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 19 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The CORRECT statement among the following regarding electromagnetic waves is (A) They can travel through vacuum (B) They consist of only electric field (C) They consist of only magnetic field (D) They require a medium to propagate (E) They move with a velocity of 3×108 cms−1
›Reveal solutionSolution
EM waves propagate through vacuum, carrying mutually perpendicular oscillating electric and magnetic fields at speed c.
Examining the options:
- (A) True — a changing electric field creates a magnetic field and vice versa, so the wave is self-sustaining and needs no material medium (sunlight reaching Earth through space proves it).
- (B) False — it has both fields, not just electric.
- (C) False — it has both fields, not just magnetic.
- (D) False — they do not require a medium.
- (E) False — the speed is 3×108 m s−1, not cm s−1.
✓Final answerThe correct option is (A).
- KEAM 2026Set eng-2026-04184 marksMCQQ.A radio can tune in to any station in the 7.5 MHz to 12 MHz band. The corresponding wavelength band is (A) 75m-12m (B) 22.5m-36m (C) 40m-25m (D) 20m-45m (E) 15m-24m
›Reveal solutionSolution
Using λ=c/f, the 7.5 MHz end maps to 40 m and the 12 MHz end to 25 m, giving a band of 40m–25m.
λ=fc,c=3×108ms−1
For f=7.5MHz=7.5×106Hz:
λ=7.5×1063×108=40m
For f=12MHz=12×106Hz:
λ=12×1063×108=25m
Wavelength band: 40m to 25m.
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04204 marksMCQQ.If the total energy transferred to a completely absorbing surface by an EM wave in unit time is 3.6 J, then the radiation pressure exerted by the wave on the surface is (A) 1×108Nm−2 (B) 1.8×108Nm−2 (C) 1×107Nm−2 (D) 1.2×107Nm−2 (E) 1.2×10−8Nm−2
›Reveal solutionSolution
For a completely absorbing surface the momentum delivered per unit time is cU/t.
For an electromagnetic wave completely absorbed by a surface, the force (rate of momentum transfer) equals the power divided by the speed of light. With energy delivered per unit time tU=3.6 J s−1:
cU/t=3×1083.6=1.2×10−8 N.
This matches the tabulated value of 1.2×10−8 Nm−2.
✓Final answerThe correct option is (E).
- KEAM 2026Set eng-2026-04214 marksMCQQ.Microwaves are (A) used in radio and television communications (B) having frequency range from 54 MHz to 890 MHz (C) short wavelength radio waves (D) produced by hot bodies and molecules (E) absorbed by ordinary glass
›Reveal solutionSolution
Microwaves = short-wavelength radio waves.
Microwaves occupy wavelengths of about 1,mm to 0.3,m (frequencies ∼109–1011,Hz), i.e. the short-wavelength end of the radio band, generated by devices such as klystrons and magnetrons. Options describing radio/TV frequency ranges, hot-body emission (infrared) or absorption by glass (infrared/UV) are incorrect.
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04224 marksMCQQ.In a plane electromagnetic wave if the amplitude of oscillating electric field is 45 Vm−1 then the amplitude of the oscillating magnetic field is (A) 2.5×10−8 T (B) 1.5×10−7 T (C) 3×10−8 T (D) 1.5×10−8 T (E) 2.5×10−7 T
›Reveal solutionSolution
In an EM wave the field amplitudes satisfy B0=E0/c.
B0=cE0=3×10845=1.5×10−7T.
✓Final answerThe correct option is (B).
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.Identify the two electromagnetic waves A and B having respective wavelengths 2 cm and 580 nm (A) A is microwave and B is visible light (B) A is infrared and B is ultraviolet ray (C) A is radio wave and B is visible light (D) A is infrared and B is visible light (E) A is radio wave and B ultraviolet ray
›Reveal solutionSolution
λ=2cm is microwave, λ=580nm is visible light.
The electromagnetic spectrum ranges: microwaves ~1 mm to 1 m, infrared ~700 nm to 1 mm, visible ~400-700 nm, ultraviolet ~10-400 nm. A wavelength of 2 cm (2×10−2m) falls in the microwave band, and 580 nm (5.8×10−7m) is in the visible (yellow) region. Hence A is a microwave and B is visible light.
✓Final answerThe correct option is (A).
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.If an EM wave travels in a medium with εr=4, μr=1, its speed (in ms−1) in terms of c (c = velocity of light in free space) is (A) c (B) 2c (C) 2c (D) 3c (E) 4c
›Reveal solutionSolution
v=c/εrμr=c/4⋅1=c/2.
The speed of an electromagnetic wave in a medium is
v=εrμrc.
With εr=4 and μr=1:
v=4×1c=2c.
✓Final answerThe correct option is (C).
- KEAM 2025Set eng-2025-04234 marksMCQQ.When a ray of light moves from one medium to another medium, (A) its frequency remains unchanged (B) its frequency alone changes (C) its wavelength remains unchanged (D) both its frequency and wavelength change (E) its velocity remains constant
›Reveal solutionSolution
When light passes into another medium its frequency stays unchanged.
Concept and Intuition
Frequency is fixed by the source and is conserved across a boundary (the fields must oscillate continuously at the interface). Speed and wavelength change with the medium's refractive index, but frequency does not.
Step-by-Step Solution
- v = f*lambda; on entering a new medium v changes.
- f is set by the source and is conserved at the boundary.
- Therefore lambda changes to keep f constant, while f remains the same.
Common Mistakes
- Thinking frequency changes; in fact it is wavelength and speed that change.
✓Final answerThe correct option is (A) — its frequency remains unchanged.
ANSWER: A
- KEAM 2025Set eng-2025-04264 marksMCQQ.An electromagnetic wave travelling in vacuum has its electric field component, E=15sin[1.57y+5.4t]j^. The wavelength of the wave is (A) 4.0 m (B) 3.0 m (C) 2.5 m (D) 2.0 m (E) 1.0 m
›Reveal solutionSolution
The propagation constant is k=1.57 m−1, giving wavelength λ=2π/k=4.0 m.
The wave is E=15sin(1.57y+5.4t)j^, of the form E=E0sin(ky+ωt).
The angular wavenumber is k=1.57 m−1=λ2π.
λ=k2π=1.572×3.14=1.576.28=4.0 m
✓Final answerThe correct option is (A).
- KEAM 2025Set eng-2025-04274 marksMCQQ.The speed of electromagnetic waves in a medium depends on the (A) intensity of the wave (B) initial phase of the wave (C) permittivity and permeability of the medium (D) energy it carries (E) reflectivity of the medium
›Reveal solutionSolution
Electromagnetic wave speed is fixed by the medium's electric and magnetic properties: v=με1.
From Maxwell's equations, the speed of an electromagnetic wave in a medium is determined solely by its permittivity ε and permeability μ:
v=με1.
In free space this gives c=μ0ε01≈3×108 m s−1. The speed is independent of the wave's intensity, phase, or energy content.
✓Final answerThe correct option is (C).
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.In a plane electromagnetic wave, the magnetic field is given by B=400×10−6sin[(4.0×10−4)(t−x/c)] T. The peak value of electric field (in Vm−1) is (A) 8×104 (B) 6×104 (C) 4×104 (D) 3×104 (E) 12×104
›Reveal solutionSolution
In an EM wave E0=cB0=3×108×400×10−6=12×104 V m−1.
In a plane electromagnetic wave the peak electric and magnetic fields are related by
E0=cB0.
Here the amplitude of the magnetic field is B0=400×10−6T, so
E0=(3×108)(400×10−6)=1.2×105=12×104 V m−1.
✓Final answerThe correct option is (E).
- KEAM 2025Set pha-2025-0429F4 marksMCQQ.If the frequency of an electromagnetic wave is 2 MHz, then the time period of oscillation of the accelerated charge is (A) 2.5×10−7s (B) 1×10−7s (C) 5×10−7s (D) 6×10−7s (E) 2×10−7s
›Reveal solutionSolution
The oscillation period equals the reciprocal of the frequency: T=1/(2×106Hz)=5×10−7s.
The accelerated charge oscillates at the same frequency as the electromagnetic wave it radiates. With f=2MHz=2×106Hz,
T=f1=2×1061=0.5×10−6=5×10−7s.
✓Final answerThe correct option is (C).
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