 MHz=50.0×106 Hz
- Then use c=3.00×108 m/s
Mistake 2: Confusing B0 formula sign or missing the factor
The error: Using B0=E0×c instead of B0=cE0.
How to avoid: Remember the fundamental relation:
c=B0E0⇒B0=cE0
Correct calculation:
B0=3.00×108120=4.00×10−7 T
Mistake 3: Mixing up ω and ν (angular vs ordinary frequency)
The error: Writing ω=ν or ω=ν1.
How to avoid: Memorise the exact relation:
ω=2πν
Correct:
ω=2π(50.0×106)=3.14×108 rad/s
Mistake 4: Using wrong formula for k (wave number)
The error: Using k=ν2π or k=cω incorrectly.
How to avoid: Two equivalent correct formulas:
k=λ2πork=cω
First find λ:
λ=νc=50.0×1063.00×108=6.00 m
Then:
k=6.002π=1.05 rad/m
Mistake 5: Writing E and B expressions with wrong phase or direction
The error: Writing E and B as if they are in the same direction, or forgetting the k^ (direction of propagation).
How to avoid: Remember the right-hand rule:
- E×B gives the direction of propagation
- For a wave travelling along +x:
- If E is along j^ (say y-direction) …
Showing the 12 most recent of 32 on this concept.
- CBSE 2026Set 55/3/11 markMCQQ.A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by B=5×10−8sin(3×1010t−150x)T, where x is in metres and t is in seconds. The velocity of the wave is : (A) 2.0×108 ms−1 (B) 4.5×107 ms−1 (C) 3.5×107 ms−1 (D) 2.5×108 ms−1
›Reveal solutionSolution
The wave velocity is found from the ratio ω/k in the given sinusoidal form. Here ω=3×1010 rad/s and k=150 rad/m, giving v=ω/k=2.0×108 m/s. The correct option is (A).
The magnetic field is given as B=5×10−8sin(3×1010t−150x) T. This is a standard travelling wave expression of the form B=B0sin(ωt−kx), where ω is the angular frequency and k is the wave number. For any wave, the phase velocity is v=ω/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 ω and k from the equation.
The term multiplying t is ω=3×1010 rad/s.
The term multiplying x is k=150 rad/m.
-
Apply the wave velocity formula:
v=kω=1503×1010
- Simplify: v=1.5×1023×1010=2×108 m/s …
-
- CBSE 2026Set A1 markMCQQ.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 uB=2μ0B2.
Apart from the numerical factor ½, the quantity μ0B2 therefore has the dimensions of energy density (J/m³).
…
- CBSE 2026Set ANNUAL1 markMCQQ.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/μ0ε0, which numerically matches the measured speed of light.
Starting from Maxwell's equations in free space (no charges or currents), the wave equations for E and B both take the standard wave-equation form with wave speed v=1/μ0ε0. Substituting μ0=4π×10−7 T m/A and ε0=8.85×10−12 C2N−1m−2 gives v≈3×108 m/s, exactly the speed of light — this a …
- CBSE 2026Set ANNUAL1 markMCQQ.In a plane electromagnetic wave the electric field oscillates sinusoidally with a frequency 2.5×1010Hz and amplitude 480V/m. The amplitude of the oscillating magnetic field will be –(a) 1.52×10−8 weber/m2(b) 1.52×10−7 weber/m2(c) 1.6×10−6 weber/m2(d) 1.6×10−7 weber/m2
›Reveal solutionSolution
B0=E0/c.
In a plane electromagnetic wave, the amplitudes of the electric and magnetic fields are related by B0=cE0:
…
- CBSE 2025Set 55/4/11 markMCQQ.The amplitude of the electric field in an electromagnetic wave in free space is 1000 Vm−1. The amplitude of the magnetic field in this electromagnetic wave is: (A) 3.0×10−3 T (B) 3.33×10−8 T (C) 3.0×1011 T (D) 3.33×10−6 T
›Reveal solutionSolution
In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light: E0=cB0. With E0=1000 V/m, we find B0=3.33×10−6 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:
E0=cB0
where E0 is the amplitude of the electric field, B0 is the amplitude of the magnetic field, and c=3×108 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:
E0=cB0
- Rearrange to solve for B0:
B0=cE0
- Substitute the given values: …
- CBSE 2025Set ANNUAL1 markMCQQ.A plane electromagnetic wave travels in free space along x-direction. At a point in space and time, E=9.0j^ V/m. Magnitude of B at this point is-(a) 3×10−8 T(b) 9×10−8 T(c) 27×109 T(d) 2.1×10−8 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=cE …
- CBSE 2025Set D1 markMCQQ.Unit of √(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
Z0=ε0μ0
is the impedance of free space. Numerically
…
- CBSE 2025Set ANNUAL1 markQ.The magnetic field in a plane electromagnetic wave is given by By=2×10−7sin(0.5×103x+1.5×1011t) T. Find the wavelength of the given electromagnetic wave.
›Reveal solutionSolution
λ=2π/k, with k=0.5×103 rad/m read off the wave equation.
Comparing By=2×10−7sin(0.5×103x+1.5×1011t) with the standard form By=B0sin(kx+ωt), the wave number is k=0.5×103 rad/m. The wavelength is
…
- CBSE 2024Set A1 markMCQQ.The value of (μ₀ε₀)^-1/2 is (A) 3 × 10^8 cm/second (B) 3 × 10^10 cm/second (C) 3 × 10^9 cm/second (D) 3 × 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=μ0ε01=3×108 m/s
…
- CBSE 2024Set ANNUAL1 markMCQQ.If the amplitude of the magnetic field is 3×10−6 T, then the amplitude of the electric field for a electromagnetic wave is :(a) 600 Vm−1(b) 100 Vm−1(c) 900 Vm−1(d) 300 Vm−1
›Reveal solutionSolution
Using E0=cB0 with c=3×108 ms−1 and B0=3×10−6 T gives E0=900 Vm−1.
Working
For an electromagnetic wave travelling in free space, the peak electric and magnetic field amplitudes are related by
E0=cB0
Given B0=3×10−6 T: …
- CBSE 2023Set TERM21 markMCQQ.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: E0=cB0.
For a plane electromagnetic wave travelling in vacuum, Maxwell's equations require the electric and magnetic field amplitudes to be related by
E0=cB0
where c=3×108 m/s is the speed of light in vacuum.
…
- CBSE 2023Set ANNUAL1 markMCQQ.If the magnitude of the magnetic field is 3×10−6 T, then the magnitude of the electric field for a electromagnetic wave is :(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=3×108 m/s and B=3×10−6 T gives E=900 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=3×108 ms−1 is the speed of light.
…
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