Q.The magnetic field in a plane electromagnetic wave is given by By=(2×10−7) Tsin(0.5×103x+1.5×1011t).
🔒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 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 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, …
🎓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.