Q.The 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?
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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 …
The key idea is that in vacuum, the electric and magnetic field amplitudes of an electromagnetic wave are related by the speed of light: E0=cB0.
Given B0=510 nT=510×10−9 T and c=3.0×108 m/s:
- Use the relation E0=cB0.
- Substitute: E0=(3.0×108)×(510×10−9). …
For an electromagnetic wave in vacuum, the electric and magnetic field amplitudes are related by E0=cB0. Substituting B0=510 nT gives E0=153 V/m.
The key idea here is that in vacuum, electromagnetic waves travel at the speed of light c, and the electric and magnetic fields are not independent — they are linked by Maxwell’s equations. For a plane harmonic wave, the amplitudes satisfy E0=cB0. This is a direct consequence of Faraday’s law and Ampère’s law in free space, where the rate of change of one field generates the other.
Why does this relation hold? Imagine a wave moving along the x-axis, with the electric field oscillating along y and the magnetic field along z. Faraday’s law tells us that a changing magnetic field produces an electric field, and the magnitudes are tied by the wave speed. In vacuum, that speed is c=3.00×108 m/s. So once you know B0, multiplying by c gives E0 directly — no extra constants needed.
Let’s work it out step by step.
- Write the given data. The amplitude of the magnetic field is B0=510 nT. Recall that 1 nT=10−9 T, so
B0=510×10−9 T=5.10×10−7 T.
- Recall the fundamental relation. For an electromagnetic wave in vacuum, the amplitudes of the electric and magnetic fields are related by
E0=cB0,
where c=3.00×108 m/s is the speed of light in vacuum.
E0=cB0
This is not an approximation — it follows exactly from Maxwell’s equations for a plane wave in free space. The ratio E/B equals c at every point and at every instant for a wave in vacuum.
- Substitute the values.
E0=(3.00×108 m/s)×(5.10×10−7 T).
Multiply the numbers:
3.00×5.10=15.3.
Multiply the powers of ten:
108×10−7=101=10.
So
E0=15.3×10 V/m=153 V/m.
(Recall that 1 T⋅m/s=1 V/m, so the units work out perfectly.) …
Method: Using the Speed of Light Relation for EM Waves
This problem uses the fundamental relation between electric and magnetic fields in an electromagnetic wave — the fields are linked by the speed of light.
Steps
- Recall the key relation In vacuum, for any electromagnetic wave:
E0=cB0
where
- E0 = amplitude of electric field (V/m)
- B0 = amplitude of magnetic field (T)
- c = speed of light in vacuum = 3.0×108 m/s
- Convert units Given B0=510 nT
1 nT=10−9 T
So:
B0=510×10−9 T=5.10×10−7 T
- Apply the formula
E0=(3.0×108)×(5.10×10−7)
- Calculate E0=3.0×5.10×108−7=15.3×101 …
Here are the most common mistakes students make when solving this problem, along with how to avoid each one.
1. Forgetting the Speed of Light Constant
The Mistake:
Students often try to use E0=cB0 but write c as 3×108 without checking units, or they confuse c with the speed of sound or another constant.
How to Avoid:
- Memorize c=3.0×108 m/s for vacuum.
- Always write the relation clearly:
E0=cB0
This is derived from Maxwell’s equations — in vacuum, the electric and magnetic field amplitudes are linked by the speed of light.
2. Unit Conversion Errors (nT to T)
The Mistake:
Given B0=510 nT, students plug in 510 directly without converting to tesla (T). This gives a wildly wrong answer.
How to Avoid:
- Always convert nano (n) to base SI units:
1 nT=10−9 T
So:
B0=510×10−9 T=5.10×10−7 T
3. Using the Wrong Formula (Mixing RMS and Amplitude)
The Mistake:
Some students use Erms=cBrms but then treat B0 as RMS, or vice versa. The problem explicitly asks for amplitude, so you must use the amplitude form.
How to Avoid:
- Read the question carefully: “amplitude of the magnetic field part” → B0.
- The correct relation for amplitudes is:
E0=cB0
(Not Erms=cBrms unless the problem gives RMS values.)
4. Arithmetic Slip-Ups in Scientific Notation
The Mistake:
When multiplying 3×108 by 5.10×10−7, students misplace the decimal or add exponents incorrectly.
How to Avoid:
- Do the multiplication step-by-step:
E0=(3.0×108)×(5.10×10−7)
Multiply coefficients: 3.0×5.10=15.3
Add exponents: 108+(−7)=101
So:
E0=15.3×101=153 V/m …
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. …
- 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: …
- 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: …
- 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 ran …
- 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. …
- 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. …
- 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: …
- 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. …
- 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. …
- 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. …
- 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 …
- 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, …
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