Q.A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E=6.3 j^ V/m. What is B at this point?
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.
λ=fc=2.45×1093×108=0.122 m=12.2 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=fλ 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=fλ — 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=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
Key result: In vacuum, the magnitudes are related by:
E=cB
This means:
- E and 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:
- μ0 measures how strongly a current creates a magnetic field
- ε0 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 μ0) or more "electric" (larger ε0), EM waves would travel slower. The actual value c≈3×108 m/s emerges from the measured values of these constants.
Summary: The Core Relations
| Quantity | Formula | Why |
|---|---|---|
| Wave speed | c=μ0ε01 | From wave equation derived from Maxwell's equations |
| Field ratio | E=cB | From Faraday's law applied to plane waves |
| Direction | E⊥B⊥ propagation | From cross-product structure of Maxwell's equations |
Exam tip: Never just quote c=1/μ0ε0 — be ready to show it comes from taking curls of Maxwell's equations and identifying the wave equation form.
Concept: Electromagnetic Wave Relation — in free space, E and B are perpendicular, in phase, and related by c=E/B.
Step 1: The wave travels along x, and E is along j^. For a plane wave, B must be perpendicular to both the direction of propagation and E, so B is along k^.
Step 2: The magnitude relation is B=E/c, where c=3×108 m/s.
Step 3:
B=3×1086.3=2.1×10−8 T
Step 4: The direction is k^, so B=2.1×10−8 k^ T.
The magnetic field is 2.1×10−8 k^ T.
For an EM wave, E and B are perpendicular, in phase, and related by c=E/B. Here B=2.1×10−8 k^ T.
The key idea is that in a plane electromagnetic wave, the electric and magnetic fields are not independent — they are linked by Maxwell’s equations. For a wave traveling in free space, the ratio of their magnitudes is fixed by the speed of light, and their directions are perpendicular to each other and to the direction of propagation.
The wave moves along the x-direction. The electric field is given as E=6.3 j^ V/m, which points along the y-axis. For the wave to travel along x, the magnetic field must lie along the z-axis — that’s the only remaining perpendicular direction. The sign (whether +k^ or −k^) is determined by the fact that E×B must point in the direction of wave travel, which is +i^.
Let’s work through it step by step.
- Recall the fundamental relation In free space, the magnitudes of E and B in an electromagnetic wave satisfy
c=BE
where c=3×108 m/s is the speed of light. This comes directly from Maxwell’s equations — the wave equation for E and B gives the same speed c, and the fields are in phase with this ratio.
- Find the magnitude of B Given E=6.3 V/m, we have
B=cE=3×1086.3=2.1×10−8 T
- Determine the direction The wave travels along +i^. The electric field is along +j^. For the Poynting vector S=μ01(E×B) to point along +i^, we need E×B to be along +i^. Using the right-hand rule: j^×k^=i^. So B must be along +k^.
A quick check: if you ever forget the cross product direction, use the cyclic order x→y→z→x. Here x is propagation, y is E, so z must be B — and the sign follows from E×B∝propagation direction.
- Write the final vector Therefore,
B=2.1×10−8 k^ T
A common mistake is to forget that the frequency 25 MHz is irrelevant here — it only tells you the wave is in the radio band, but the relation E/B=c holds for any frequency in free space. Don’t let extra data distract you.
The magnetic field at that point is B=2.1×10−8 k^ T.
Method: The Right-Hand Rule and the Wave Relation for EM Waves
This problem uses the plane wave relation between electric and magnetic fields in free space, combined with the direction rule for electromagnetic waves.
Key Concept
For a plane EM wave traveling in free space:
- E, B, and the direction of propagation k^ are mutually perpendicular.
- The magnitudes are related by:
∣B∣=c∣E∣
where c=3×108 m/s.
Steps
-
Identify the direction of propagation
The wave travels along the x-direction. So k^=i^.
-
Identify the direction of E
Given: E=6.3 j^ V/m. So E points along +y.
-
Apply the right-hand rule
For a wave traveling in the +k^ direction:
k^=E^×B^
Here k^=i^, E^=j^.
Using i^=j^×B^, we get B^=k^ (the +z direction).
- Calculate the magnitude of B
B=cE=3×1086.3=2.1×10−8 T
- Write the final vector
B=2.1×10−8 k^ T
Quick Check
- Frequency 25 MHz is not needed here — it only confirms the wave is in the radio band, but the relation E=cB is frequency-independent in free space.
- The direction matches: x-propagation, y-electric field, z-magnetic field.
This is a classic problem from the Electromagnetic Waves chapter in NCERT Class 12 Physics. Here's a breakdown of the common mistakes students make on it, and how to avoid them.
🔍 The Correct Approach First
For an EM wave in free space:
- E, B, and direction of propagation are mutually perpendicular.
- Relation: ∣B∣=c∣E∣, where c=3×108 m/s.
- Direction: E×B gives the direction of wave travel.
Here:
- Wave travels along +x.
- E=6.3 j^ V/m (along +y).
- So B must be along +z (since j^×k^=i^).
Calculation:
∣B∣=3×1086.3=2.1×10−8 T
Final answer:
B=2.1×10−8 k^ T
✗ Common Mistake #1: Forgetting the Direction Rule
What students do wrong:
They calculate magnitude correctly but write B along +y or +x, or just give magnitude.
Why it happens:
They memorise "E and B are perpendicular" but don't apply the right-hand rule or the cross-product relation E×B∥propagation direction.
How to avoid:
- Always write: propagation direction = E×B direction.
- Use unit vectors: i^×j^=k^, j^×k^=i^, etc.
- Practice with all three axes.
✗ Common Mistake #2: Using Wrong Value of c
What students do wrong:
They use c=3×108 m/s but sometimes mistakenly use 3×108 km/s or forget the exponent.
Why it happens:
Rushing or not writing the formula clearly.
How to avoid:
- Always write c=3×108 m/s at the top.
- Double-check units: E in V/m, B in T.
- If frequency is given, it's a distractor — you don't need it here.
✗ Common Mistake #3: Using Frequency Unnecessarily
What students do wrong:
They try to use c=fλ or B=cE with frequency, leading to wrong numbers.
Why it happens:
The problem gives frequency (25 MHz) — students think it must be used.
How to avoid:
- Recognise: For a plane wave in free space, B=E/c is always true, independent of frequency.
- Frequency is only needed if they ask for wavelength or wave number.
✗ Common Mistake #4: Unit Confusion
What students do wrong:
They write B in Gauss instead of Tesla, or forget to convert MHz.
Why it happens:
Mixing CGS and SI units.
How to avoid:
- Stick to SI: E in V/m, B in T, c in m/s.
- 1 T = 104 G — but NCERT uses Tesla.
- Frequency in Hz: 25 MHz = 25×106 Hz (but again, not needed here).
✓ Quick Checklist to Avoid All Mistakes
| Step | Action |
|---|---|
| 1 | Identify propagation direction (given: +x) |
| 2 | Identify E direction (given: +y) |
| 3 | Use E×B∥ propagation to find B direction |
| 4 | Compute B=E/c with c=3×108 |
| 5 | Write final vector: magnitude + unit vector |
Final takeaway:
In free space, B=E/c always. The direction is the only tricky part — use the cross-product rule carefully.
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
Watch outA common mistake is to confuse k with wavelength λ or to use v=fλ without first finding f and λ correctly. Here f=ω/(2π) and λ=2π/k, so v=fλ=(ω/2π)(2π/k)=ω/k — same result. But the direct ratio is faster and less error-prone.
TipIn an exam, whenever you see a wave written as sin(ωt−kx) or cos(ωt−kx), immediately read off ω and k and compute v=ω/k. This works for any sinusoidal wave — mechanical or electromagnetic.
✓Final answerThe velocity of the wave is 2.0×108 m/s, which corresponds to option (A).
-
- 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).
Checking: [B]= T = kg·s⁻²·A⁻¹, [μ0]= kg·m·s⁻²·A⁻². Then μ0B2 = kg·m⁻¹·s⁻² = J/m³ = energy density. ✓
✓Final answer(A) energy density.
- 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 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(a) c=μ0ε01
- 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:
B0=3×108480=1.6×10−6 T=1.6×10−6 weber/m2
✓Final answer(c) 1.6×10−6 weber/m2.
- 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:
- E0=1000 V/m
- c=3×108 m/s
B0=3×1081000
- Simplify the fraction:
B0=3×1081000=3×108103=31×10−5
B0=0.333...×10−5=3.33×10−6 T
TipA 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 B0>E0 in magnitude, something has gone wrong.
✓Final answerThe correct option is (D) 3.33×10−6 T.
- 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
Here E=9.0 V/m and c=3×108 m/s, so
B=3×1089.0=3×10−8 T
✓Final answer(a) 3×10−8 T
- 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
Z0=8.85×10−124π×10−7≈377Ω
Since impedance is measured in ohms, the unit of √(μ₀/ε₀) is the ohm.
✓Final answer(B) ohm.
- 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
λ=k2π=5002π≈1.257×10−2 m≈1.26 cm
✓Final answerλ≈1.26×10−2 m ≈1.26 cm.
- 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
Converting to CGS: 3×10⁸ m/s × 100 cm/m = 3×10¹⁰ cm/s.
✓Final answer(B) 3 × 10¹⁰ cm/second.
- 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:
E0=(3×108)×(3×10−6)=9×102=900 Vm−1
✓Final answerThe correct option is (c): E0=900 Vm−1
- 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.
Given B0=510 nT=510×10−9 T:
E0=(3×108)(510×10−9)=153 NC−1
✓Final answer(c) 153 NC−1.
- 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.
Given B=3×10−6 T:
E=(3×108)×(3×10−6)=9×102=900 Vm−1
✓Final answerThe correct option is (c): E=900 Vm−1
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