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 15 on this concept.
- COMEDK 2025Set 2025-E1 markMCQQ.A plane electromagnetic wave with frequency 40 MHz travels in free space. At a particular point in space and time, the magnetic field is 2×10−8T. What will be the electric field at this point? (A) 16Vm−1 (B) 6Vm−1 (C) 8Vm−1 (D) 18Vm−1
›Reveal solutionSolution
In free space, the electric and magnetic fields of an electromagnetic wave are related by E=cB. Given B=2×10−8T and c=3×108m/s, the electric field is 6V/m, so the correct option is (B).
The key concept here is the intrinsic relationship between the electric and magnetic fields in an electromagnetic wave in free space. Unlike in circuits or static fields, where E and B are independent, in a traveling EM wave they are locked together: their magnitudes are proportional, and the constant of proportionality is the speed of light c. This comes directly from Maxwell’s equations — specifically, Faraday’s law and Ampère’s law — which show that a changing magnetic field creates an electric field, and vice versa, with the ratio fixed by c=1/μ0ε0.
The frequency given (40 MHz) is a red herring here: it tells us the wave is in the radio band, but the instantaneous relation E=cB holds at every point and time for a plane wave in vacuum, regardless of frequency. So we don’t need the frequency at all.
Let’s work it through:
- Recall the fundamental relation for a plane electromagnetic wave in free space:
BE=c
where c=3×108m/s is the speed of light. This holds because the wave’s energy is equally shared between the fields, and the wave equation forces the ratio.
- Plug in the given magnetic field:
B=2×10−8T
So
- COMEDK 2025Set 2025-M1 markMCQQ.The electric and magnetic fields associated with an electromagnetic wave propagating along +z axis, can be represented by (A) E=E0i,B=B0j (B) E=E0i,B=B0k (C) E=E0k,B=B0j (D) E=E0j,B=B0k
›Reveal solutionSolution
For an electromagnetic wave propagating along the +z axis, the electric and magnetic fields must be perpendicular to each other and to the direction of propagation. The only option satisfying this is (A): E=E0i^, B=B0j^.
The key concept here is the mutual perpendicularity of E, B, and the direction of propagation in an electromagnetic wave. In free space, electromagnetic waves are transverse: the electric field, magnetic field, and wave vector k (pointing in the propagation direction) form a right-handed orthogonal set. This means:
- E⊥k
- B⊥k
- E⊥B
- And the direction of E×B gives the direction of propagation.
Let’s check each option step by step.
-
Identify the propagation direction. The wave propagates along the +z axis, so the wave vector k is along k^ (the unit vector in the z-direction).
-
Check option (A): E=E0i^, B=B0j^.
- E is along x-axis (i^), B is along y-axis (j^).
- Both are perpendicular to k^ (z-axis) because i^⋅k^=0 and j^⋅k^=0.
- E×B=E0B0(i^×j^)=E0B0k^, which points along +z. This matches the propagation direction.
- So (A) is valid.
-
Check option (B): E=E0i^, B=B0k^.
- E is along x-axis, B is along z-axis.
- B is parallel to the propagation direction, not perpendicular. This violates the transverse nature of EM waves in free space.
- Invalid.
-
Check option (C): E=E0k^, B=B0j^.
- E is along z-axis, parallel to propagation. Again, not transverse. …
- COMEDK 2024Set 2024-E1 markMCQQ.Column - I lists the waves of the electromagnetic spectrum. Column - II gives approximate frequency range of these waves. Match Column - I and Column - II and choose the correct match from the given choices. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Column I Column II (A) Radiowaves (P) 1018 to 1020 Hz (B) Microwaves (P) 1011 to 5×1014 Hz (C) Infrared (R) 104 to 108 Hz (D) X-rays (S) 109 to 1012 Hz (A) (A)-(R) (B)-(P) (C)-(S) (D)-(Q) (B) (A)-(R) (B)-(S) (C)-(Q) (D)-(P) (C) (A)-(R) (B)-(S) (C)-(P) (D)-(Q) (D) (A)-(R) (B)-(Q) (C)-(S) (D)-(P)
›Reveal solutionSolution
A-R, B-S, C-Q, D-P by increasing frequency.
Match each band by frequency:
- Radiowaves (A): lowest frequency 104–108 Hz = R.
- Microwaves (B): 109–1012 Hz = S.
- Infrared (C): 1011–5×1014 Hz = Q (the entry misprinted as a second P). …
- COMEDK 2023Set 2023-E1 markMCQQ.What feature of the infrared waves make it useful for the haze photography? (A) Since it is invisible (B) Since it has large wave length (C) Since it is absorbed by the medium (D) Since it has high frequency
›Reveal solutionSolution
Infrared's long wavelength means little scattering by haze particles, so it passes through and produces clearer distant images.
Scattering of light by small particles (Rayleigh scattering) falls off strongly with increasing wavelength (∝1/λ4). Infrared radiation has a much larger wavelength than visible light, so it is scattered far less by the fine particles of haze, mist and smoke. It therefore penetrates the haze and reaches the camera, allowing clear photograp …
- COMEDK 2023Set 2023-M1 markMCQQ.The ratio of amplitude of magnetic field to the amplitude of electric field of an electromagnetic wave propagating in vacuum is (A) reciprocal of speed of light in vacuum (B) the speed of light in vacuum (C) proportional to frequency of the electromagnetic wave (D) inversely proportional to the frequency of the electromagnetic wave
›Reveal solutionSolution
Maxwell's relation E0=cB0 makes B0/E0=1/c, i.e. the reciprocal of the speed of light, independent of frequency.
For a plane electromagnetic wave in vacuum, the field amplitudes satisfy:
B0E0=c⇒E0B0=c1. …
- COMEDK 2023Set 2023-M1 markMCQQ.A plane electromagnetic wave of frequency 20 MHz travels through a space along x-direction. If the electric field vector at a certain point in space is 6 Vm−1, then what is the magnetic field vector at that point? (A) 2×10−8 T (B) 21×10−8 T (C) 2 T (D) 21 T
›Reveal solutionSolution
In an EM wave E and B are related by B=E/c, giving B=6/(3×108)=2×10−8T.
For a plane electromagnetic wave, the field magnitudes satisfy
B=cE=3×108 m s−16 V m−1=2×10−8 T. …
- KCET 2022Set B-31 markMCQQ.A fully charged capacitor ‘C’ with initial charge ‘q_0’ is connected to a coil of self inductance ‘L’ at t=0. The time at which the energy is stored equally between the electric and the magnetic field is (A) πLC (B) 4πLC (C) 2πLC (D) LC
›Reveal solutionSolution
In an LC oscillation, energy sloshes between capacitor and inductor. Equal sharing happens when the charge on the capacitor is q0/2, which occurs at t=4πLC.
The problem is about an ideal LC circuit — a capacitor charged to q0 connected to an inductor at t=0. No resistance, so total energy is conserved. At any instant, the capacitor stores electric energy UE=2Cq2 and the inductor stores magnetic energy UB=21Li2. Their sum is constant: 2Cq02.
The question asks: when are these two equal? That means UE=UB, so each is half the total energy. That gives 2Cq2=21⋅2Cq02, i.e. q2=2q02, so q=±2q0.
Now we need the time when the charge first reaches this value. The charge on the capacitor in an LC circuit oscillates sinusoidally. Since the capacitor is fully charged at t=0 and then begins to discharge, the charge follows a cosine function:
q(t)=q0cos(ωt)
where ω=LC1 is the angular frequency of the LC oscillation.
- Set q(t)=2q0.
q0cos(ωt)=2q0⇒cos(ωt)=21
- The smallest positive angle whose cosine is 1/2 is π/4 radians. So: ωt=4π …
- COMEDK 2022Set 20221 markMCQQ.Speed of electromagnetic wave in a medium having relative permittivity εr and relative permeability μr is (speed of light in air, c=3×108 m/s) (A) μrεr1 (B) μrεrc (C) cεrμr (D) μrεrc
›Reveal solutionSolution
(Option A is dimensionally wrong — it is a pure number, not a speed; the refractive index is n = √(μ_r ε_r), and v = c/n.)
Concept: Speed of an EM wave in a medium, v = 1/√(με), where μ = μ₀μ_r and ε = ε₀ε_r.
v = 1/√(μ₀μ_r ε₀ε_r) = [1/√(μ₀ε₀)] × 1/√(μ_r ε_r) = c/√(μ_r ε_r). …
- COMEDK 2022Set 20221 markMCQQ.A 30 mW laser beam has a cross-sectional area of 15 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Permuittivity of space, ε0=9×10−12 Speed of light, c=3×108 m/s] (A) 1.22 kV/m (B) 12 kV/m (C) 10 kV/m (D) 201 kV/m
›Reveal solutionSolution
Solve for E0: E0 = sqrt( 2I / (epsilon0 c) ) = sqrt( 2 x 2000 / (9 x 10^-12 x 3 x 10^8) ) = sqrt( 4000 / (2.7 x 10^-3) ) = sqrt( 1.481 x 10^6 ) = 1.22 x 10^3 V/m = 1.22 kV/m
Concept: intensity of an EM wave in terms of the peak electric field:
I = (1/2) epsilon0 c E0^2
Intensity of the beam:
I = P/A = 30 x 10^-3 W / (15 x 10^-6 m^2) = 2000 W/m^2
Solve for E0:
E0 = sqrt( 2I / (epsilon0 c) ) …
- COMEDK 2021Set 20211 markMCQQ.The correct arrangement in increasing order of wavelength of X-rays, UV rays, microwave is (A) microwave, X-rays, UV rays (B) UV rays, X-rays, microwave (C) X-rays, UV rays, microwave (D) microwave, UV rays, X-rays
›Reveal solutionSolution
Increasing wavelength: X-rays < UV rays < microwave.
Concept: electromagnetic spectrum ordering.
Typical wavelengths:
- X-rays: ~1e-10 m (0.01-10 nm)
- Ultraviolet: ~1e-8 m (10-400 nm)
- Microwave: ~1e-2 m (1 mm - 1 m) …
- COMEDK 2021Set 20211 markMCQQ.Which of the following waves are used to treatment of muscles ache? (A) Ultraviolet (B) Infrared (C) Microwave (D) X-rays
›Reveal solutionSolution
Infrared radiation has a heating effect (it excites molecular vibrations) and is used in heat lamps / IR therapy to relieve muscular aches and sprains. UV is used for sterilisation, microwaves for cooking and radar, X-rays for imaging.
Concept: uses of electromagnetic waves. …
- COMEDK 2021Set 2021-B1 markMCQQ.The radiations used in treatment of muscles ache are (A) Microwave (B) Ultraviolet (C) Infrared (D) X-rays
›Reveal solutionSolution
Infrared radiation is the heat radiation used to treat muscle aches.
Infrared waves are absorbed by tissue and produce a warming effect, which relaxes muscles, dilates blood vessels and eases pain. This is the basis of infrared physiotherapy lamps used for muscular aches. Microwaves and ultraviolet are not used this way f …
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