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?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 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 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.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.