Q.An infinitely long thin wire carrying a uniform linear static charge density λ lies along the z-axis. The wire is set into motion along its own length with a uniform velocity v=vk^, i.e. directed along the +z-axis. Consider a field point at perpendicular (radial) distance r from the wire, with r^ the radially outward unit vector and ϕ^ the azimuthal unit vector circling the wire. Calculate the Poynting vector S=μ01(E×B).
🔒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 moving charged wire is both a line charge (radial E) and a current I=λv (azimuthal B). Crossing them makes energy flow along the wire. …
A moving charged wire acts simultaneously as a static line charge — producing a radial electric field — and as a current I=λv — producing a circular magnetic field. The cross product of these two fields yields a Poynting vector directed along the wire's motion (+z) whose magnitude decreases as 1/r2: S=4π2ε0r2λ2vk^.
Concept
Energy transport in electromagnetism is given by the Poynting vector S=μ01(E×B). Here we need E and B separately at a point a perpendicular distance r from the wire.
Electric field of the line charge
By Gauss's law, an infinite line charge of linear density λ has a purely radial field at perpendicular distance r:
E=2πε0rλr^
Magnetic field of the moving charge
The charge sliding along +z with speed v constitutes a steady current I=λv. By Ampère's law this current produces an azimuthal (circular) magnetic field:
B=2πrμ0Iϕ^=2πrμ0λvϕ^
Forming the Poynting vector …
Method: Combining a Static-Charge Field and a Current-Produced Field to Find the Poynting Vector
Use this for problems that ask for the Poynting vector (or energy flow) around a setup that is not itself a propagating EM wave — e.g. a charged conductor carrying current — by separately finding its E and B fields via familiar static/steady-current formulas, then combining them.
Steps
Step 1: Recognise that a moving charged object is simultaneously two things.
A static charge distribution (produces an E field via Gauss's/Coulomb's law) and a current (produces a B field via Ampere's law). Treat these two contributions completely independently in Steps 2–3, only combining them at the very end.
Step 2: Find E using the appropriate static-charge formula for the given geometry.
For an infinite line charge of linear density λ, Gauss's law gives a purely radial field:
E=2πε0rλr^
Step 3: Find B using the appropriate steady-current formula for the SAME geometry. …
- GUJCET 2026Set x1 markMCQQ.A charged particle oscillates about its mean equilibrium position with a frequency of 8×109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? (A) 4×109 Hz (B) 1.6×1010 Hz (C) 8×109 Hz (D) 2×109 Hz
›Reveal solutionSolution
[!TLDR]
Using the NCERT/CBSE list of insect-resistant crop varieties: X = flat bean, Y = Pusa A-4, Z = Aphids.
Concept
Under 'Strategies for Enhancement in Food Production', plant breeders release crop varieties resistant to specific insect pests. A standard table pairs each crop with its released variety and the pest it resists. This GSEB/NCERT-aligned question tests recall of three of those pairings.
Solution
- Row (i): Rape-seed mustard, variety Pusa Gaurav. This mustard variety is bred for resistance to the mustard aphid, so Z = Aphids.
- Row (ii): variety Pusa Sem 2, resisting Jassids. 'Pusa Sem' varieties belong to flat bean, so X = flat bean.
- Row (iii): Okra (bhindi) resisting shoot borer. The released okra variety here is Pusa A-4 (resistant to shoot and fruit borer), so Y = Pusa A-4. …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.The amplitude of the magnetic field of Electromagnetic wave is B_0 = 510 nT, then amplitude of electric field of Electromagnetic wave is E_0 = ___.(a) 143 V/m(b) 153 V/m(c) 135 V/m(d) 170 V/m
›Reveal solutionSolution
In an electromagnetic wave, the electric and magnetic field amplitudes are related by E_0 = c B_0, where c is the speed of light.
Given B_0 = 510 nT = 510 x 10^-9 T, c = 3 x 10^8 m/s.
…
- GUJCET 2023Set 091 markMCQQ.If E and B represent electric and magnetic field vectors of electromagnetic wave, the direction of propagation of electromagnetic wave is along ______. (A) B (B) E (C) B×E (D) E×B
›Reveal solutionSolution
The Poynting direction E×B gives the wave's propagation direction.
Concept: In an electromagnetic wave, E, B, and the propagation direction form a right-handed triad, with p …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.For a given electromagnetic waves the magnitude of electric field is 6.6 V/m at a point in space. The magnitude of magnetic field at this point is ___ T.(a) 2.1 x 10^-8(b) 6.6 x 10^-8(c) 19.8 x 10^-8(d) 2.2 x 10^-8
›Reveal solutionSolution
In an EM wave the field magnitudes obey E = cB, so B = E/c = 6.6/(3x10^8) = 2.2 x 10^-8 T.
…
- GUJCET 2022Set 171 markMCQQ.A radio can tune into any station in the 6 MHz to 12 MHz band. What is the corresponding wavelength band? (c=3×108 m/s) (A) 40 m to 60 m (B) 25 m to 50 m (C) 20 m to 30 m (D) 10 m to 20 m
›Reveal solutionSolution
λ=c/f; higher frequency gives shorter wavelength.
Steps.
- At f=6 MHz: λ=6×1063×108=50 m. …
- GUJCET 2022Set 171 markMCQQ.A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? (A) 1018 Hz (B) 109 Hz (C) 10−9 Hz (D) 1010 Hz
›Reveal solutionSolution
An oscillating charge radiates EM waves at exactly its own oscillation frequency.
Concept. An accelerating/oscillating charge produces electromagnetic waves whose frequency equals the frequency of oscillation of the charge. …
- GUJCET 2021Set 151 markMCQQ.A plane electromagnetic wave of frequency 25 MHz travels in free space along the X-direction. At a particular point in space and time, where B=2.1×10−8k^ T then find E at this point? (A) −2.1j^mV (B) 6.3j^mV (C) 4.2j^mV (D) −3.2j^mV
›Reveal solutionSolution
For an EM wave E=cB, with E,B and the propagation direction mutually perpendicular (E×B points along propagation).
Concept. E=cB and E^×B^=propagation^. …
- GUJCET 2019Set 131 markMCQQ.At large distances from source E and B are in phase and the decrease in their magnitude is comparitively slower with distance r as per. (A) r2 (B) r−3 (C) r (D) r−1
›Reveal solutionSolution
Far from the source, radiated E and B are in phase and decrease as r1.
Concept: The radiation (far) field of an accelerating charge dominates at large distances because it decays only as 1/r, unlike the static (1/r2) or induction (1/r3) terms. This slow decay is why radiated energy reaches far away.
Steps: …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.The maximum value of E in an electromagnetic wave is equal to 1.8 Vm^-1. Thus the maximum value of B is ___.(a) 6 x 10^-8 T(b) 3 x 10^-6 T(c) 6 x 10^-9 T(d) 2 x 10^-10 T
›Reveal solutionSolution
In an electromagnetic wave, the peak electric and magnetic fields are related by B0=E0/c.
Given E0=1.8 V/m, c=3×108 m/s.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.For a radiation of 6 GHz passing through air, the wave number (number of waves) per 1 m length is ___ (1 GHz = 10^9 Hz).(a) 5(b) 3(c) 20(d) 30
›Reveal solutionSolution
The number of complete waves per unit length (wave number in this sense) equals f/c, the reciprocal of wavelength.
Given f=6 GHz =6×109 Hz, c=3×108 m/s.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.In the region closer to the oscillating charges, the phase difference between E (vector) and B (vector) fields is ___ and their magnitude quickly decreases as ___ with distance r from the source.(a) 0, r^-1(b) pi/2, r^-3(c) pi/2, r^-1(d) 0, r^-3
›Reveal solutionSolution
Close to the oscillating charges (the near field), E and B are pi/2 out of phase and their amplitudes decrease steeply, as r^-3.
Near an oscillating charge (the induction/near-field zone, distances small compared with the wavelength):
- The electric and magnetic fields are out of phase by pi/2 (90 degree). …
- GUJCET 2015Set C1 markMCQQ.To transmit a signal of 3 KHz frequency, the minimum length of antenna is _____ km (A) 25 (B) 20 (C) 50 (D) 75
›Reveal solutionSolution
[!TLDR]
λ=c/f=100 km; minimum antenna length =λ/4=25 km. Answer: (A).
Concept
To radiate a signal efficiently, an antenna should have a length of at least about a quarter of the signal wavelength, Lmin=λ/4, where λ=c/f (NCERT/CBSE communication systems).
Solution
Wavelength of the 3 kHz signal: …
🎓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.