Q.The magnetic field in a plane electromagnetic wave is given by By=(2×10−7) Tsin(0.5×103x+1.5×1011t).
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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 wave is By=(2×10−7) Tsin(0.5×103x+1.5×1011t). Compare with B=B0sin(kx+ωt).
(a) Wavelength and frequency.
- k=0.5×103=500 rad/m⇒λ=k2π=5002π≈1.26×10−2 m (=1.26 cm).
- ω=1.5×1011 rad/s⇒f=2πω=2π1.5×1011≈2.39×1010 Hz. …
Reading k=500 rad/m and ω=1.5×1011 rad/s from the wave gives λ=2π/k≈1.26 cm and f=ω/2π≈2.39×1010 Hz; since the argument is kx+ωt the wave travels along −x, and the electric field is Ez=(60 V/m)sin(0.5×103x+1.5×1011t).
Compare with the standard form. Writing By=B0sin(kx+ωt) with B0=2×10−7 T, we read off
k=0.5×103=500 rad/m,ω=1.5×1011 rad/s.
Step 1 — Wavelength.
λ=k2π=5002π≈1.26×10−2 m=1.26 cm.
Step 2 — Frequency.
f=2πω=2π1.5×1011≈2.39×1010 Hz.
Cross-check with c=fλ=(2.39×1010)(1.26×10−2)≈3.0×108 m/s — the expected speed of light.
Step 3 — Direction of propagation. In sin(kx+ωt) the x and t terms carry the same sign, so a point of constant phase satisfies kx+ωt=const; as t increases, x must decrease. Hence the wave travels along the −x direction.
Step 4 — Amplitude of the electric field. For a plane EM wave in vacuum E0=cB0:
E0=(3×108)(2×10−7)=60 V/m. …
Method: Standard Wave Analysis for EM Waves
We use the general wave equation comparison method — matching the given wave to the standard form to extract parameters, then applying the EM wave relation E=Bc.
Step 1: Identify the wave form
The given magnetic field:
By=(2×10−7) Tsin(0.5×103x+1.5×1011t)
Standard form for a wave traveling along x:
B=B0sin(kx+ωt)
Here the + sign means the wave travels in the negative x-direction.
Step 2: Extract k and ω
From comparison:
- Wave number: k=0.5×103=500 rad/m
- Angular frequency: ω=1.5×1011 rad/s
Step 3: Find wavelength λ and frequency f
Wavelength:
λ=k2π=5002π=250π m
λ=1.26×10−2 m
Frequency:
f=2πω=2π1.5×1011
f=2.39×1010 Hz
Step 4: Write the electric field expression
For an EM wave in vacuum:
E0=B0c
Given B0=2×10−7 T and c=3×108 m/s: …
Common Mistakes & How to Avoid Them
Mistake 1: Confusing the sign in the wave equation
The error:
Students see By=(2×10−7)sin(0.5×103x+1.5×1011t) and assume the wave travels along +x because the coefficient of t is positive.
Why it’s wrong:
The general form is sin(kx−ωt) for a wave traveling in +x direction. Here we have sin(kx+ωt), which means the wave travels in −x direction.
How to avoid:
- Always compare with the standard form:
- sin(kx−ωt) → wave moves along +x
- sin(kx+ωt) → wave moves along −x
- Memorise: The sign of the t term tells you the direction — opposite to what intuition might suggest.
Mistake 2: Using the wrong formula for wavelength
The error:
Students write k=λ2π but then plug k=0.5×103 without checking units.
Why it’s wrong:
The given k=0.5×103m−1 is correct, but students sometimes forget to convert or misplace the decimal.
How to avoid:
- Always write:
k=λ2π⇒λ=k2π
- Substitute carefully:
λ=0.5×1032π=5002π=250πm
- Double-check: λ should be in metres — if you get a weird number, re-check k.
Mistake 3: Confusing angular frequency ω with frequency f
The error:
Students write f=ω or use f=2πω incorrectly.
Why it’s wrong:
Here ω=1.5×1011rad/s. The frequency is:
f=2πω=2π1.5×1011Hz
How to avoid:
- Remember:
- ω = angular frequency (rad/s)
- f = ordinary frequency (Hz or s−1)
- Relation: ω=2πf
- Always write the unit — if you get rad/s, you know it’s ω, not f.
Mistake 4: Forgetting the direction of the electric field
The error:
Students write Ex or Ez without checking the cross-product relation.
Why it’s wrong:
For an EM wave, E, B, and direction of propagation k^ are related by:
E×B∥direction of propagation
Here:
- B is along +y
- Wave travels along −x
- So E must be along +z (use right-hand rule)
How to avoid:
- Use the right-hand rule:
- Point fingers along E
- Curl them toward B
- Thumb points in direction of wave travel
- Check: If wave goes in −x, B along +y, then E must be along +z.
Mistake 5: Using the wrong relation between E0 and B0
The error:
Students write E0=cB0 but forget that c=3×108m/s.
Why it’s wrong:
The formula is correct, but students sometimes use c=3×108 incorrectly or forget to multiply.
How to avoid:
- Always write:
E0=cB0
- Substitute: …
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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