You have already met half of this idea. In Electromagnetic Induction you learned Faraday’s discovery: change the magnetic flux through a loop and an electric field appears, pushing charge around the loop. Chapter 8 hands you the missing mirror image — change the electric flux through a surface and a magnetic field appears, whether or not there is any wire there. Maxwell spotted that symmetry, and the moment you put the two halves together, a self-sustaining travelling wave falls out of them for free.
This page is a full tutor-style set of Electromagnetic Waves class 12 physics notes for the CBSE 2026-27 session: displacement current explained from the leak it was invented to plug, the characteristics and the transverse nature of electromagnetic waves, the electromagnetic spectrum with a table of uses you can actually revise from, a full formula list, twenty-nine solved examples, and a set of electromagnetic waves numericals with solutions plus important questions with answers hidden behind tap-to-open panels.
Exactly what the 2026-27 syllabus asks for. The prescribed scope for Chapter 8 is the basic idea of displacement current; electromagnetic waves, their characteristics and their transverse nature (qualitative idea only); and the electromagnetic spectrum — radio waves, microwaves, infrared, visible, ultraviolet, X-rays and gamma rays — including elementary facts about their uses. Notice the two words in brackets: qualitative only. Maxwell’s equations as a formal set of four, and the derivation of the wave speed from them, are not examinable material for you. Where this page goes a step past the bare syllabus wording (energy density, intensity, radiation pressure), it says so in an amber note so you always know which side of the line you are standing on.
Meet Your Tutor
Electromagnetic Waves is a compact chapter, but its spectrum can become a blur of orders and uses. I will help you rebuild it from frequency, wavelength and energy, connect each band to a physical source and application, and use quick checks so you can reason out the order instead of reciting it nervously.
What You’ll Learn
- The Gap in Ampere’s Law — Why a New Idea Was Needed
- Displacement Current — The Basic Idea
- What an Electromagnetic Wave Actually Is
- Characteristics of Electromagnetic Waves
- The Transverse Nature of Electromagnetic Waves
- Speed of Light from Free-Space Constants
- Energy, Intensity and Momentum Carried by EM Waves
- The Electromagnetic Spectrum
- Uses of Each Part of the Spectrum — Elementary Facts
- Electromagnetic Waves Class 12 Physics Numericals With Solutions
- Formula List and Quick Revision
Your Game Plan
Six sittings, roughly forty minutes each. Do them in order — this chapter is a chain of reasoning, and skipping ahead to the spectrum table makes displacement current look like an arbitrary rule someone invented on a Tuesday.
- Sitting 1 — find the hole. Work through the charging capacitor until you can say out loud, without notes, exactly why Ampere’s circuital law gives two different answers for the same loop.
- Sitting 2 — plug the hole. Learn Id = ε₀ dΦE/dt and prove to yourself that in a capacitor it comes out exactly equal to the wire current. Do Examples 2 to 4.
- Sitting 3 — build the wave. The regeneration picture, the plane-wave equations, E₀/B₀ = c, and the “E cross B points the way you go” rule. Sketch the wave diagram from memory three times.
- Sitting 4 — the numbers. c = 1/√(μ₀ε₀) as a stated result you can evaluate, speed in a medium, then energy density, intensity and radiation pressure.
- Sitting 5 — the spectrum. Memorise the order with the mnemonic, then learn the table row by row: band, wavelength, how it is made, what it is used for. This is where the easy marks live.
- Sitting 6 — test yourself. Cover the answers and do the ten worksheet questions cold. Anything you miss, go back to that section, not to the answer.
Study Notes
The Gap in Ampere’s Law — Why a New Idea Was Needed
Start with the law you already trust. In Moving Charges and Magnetism you met Ampere’s circuital law: walk right round any closed loop, add up B·dl as you go, and the total equals μ₀ times the current threading that loop. Written out: ∮B·dl = μ₀I. It works beautifully for a long straight wire, a solenoid, a toroid. For decades nobody had reason to doubt it.
Now here is the experiment that broke it. Take a parallel-plate capacitor and connect it to a battery so it is charging. Charge flows along the wires. It piles up on the plates. It does not cross the gap — there is nothing there to carry it. And yet if you put a tiny magnetometer in the gap while the capacitor is charging, you measure a magnetic field. Circling the gap, exactly as if a current were flowing through it.
Why is that fatal? Because a closed loop does not bound just one surface — it bounds infinitely many. Take a loop that encircles the wire feeding the left plate. You could stretch a flat drum-skin across it, and the wire punches through: current threaded = I, so Ampere says ∮B·dl = μ₀I. Or you could stretch a baggy, balloon-shaped surface across the same loop, bulging out sideways so that it slips past the edge of the plate and passes through the gap instead. Now no wire pierces it at all: current threaded = 0, so Ampere says ∮B·dl = 0. Same loop. Same instant. Two different answers.
That is the whole detective story, and it is worth telling yourself in exactly that order, because the fix is then obvious rather than arbitrary. Something must be flowing through the gap in the “current” column of the equation — something that is not moving charge, because there is no charge in the gap. What is in the gap? An electric field, and it is getting stronger every second as the plates charge up. Maxwell’s guess: the changing electric field is the missing current.
Ampere alone. The loop lies entirely in the gap. No conduction current pierces any flat surface stretched across it, so I = 0 and the law gives ∮B·dl = 0, hence B = 0.
What is really there. The changing electric field fills the whole plate area πR², so the displacement current is spread uniformly over it. Inside radius r the enclosed part is Id(enclosed) = I × (r/R)² = 2.0 × (3.0/6.0)² = 2.0 × 0.25 = 0.50 A.
Then μ₀Id(enclosed) = B × 2πr gives
B = μ₀ × 0.50 / (2π × 0.030) = (4π×10⁻⁷ × 0.50) / 0.1885 = 3.3×10⁻⁶ T = 3.3 µT.
Why it works. A field of a few microtesla is small but thoroughly measurable — it is about a fifteenth of the Earth’s magnetic field. Ampere’s law was not off by a rounding error. It predicted zero where nature gives you microtesla. That is why a new term was needed.
Displacement Current — The Basic Idea
Maxwell’s repair is a single extra term. Define the displacement current through a surface as
where ΦE = ∮E·dA is the electric flux through that surface. Read it as a sentence: a rate of change of electric flux counts as a current. The constant ε₀ is there to make the units come out in amperes, and nothing else. Then the repaired law — the Ampere–Maxwell law — is
Now check that it actually cures the disease. For the charging capacitor, the field between the plates is uniform and E = σ/ε₀ = q/(ε₀A), a result you derived in Electrostatic Potential and Capacitance. The flux through a surface filling the gap is therefore
The ε₀ cancels and out drops dq/dt — which is precisely the current in the wire. The displacement current in the gap exactly equals the conduction current in the wire. So whichever surface you stretch across your loop, the total (Ic + Id) threading it is the same, and ∮B·dl has one unambiguous value. The bookkeeping is fixed, and current is now continuous everywhere in the circuit — conduction current in the wires handing over to displacement current in the gap, like a relay baton.
| Feature | Conduction current Ic | Displacement current Id |
|---|---|---|
| What it is | Actual drift of free charge carriers — electrons in a metal, ions in a solution | A term arising from a changing electric field; no charge moves |
| Formula | I = dq/dt = nAvde | I = ε₀ dΦE/dt |
| Needs a medium? | Yes — needs free charges, so a conductor or electrolyte | No — happens quite happily in vacuum |
| When is it zero? | In an insulator or a vacuum gap | When the electric field is constant in time |
| Produces a magnetic field? | Yes | Yes — and of exactly the same kind |
| Heating effect | Yes, I²R heating in the conductor | No resistive heating; it is not a flow of charge |
| In a steady DC circuit | The whole of the current | Zero, because the fields are not changing |
| Inside a charging capacitor | Zero in the gap, I in the wires | Equal to I in the gap, zero in the wires |
Charge and voltage are linked by q = CV, so the current is I = dq/dt = C dV/dt. Since Id = Ic for a capacitor, that is the displacement current as well:
Id = C dV/dt = (100×10⁻¹² F)(5.0×10⁴ V s⁻¹) = 5.0×10⁻⁶ A = 5.0 µA.
Why it works. You never had to touch ε₀ or the plate area. The identity Id = Ic means any ordinary circuit method for finding the capacitor current gives you the displacement current for free. This is the fastest route in almost every exam question of this type.
Here the flux is ΦE = EA with A fixed, so dΦE/dt = A dE/dt and
Id = ε₀A dE/dt = (8.854×10⁻¹²)(1.0×10⁻⁴ m²)(6.0×10¹²)
= 8.854×10⁻¹² × 6.0×10⁸ = 5.31×10⁻³ A = 5.3 mA.
Sanity check on the size. 6×10¹² V m⁻¹ s⁻¹ sounds enormous, but it only means the field climbs by 6 kV mm⁻¹ in a microsecond — entirely ordinary for a fast pulse. A few milliamps is the right order of magnitude.
The displacement current equals the conduction current, so ε₀A dE/dt = 0.25 A, giving
dE/dt = 0.25 / [(8.854×10⁻¹²)(40×10⁻⁴)] = 0.25 / (3.542×10⁻¹⁴) = 7.1×10¹² V m⁻¹ s⁻¹.
Why it works. Same equation, read right to left. Note how the plate area does the heavy lifting: squeeze the same current into smaller plates and the field has to race up faster to keep the flux changing at the required rate.
Inside a good conductor carrying a steady current, the electric field is constant in time (it is E = J/σ, and J is not changing). Constant E means constant flux, so dΦE/dt = 0 and Id = ε₀ dΦE/dt = 0.
Why it works. Displacement current is not a property of wires; it is a property of changing fields. In steady DC there is nothing changing, which is exactly why Ampere’s original law survived unchallenged for so long. Bring in an AC source, as in Alternating Current, and a small displacement current appears inside the wire too — utterly negligible next to the conduction current at mains frequency, but no longer zero.
What an Electromagnetic Wave Actually Is
Once you accept displacement current, something extraordinary follows, and it costs nothing extra. Line up the two laws side by side and read them as instructions.
Ampere–Maxwell law: an electric field that is changing in time creates a magnetic field around it. (No wire required either.)
Read them one after the other and the loop closes. Suppose at some point in empty space you manage to make an electric field that is changing. It creates a magnetic field a little further along. That new magnetic field is itself growing from nothing, so it is changing — and so it creates an electric field a little further along still. Which is also changing, so it creates a magnetic field… The pattern does not need a rope to travel along, or air, or aether. Each field lays the track for the other. That is what I mean by the wave that builds its own road, and it is the single most useful sentence in this chapter.
Where do they come from? A charge sitting still makes a static electric field — no wave. A charge moving at constant velocity makes a steady current and a steady magnetic field — still no wave. It takes an accelerating charge to launch an electromagnetic wave, because only then are the fields changing in a way that keeps changing. The practical version is an oscillating charge: drive charge back and forth in an aerial at frequency f and it radiates an electromagnetic wave of exactly that frequency f. An LC combination or any AC source, of the kind you studied in Alternating Current, is the standard way of doing it — and the frequency of the wave is set by the frequency of the oscillating source, nothing else.
What the wave looks like written down. For a plane wave travelling along the x-axis with its electric field oscillating along y and its magnetic field along z:
Bz = B₀ sin(kx − ωt)
with k = 2π/λ the propagation constant (or wave number, unit rad m⁻¹), ω = 2πf the angular frequency (rad s⁻¹), and the wave speed given by c = ω/k = fλ. Both fields carry the same sine factor, which is the mathematical statement that E and B are in phase: they reach their maxima at the same instant and pass through zero at the same instant.
Ey = 45 sin[2π(x/0.60 − t/2.0×10⁻⁹)] V m⁻¹,
x in metres, t in seconds. Find the wavelength, period, frequency, the wave speed, the peak magnetic field, and the direction of B.
Step 1 — match the pattern. Writing it as sin[2π(x/λ − t/T)] we read straight off λ = 0.60 m and T = 2.0×10⁻⁹ s = 2.0 ns.
Step 2 — frequency. f = 1/T = 1/(2.0×10⁻⁹) = 5.0×10⁸ Hz = 500 MHz.
Step 3 — check the speed. v = fλ = (5.0×10⁸)(0.60) = 3.0×10⁸ m s⁻¹. It is c, as it must be in vacuum. Always do this check — it catches arithmetic slips instantly.
Step 4 — peak B. B₀ = E₀/c = 45/(3.0×10⁸) = 1.5×10⁻⁷ T = 150 nT.
Step 5 — direction. The wave moves along +x (the sign of x and t are opposite inside the bracket) and E is along +y. Since E×B must point along the travel direction, and (+y)×(+z) gives (+x), B oscillates along the +z direction.
Why it works. Everything came from pattern-matching, not from memorising a derivation. If instead the wave is given in the k and ω form, use λ = 2π/k and f = ω/2π; here that would be k = 2π/0.60 = 10.5 rad m⁻¹ and ω = 2π/(2.0 ns) = 3.14×10⁹ rad s⁻¹.
(a) The radiated wave has the same frequency as the source: 10 MHz = 1.0×10⁷ Hz.
(b) λ = c/f = (3.0×10⁸)/(1.0×10⁷) = 30 m.
(c) λ = (3.0×10⁸)/(1.5×10⁶) = 200 m.
Why it works. A transmitting aerial has to be a decent fraction of a wavelength long to radiate efficiently, which is exactly why medium-wave masts are enormous and a mobile-phone antenna hides inside the case — at 900 MHz the wavelength is only 33 cm.
Characteristics of Electromagnetic Waves
The syllabus asks for the characteristics of electromagnetic waves by name, so here they are as a list you can reproduce under pressure. Almost every one of them is a direct consequence of the road-building picture.
- No material medium is needed. They propagate through vacuum, because the fields regenerate each other rather than shaking any substance.
- They travel at c = 3×10⁸ m s⁻¹ in vacuum — the same speed for every frequency, from radio to gamma. In a material medium the speed drops to v = c/n.
- They are transverse. Both E and B are perpendicular to the direction of travel.
- E, B and the direction of propagation are mutually perpendicular and form a right-handed set in the order E, B, direction.
- E and B oscillate in phase — same frequency, same phase, peaks together.
- The ratio of the field magnitudes is fixed: E₀/B₀ = E/B = c. The electric field is numerically vastly larger, which is why the electric part dominates most interactions with matter.
- They are produced by accelerating (in practice, oscillating) charges, and the radiated frequency equals the frequency of oscillation of the source.
- They carry energy and momentum, so they can exert a pressure on a surface they strike.
- They are electrically neutral — they carry no charge, so they are not deflected by external electric or magnetic fields.
- They obey the superposition principle, and can be reflected, refracted, diffracted, interfered and polarised like any other wave.
- Frequency is the fingerprint. When a wave crosses into a new medium its frequency is unchanged (the source has not changed); the speed and hence the wavelength change instead.
E₀ = cB₀ = (3.0×10⁸ m s⁻¹)(510×10⁻⁹ T) = (3.0×10⁸)(5.10×10⁻⁷) = 153 V m⁻¹.
Sanity check. The electric number should come out about 10⁸ times bigger than the magnetic one, and 153 versus 5.1×10⁻⁷ is a factor of 3×10⁸. Correct. If your electric field comes out smaller than your magnetic field, you have divided instead of multiplied.
Frequency in air: f = c/λ = (3.0×10⁸)/(600×10⁻⁹) = 5.0×10¹⁴ Hz.
Speed in glass: v = c/n = (3.0×10⁸)/1.5 = 2.0×10⁸ m s⁻¹.
Wavelength in glass: the frequency cannot change, so λ′ = v/f = (2.0×10⁸)/(5.0×10¹⁴) = 4.0×10⁻⁷ m = 400 nm. Equivalently λ′ = λ/n = 600/1.5 = 400 nm.
Why it works. The source is still wobbling at 5.0×10¹⁴ times a second, and the glass cannot argue with that. So f is fixed and the wavelength must shrink to match the slower speed. This is the reason a ray bends at a boundary, as you saw in Light: Reflection and Refraction. Note the colour of the light — which is set by frequency, not wavelength — does not change under water either.
The relation E = cB holds instant by instant, not just at the peaks:
B = E/c = 24/(3.0×10⁸) = 8.0×10⁻⁸ T = 80 nT.
Why it works. Because E and B are in phase, their ratio is the same constant everywhere in the wave. Had they been out of step, no such instant-by-instant ratio could exist — which is a neat way of seeing why the in-phase property matters.
The Transverse Nature of Electromagnetic Waves
“Transverse” means the thing that is oscillating points across the direction of travel, not along it. In a wave on a string, the string moves up and down while the wave runs sideways — transverse. In a sound wave in air, the air oscillates back and forth along the direction the sound is going — longitudinal. Electromagnetic waves are firmly in the first camp: both E and B point at right angles to the direction of propagation, and at right angles to each other.
The reasoning, in words. Think again about how the wave sustains itself. The changing electric field at one place has to produce a magnetic field encircling it — that is what the Ampere–Maxwell law says, exactly as a current produces a field that loops around the current rather than pointing along it. So B comes out perpendicular to E. Then that changing B has to produce an electric field encircling it, which brings you back to a direction perpendicular to B. The only way to keep handing the baton forward and still keep both fields perpendicular to each other is for the whole arrangement to march along the third, mutually perpendicular direction. There is no room left for a component of either field along the direction of travel. That is the qualitative argument, and it is all you need.
The experimental proof that light is transverse is polarisation — you can filter out one of the two possible transverse orientations, which would be impossible for a longitudinal wave since there is only one direction along the axis. You will meet polarisation properly in the Optics unit; here it is enough to know that it is the evidence.
We need E×B to point along +x, the direction of travel. Set up the right-handed triple: (+y)×(+z) = (+x). So with E along +y, B must be along +z.
Check with your hand. Right hand, fingers along +y, curl them towards +z; the thumb points along +x. Correct.
We need E×B to point along −y. Try (+z)×(+x) = (+y) — wrong sign. So flip the second vector: (+z)×(−x) = −(+y) = −y. Correct. Therefore B points along −x.
Why it works. The three cyclic products worth memorising are (+x)×(+y) = +z, (+y)×(+z) = +x and (+z)×(+x) = +y. Any other combination is one of these with a sign flipped. Two minutes learning those three lines will bank you every direction mark in this chapter.
Compute E×B in direction terms: (−z)×(+y) = −[(+z)×(+y)] = −[−x] = +x. The wave travels along +x.
Trick used. (+z)×(+y) is the reverse of (+y)×(+z) = +x, so it equals −x. Reversing the order of a cross product reverses its sign. Then the extra minus sign from E flips it back to +x.
Speed of Light from Free-Space Constants
Here is the result that made Maxwell famous, and the moment electricity and light became the same subject. Work out how fast the self-regenerating field pattern must travel, and the answer comes out in terms of two constants that had been measured in laboratories with batteries and coils, with nothing to do with light at all:
Both constants you already know. μ₀ = 4π×10⁻⁷ T m A⁻¹ is the permeability of free space, which turned up in every magnetic-field formula in Moving Charges and Magnetism. ε₀ = 8.854×10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space, from Coulomb’s law. Neither was measured using light. Put the numbers in:
1/(μ₀ε₀) = 8.9877×10¹⁶
c = √(8.9877×10¹⁶) = 2.998×10⁸ m s⁻¹ ≈ 3×10⁸ m s⁻¹
That is the measured speed of light to four significant figures, obtained from a coil constant and a capacitor constant. Maxwell’s conclusion was the only one available: light is an electromagnetic wave. Every other band in the spectrum was then predicted before it was found.
In a material medium, replace the vacuum constants by the medium’s own:
and the refractive index is n = c/v = √(μrεr)
For most transparent materials μr is very close to 1, so this simplifies to the handy n ≈ √εr. Since v is always less than c in matter, n is always greater than 1.
Step 1. 4π = 12.566, so μ₀ = 1.2566×10⁻⁶.
Step 2. Multiply: (1.2566)(8.854) = 11.126, and 10⁻⁶×10⁻¹² = 10⁻¹⁸, so μ₀ε₀ = 11.126×10⁻¹⁸ = 1.1126×10⁻¹⁷.
Step 3. Reciprocal: 1/1.1126 = 0.8988, and 1/10⁻¹⁷ = 10¹⁷, so we get 0.8988×10¹⁷ = 8.988×10¹⁶.
Step 4. Square root. Write the exponent as even: 89.88×10¹⁵ is awkward, so use 8.988×10¹⁶ directly: √8.988 = 2.998 and √10¹⁶ = 10⁸.
c = 2.998×10⁸ ≈ 3.0×10⁸ m s⁻¹.
Exam-craft note. The step people fumble is the square root of a power of ten. Always rewrite the number so that the exponent is even before taking the root; then halve the exponent and root the mantissa. Here 10¹⁶ is already even, giving 10⁸ cleanly.
√(μrεr) = √(1.0 × 4.0) = 2.0.
v = c/2.0 = (3.0×10⁸)/2.0 = 1.5×10⁸ m s⁻¹, and n = c/v = 2.0.
Sanity check. n = 2.0 is a stiff but perfectly real refractive index — diamond is about 2.42. If you ever compute n < 1 or v > c for an ordinary medium, you have inverted the formula.
Energy, Intensity and Momentum Carried by EM Waves
An electromagnetic wave carries energy — you can feel it as warmth on your face in sunlight. That energy is stored in the two fields, and it is worth seeing that the two shares are exactly equal.
magnetic energy density uB = B² / 2μ₀
Substitute B = E/c and then c² = 1/(μ₀ε₀):
So the electric and magnetic halves of an electromagnetic wave carry exactly equal energy, at every instant. That is a satisfying result and a favourite one-mark question. The total instantaneous density is therefore u = uE + uB = ε₀E² = B²/μ₀.
Averaging sin² over a cycle gives ½, so the time-averaged densities are ⟨uE⟩ = ¼ε₀E₀² and ⟨uB⟩ = B₀²/4μ₀, and the average total energy density is
Intensity is the average power crossing unit area held perpendicular to the beam. Since the energy in a column of length c sweeps past in one second, I = ⟨u⟩c:
Momentum and radiation pressure. An electromagnetic wave carries momentum as well as energy, in the ratio p = U/c for energy U. If a surface absorbs the wave completely it takes all that momentum, and the pressure it feels is
The factor of two for a mirror is the same idea as a ball bouncing off a wall rather than sticking to it: the momentum change is doubled, so the force is doubled.
B₀ = E₀/c = 48/(3.0×10⁸) = 1.6×10⁻⁷ T = 160 nT.
uE(peak) = ½ε₀E₀² = ½(8.854×10⁻¹²)(48)² = ½(8.854×10⁻¹²)(2304) = 1.02×10⁻⁸ J m⁻³.
uB(peak) = B₀²/2μ₀ = (1.6×10⁻⁷)²/[2(1.2566×10⁻⁶)] = (2.56×10⁻¹⁴)/(2.513×10⁻⁶) = 1.02×10⁻⁸ J m⁻³.
Honest note on the last digit. Work to more figures and the two come out 1.0200×10⁻⁸ and 1.0186×10⁻⁸ — a gap of 0.14%. That is not physics; it is purely because we rounded c to 3.0×10⁸ instead of 2.998×10⁸. Use the exact c and they agree to every digit your calculator will show. In an exam, quote both as 1.02×10⁻⁸ J m⁻³ and state that they are equal.
Route 1, peak field: I = ½ε₀cE₀² = ½(8.854×10⁻¹²)(3.0×10⁸)(2304) = 3.06 W m⁻².
Route 2, rms field: Erms = 48/√2 = 33.9 V m⁻¹, so I = ε₀cErms² = (8.854×10⁻¹²)(3.0×10⁸)(33.9)² = 3.06 W m⁻². Same answer, as it must be.
Sanity check. 3 W m⁻² is roughly what you get a couple of metres from a bright reading lamp — about 1/500 of full sunlight. Reasonable.
(a) From I = ½ε₀cE₀², rearranged: E₀ = √(2I/ε₀c) = √[2(1400)/((8.854×10⁻¹²)(3.0×10⁸))] = √(2800/2.656×10⁻³) = √(1.054×10⁶) = 1.03×10³ V m⁻¹.
(b) B₀ = E₀/c = (1.027×10³)/(3.0×10⁸) = 3.4×10⁻⁶ T = 3.4 µT.
(c) Prad = I/c = 1400/(3.0×10⁸) = 4.7×10⁻⁶ Pa.
(d) F = PradA = (4.67×10⁻⁶)(1.0) = 4.7×10⁻⁶ N.
Two sanity anchors worth remembering. The peak magnetic field of sunlight, 3.4 µT, is about a fifteenth of the Earth’s magnetic field (roughly 50 µT) — small, but the same ballpark. And the radiation pressure, 4.7×10⁻⁶ Pa, is about 5×10⁻¹¹ of atmospheric pressure (1.0×10⁵ Pa). That is why you cannot feel sunlight pushing you, and also why a solar sail has to be the size of a football pitch to be any use.
Intensity. A = 2.0 mm² = 2.0×10⁻⁶ m², so
I = P/A = 30/(2.0×10⁻⁶) = 1.5×10⁷ W m⁻².
Absorbing surface. Prad = I/c = (1.5×10⁷)/(3.0×10⁸) = 0.050 Pa.
Mirror. Prad = 2I/c = 0.10 Pa.
Force on the mirror. F = (2I/c)A = 2P/c = 2(30)/(3.0×10⁸) = 2.0×10⁻⁷ N = 0.20 µN.
Neat shortcut. Notice F = 2P/c fell out without needing the area at all — the area cancels between I = P/A and F = PradA. Whenever a question gives you total beam power and asks for total force, go straight to F = P/c (absorbing) or F = 2P/c (reflecting) and skip the intensity entirely.
From I = ε₀cErms²:
Erms = √[I/(ε₀c)] = √[1200/(2.656×10⁻³)] = √(4.517×10⁵) = 672 V m⁻¹.
Brms = Erms/c = 672/(3.0×10⁸) = 2.24×10⁻⁶ T.
Cross-check. The average energy density is ⟨u⟩ = I/c = 1200/(3.0×10⁸) = 4.0×10⁻⁶ J m⁻³. And ε₀Erms² = (8.854×10⁻¹²)(672)² = 4.0×10⁻⁶ J m⁻³. They match, so the arithmetic is sound.
The Electromagnetic Spectrum
There is only one kind of electromagnetic wave. What we call “radio” and “X-rays” are the same physical thing at different frequencies, exactly as a bass note and a piccolo note are both sound. They all travel at c in vacuum, they are all transverse, they all have E ⊥ B ⊥ direction, and they are all linked by the same c = fλ. The only differences are the wavelength, how we happen to produce them, and how they interact with matter.
“Really Mighty Inventions Very Useful, eXtremely Good”
and pair it with the physical trend: reading left to right the wavelength shrinks, while the frequency and the photon energy climb. Get the order right and half the spectrum questions in any paper are already answered.
The photon energy climbs with frequency, and that single fact explains the whole change in character across the table. Radio photons carry around 10⁻⁶ eV — far too little to disturb an atom. Visible photons carry 1.8 to 3.1 eV, which is just right for nudging an outer electron, which is exactly why atoms emit and absorb visible light and why our eyes evolved to use this band. X-ray photons carry thousands of eV, enough to tear electrons off atoms outright — hence “ionising radiation”, hence the lead apron at the dentist. Same wave, different punch per photon.
| Band | Wavelength range (approx.) | Frequency range (approx.) | How it is produced | Elementary facts about uses |
|---|---|---|---|---|
| Radio waves | longer than about 0.1 m (metres to kilometres) | below about 3×10⁹ Hz (a few kHz up to ~3 GHz) | Accelerated charges in conducting wires — an oscillating LC or AC circuit driving an aerial | AM and FM radio, television broadcasting, mobile telephony and Wi-Fi at the short end, radio astronomy, long-distance communication by reflection from the ionosphere |
| Microwaves | about 0.1 m down to 1 mm | about 3×10⁹ to 3×10¹¹ Hz | Special vacuum tubes — klystrons, magnetrons and Gunn diodes | Radar and aircraft navigation, police speed guns, microwave ovens (2.45 GHz, which sets water molecules rotating so food heats from within), satellite communication |
| Infrared | about 1 mm down to 700 nm | about 3×10¹¹ to 4.3×10¹⁴ Hz | Hot bodies, and vibrating or rotating molecules — every warm object radiates it | TV and appliance remote controls, thermal imaging and night-vision cameras, physiotherapy heat lamps, optical-fibre communication, weather and crop remote sensing from satellites; also the band trapped by greenhouse gases, which is why the atmosphere keeps the Earth warm |
| Visible light | about 700 nm (red) down to 400 nm (violet) | about 4.3×10¹⁴ to 7.5×10¹⁴ Hz | Rearrangement of outer electrons in atoms and molecules — the Sun, filament lamps, LEDs, flames | Human vision, photography, illumination, photosynthesis, optical instruments and everything in Ray Optics |
| Ultraviolet | about 400 nm down to 1 nm | about 7.5×10¹⁴ to 3×10¹⁷ Hz | The Sun, mercury-vapour and arc lamps, and very hot bodies | Sterilising drinking water and surgical instruments, LASIK eye surgery, checking banknotes and security marks for forgery, forming vitamin D in skin; harmful in excess (sunburn, cataracts), and mostly absorbed by the ozone layer before it reaches the ground |
| X-rays | about 1 nm down to about 10⁻³ nm (1 pm) | about 3×10¹⁷ to 3×10²⁰ Hz | Sudden deceleration of high-energy electrons striking a metal target in an X-ray tube | Medical radiography of bones and teeth, radiotherapy for cancer, detecting flaws and cracks in metal castings, airport and baggage security scanning, studying crystal structure |
| Gamma rays | shorter than about 1 pm, down to 10⁻¹⁶ m and beyond | above about 3×10²⁰ Hz | Transitions inside atomic nuclei — radioactive decay and nuclear reactions | Destroying cancer cells in radiotherapy, sterilising surgical instruments and preserving food, imaging in nuclear medicine, industrial thickness gauging |
(a) λ = c/f = (3.0×10⁸)/(1.00×10⁸) = 3.0 m — radio.
(b) λ = (3.0×10⁸)/(2.45×10⁹) = 0.122 m = 12.2 cm — microwave.
(c) λ = (3.0×10⁸)/(9.00×10⁸) = 0.333 m = 33.3 cm — the short (UHF) end of the radio band, bordering on microwaves.
Why it works. One relation, three answers. Also notice how useful the oven number is as a memory anchor: a few centimetres to a few tens of centimetres is the microwave band, which is why the holes in the mesh on the oven door (a millimetre or two) are far too small for the waves to escape through.
Frequency. f = c/λ = (3.0×10⁸)/(0.10×10⁻⁹) = (3.0×10⁸)/(1.0×10⁻¹⁰) = 3.0×10¹⁸ Hz.
Photon energy. E = hf = (6.63×10⁻³⁴)(3.0×10¹⁸) = 1.99×10⁻¹⁵ J.
In eV: (1.99×10⁻¹⁵)/(1.60×10⁻¹⁹) = 1.24×10⁴ eV ≈ 12.4 keV.
Sanity check and a shortcut. Diagnostic X-rays really do run at tens of keV, so 12 keV is right. The shortcut worth memorising: a photon of wavelength 1 nm carries about 1240 eV, and energy is inversely proportional to wavelength — so 0.1 nm gives ten times as much, 12 400 eV. Try it on visible light: 1240/500 ≈ 2.5 eV at 500 nm. Correct.
(a) 500 nm lies between 400 and 700 nm: visible — and towards the green.
(b) λ = c/f = (3.0×10⁸)/(3.0×10¹⁹) = 1.0×10⁻¹¹ m = 10 pm. That is below the 1 pm–1 nm X-ray window at the short end, so it is in the hard X-ray / gamma-ray overlap region; if the question tells you it came from a nucleus, call it gamma, and if from an electron beam, call it an X-ray.
(c) 2.0 cm = 2.0×10⁻² m, between 1 mm and 0.1 m: microwave, with f = (3.0×10⁸)/(0.020) = 1.5×10¹⁰ Hz = 15 GHz.
Why it works. Convert to metres first, every time, then compare with the powers of ten in the table. Part (b) is the honest answer to a genuinely ambiguous question — and saying so scores better than picking one at random.
Uses of Each Part of the Spectrum — Elementary Facts
The syllabus explicitly asks for elementary facts about their uses, and this is the most reliably examined material in the chapter. Questions look like “name the electromagnetic radiation used to detect fractures in bones” or “which radiation is used in remote controls, and state one other use”. Learn each band with one signature use and two backups, and always be ready to say why that band is the right tool for the job — the reason is nearly always its wavelength or its photon energy.
- Radio waves — signature use: broadcasting and communication (AM, FM, television, mobile telephony). Why radio? The wavelengths are metres to kilometres, so they diffract round hills and buildings instead of being blocked by them, and the long waves reflect off the ionosphere, which is how a short-wave signal reaches the other side of the world.
- Microwaves — signature use: radar, plus microwave ovens and satellite links. Why microwaves? Their short wavelength travels in a tight straight beam that can be aimed and timed, which is what radar needs; and 2.45 GHz happens to be absorbed efficiently by water molecules, which is what cooks food from the inside out.
- Infrared — signature use: remote controls and thermal imaging, plus physiotherapy heat lamps and optical-fibre communication. Why infrared? Every warm object emits it, so a camera sensitive to infrared can see a person in total darkness; and it is absorbed by water and organic molecules as heat, so it warms tissue gently. The infrared trapped by CO₂ and water vapour is the greenhouse effect.
- Visible light — signature use: vision, plus photography, illumination and photosynthesis. Why this narrow band? Its photon energies (about 1.8 to 3.1 eV) are exactly what it takes to trigger a chemical change in the pigment in a retina or a leaf, and it happens to be the band the Sun pours out most strongly and the atmosphere lets through. If you want to see what this band then does in lenses, mirrors and rainbows, that is Light: Reflection and Refraction and The Human Eye and the Colourful World.
- Ultraviolet — signature use: sterilisation of water and surgical instruments, plus LASIK surgery and detecting forged notes. Why UV? Its photons carry enough energy (a few eV upwards) to break the chemical bonds in bacterial DNA, which is what kills the microbes — and the same energy is what burns your skin and damages the lens of the eye. The ozone layer absorbs most of it before it reaches us.
- X-rays — signature use: medical radiography of bones and teeth, plus radiotherapy, industrial flaw detection and security scanning. Why X-rays? Photon energies of thousands of electronvolts let them pass through soft tissue while being absorbed by denser bone, so you get a shadow picture; and their wavelength is comparable to atomic spacings, which is why they can also map crystals.
- Gamma rays — signature use: cancer radiotherapy, plus sterilising medical equipment, preserving food and nuclear-medicine imaging. Why gamma? The highest photon energies of all, so they penetrate deep into tissue and destroy cells — useful when the cells you want destroyed are a tumour, dangerous otherwise.
(a) X-rays — they pass through fabric and plastic but are absorbed by dense metal, giving a shadow image.
(b) Infrared — cheap to generate with an LED, harmless, and it does not pass through walls, so it will not switch on the neighbour’s television.
(c) Ultraviolet — its photon energy is enough to break the bonds in bacterial DNA.
(d) Microwaves — a narrow beam that reflects off the car, and the Doppler shift in the returned frequency gives the speed.
Exam-craft note. Answers to “name the radiation” questions are worth one mark for the band and often one more for the reason. Always add the reason, even when it is not asked — it costs one line.
f = c/λ = (3.0×10⁸)/(940×10⁻⁹) = 3.2×10¹⁴ Hz.
Using the 1 nm → 1240 eV shortcut: E = 1240/940 = 1.3 eV.
940 nm is longer than the 700 nm red edge, so it lies just outside the visible band: near infrared. Its photon energy, 1.3 eV, is just below the 1.8 eV needed for red light — which is exactly why your eye cannot see the LED flashing.
Why it works. The remote sits deliberately just past the red edge: near enough that cheap silicon detectors respond strongly, far enough that the flashes are invisible and do not annoy you.
Electromagnetic Waves Class 12 Physics Numericals With Solutions
The numericals in this chapter are short, and they nearly all reduce to one of five moves: c = fλ, E₀ = cB₀, Id = ε₀ dΦE/dt (or C dV/dt), I = ½ε₀cE₀², and Prad = I/c. What actually loses marks is powers of ten. Work through these four with a pencil, then do the worksheet cold.
(a) Plate area A = πR² = π(0.050)² = 7.854×10⁻³ m².
Since Id = Ic = ε₀A dE/dt,
dE/dt = 0.50 / [(8.854×10⁻¹²)(7.854×10⁻³)] = 0.50/(6.954×10⁻¹⁴) = 7.2×10¹² V m⁻¹ s⁻¹.
(b) The displacement current is spread evenly over the plate area, so the fraction enclosed by a circle of radius r = 2.0 cm is (r/R)²:
Id(enclosed) = 0.50 × (2.0/5.0)² = 0.50 × 0.16 = 0.080 A.
B(2πr) = μ₀ × 0.080, so B = (4π×10⁻⁷)(0.080)/(2π×0.020) = 8.0×10⁻⁷ T = 0.80 µT.
(c) At r = R the whole displacement current is enclosed:
B = μ₀Ic/(2πR) = (4π×10⁻⁷)(0.50)/(2π×0.050) = 2.0×10⁻⁶ T = 2.0 µT.
Pattern to notice. Inside the plates B grows in proportion to r (B ∝ r), peaks at the edge, and outside falls off as 1/r — exactly the same shape as the field of a thick current-carrying wire in Moving Charges and Magnetism. That parallel is not a coincidence: displacement current behaves like a real current in every way that matters to the magnetic field.
(a) t = d/c = (3.6×10⁷ m)/(3.0×10⁸ m s⁻¹) = 0.12 s = 120 ms.
(b) The signal must go up and come down again: t = 2(0.12) = 0.24 s = 240 ms.
(c) t = (3.84×10⁸)/(3.0×10⁸) = 1.28 s.
Why it works. All electromagnetic waves travel at c, so “how long” questions are pure distance-over-speed. The 240 ms figure is real and audible — it is why satellite interviews have that awkward pause before the answer comes back.
(a) λ = c/f = (3.0×10⁸)/(9.0×10⁸) = 0.33 m.
(b) λ/2 = 0.167 m = 16.7 cm.
Sanity check against the real world. A handset antenna is a few centimetres long, and it is folded and loaded to fit inside the case, but 16.7 cm is the right ballpark for the raw half-wave length — and it explains why a 2G mast is a manageable object while an AM mast at 1 MHz (λ = 300 m) is a guyed tower you can see from the next town.
(a) P = IA = (1.0×10³)(2.0) = 2.0×10³ W = 2.0 kW.
(b) U = Pt = (2.0×10³)(3600) = 7.2×10⁶ J (= 2.0 kWh).
(c) p = U/c = (7.2×10⁶)/(3.0×10⁸) = 2.4×10⁻² kg m s⁻¹.
(d) F = p/t = (2.4×10⁻²)/3600 = 6.7×10⁻⁶ N. Check it directly: F = IA/c = (1.0×10³)(2.0)/(3.0×10⁸) = 6.7×10⁻⁶ N. Agreed.
The physics in the contrast. Two kilowatt-hours of energy, and a force smaller than the weight of a grain of sand. Energy and momentum in an electromagnetic wave differ by a factor of c, and c is a very big number — which is why light is useful for carrying energy and information, and almost useless for pushing things.
Formula List and Quick Revision
Everything in Chapter 8 that is worth writing on a revision card, in one place. Nothing here needs deriving — these are results to know, recognise and substitute into.
| Quantity | Relation | What to watch for |
|---|---|---|
| Displacement current | Id = ε₀ dΦE/dt | Unit: ampere. Zero whenever the electric field is steady. |
| Displacement current in a capacitor | Id = Ic = C dV/dt = ε₀A dE/dt | The fastest route in most problems. Area A must be in m². |
| Ampere–Maxwell law | ∮B·dl = μ₀(Ic + Id) | Always the total current threading the loop. |
| Speed in vacuum | c = 1/√(μ₀ε₀) = 3×10⁸ m s⁻¹ | A stated result — not to be derived. Rewrite the exponent as even before taking the root. |
| Speed in a medium | v = c/√(μrεr), n = c/v = √(μrεr) | n is always > 1. For most transparent media μr ≈ 1. |
| Wave relation | c = fλ in vacuum; v = fλ′ in a medium | f never changes on entering a new medium; λ and v do. λ′ = λ/n. |
| Plane wave | E = E₀ sin(kx − ωt), B = B₀ sin(kx − ωt) | Same sine factor in both ⇒ E and B are in phase. k = 2π/λ, ω = 2πf, c = ω/k. |
| Field ratio | E/B = c, so E₀ = cB₀ and Erms = cBrms | Holds at every instant, not just at the peaks. |
| Geometry | E ⊥ B ⊥ direction of propagation, right-handed in that order | “E cross B points the way you go.” |
| Energy densities | uE = ½ε₀E², uB = B²/2μ₀, and uE = uB | The two halves are exactly equal — a standard one-mark question. |
| Average energy density | ⟨u⟩ = ½ε₀E₀² = B₀²/2μ₀ = I/c | The averaging factor ½ comes from ⟨sin²⟩ = ½. |
| Intensity | I = ⟨u⟩c = ½ε₀cE₀² = ε₀cErms² | ½ with peak field; no ½ with rms field. Never both. |
| Momentum | p = U/c | U is the energy delivered. Follow with F = p/t if a force is wanted. |
| Radiation pressure | Prad = I/c (absorbed), 2I/c (perfectly reflected) | Total force on a beam target: F = P/c or 2P/c, and the area cancels. |
| Photon energy (cross-link) | E = hf = hc/λ; 1 nm ↔ about 1240 eV | Useful for comparing bands. Energy → up as wavelength → down. |
Quick revision — the ten lines that carry the chapter.
- Ampere’s law fails for a charging capacitor because two surfaces on the same loop give different currents; displacement current fixes it.
- Id = ε₀ dΦE/dt, and in a capacitor it equals the wire current exactly, so total current is continuous.
- A changing E makes a B; a changing B makes an E; together they sustain a wave in vacuum — light needs no medium.
- Electromagnetic waves are produced by accelerating (oscillating) charges, and radiate at the source frequency.
- They are transverse, with E ⊥ B ⊥ direction of travel, right-handed, and E and B in phase.
- E₀/B₀ = c, so the electric field number is always about 10⁸ times the magnetic one.
- c = 1/√(μ₀ε₀) ≈ 3×10⁸ m s⁻¹ — know it, do not derive it; v = c/n in a medium.
- Both fields carry equal energy; I = ½ε₀cE₀²; radiation pressure is I/c, doubled for a mirror.
- Spectrum order: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma — “Really Mighty Inventions Very Useful, eXtremely Good”.
- One use each: broadcasting, radar, remote controls and thermal imaging, vision, sterilisation, radiography, radiotherapy.
Practice Worksheet
Ten original questions in the style CBSE actually sets, mixing one-markers, direction reasoning and full numericals. Do them on paper with the answers covered — opening a panel before you have committed to an answer teaches you nothing. Take ε₀ = 8.854×10⁻¹² F m⁻¹, μ₀ = 4π×10⁻⁷ T m A⁻¹ and c = 3×10⁸ m s⁻¹ throughout.
Q1. The voltage across a 250 pF capacitor is rising steadily at 1.2×10⁵ V s⁻¹. Find the displacement current in the gap between its plates. (2 marks)
Id = (250×10⁻¹² F)(1.2×10⁵ V s⁻¹) = 3.0×10⁻⁵ A.
Id = 3.0×10⁻⁵ A = 30 µA.
You never need ε₀ or the plate area here, because the displacement current in the gap is equal to the conduction current in the wires, and that is just C dV/dt.
Q2. A plane electromagnetic wave in vacuum has a peak magnetic field of 2.0×10⁻⁸ T and a frequency of 6.0×10⁸ Hz. Find (a) the peak electric field and (b) the wavelength. Name the band. (3 marks)
(b) λ = c/f = (3.0×10⁸)/(6.0×10⁸) = 0.50 m.
A wavelength of 0.50 m and a frequency of 600 MHz put this in the radio band, at its short (UHF) end where it borders the microwaves — roughly where terrestrial television broadcasts sit.
Q3. A parallel-plate capacitor has circular plates of radius 5.0 cm. The electric field between them is increasing at 1.0×10¹² V m⁻¹ s⁻¹. Find the displacement current. (3 marks)
Id = ε₀A dE/dt = (8.854×10⁻¹²)(7.854×10⁻³)(1.0×10¹²)
= (6.954×10⁻¹⁴)(1.0×10¹²) = 6.95×10⁻² A.
Id = 7.0×10⁻² A ≈ 70 mA.
Watch the radius: 5.0 cm is 0.050 m, and squaring it gives 2.5×10⁻³, not 25.
Q4. A laser delivers 5.0 mW into a beam of cross-section 1.0 mm². Find (a) the intensity, (b) the peak electric field, (c) the peak magnetic field, and (d) the radiation pressure on a totally absorbing surface. (5 marks)
(b) From I = ½ε₀cE₀²: E₀ = √(2I/ε₀c) = √[(2)(5.0×10³)/(2.656×10⁻³)] = √(3.765×10⁶) = 1.9×10³ V m⁻¹.
(c) B₀ = E₀/c = (1.940×10³)/(3.0×10⁸) = 6.5×10⁻⁶ T = 6.5 µT.
(d) Prad = I/c = (5.0×10³)/(3.0×10⁸) = 1.7×10⁻⁵ Pa.
Note that ε₀c = (8.854×10⁻¹²)(3.0×10⁸) = 2.656×10⁻³ is worth computing once and reusing — it appears in every intensity problem.
Q5. A perfectly reflecting mirror of area 0.50 m² is held facing a beam of intensity 6.0×10² W m⁻². Find the radiation pressure and the total force on the mirror. How would each answer change if the mirror were painted matt black? (4 marks)
F = PradA = (4.0×10⁻⁶)(0.50) = 2.0×10⁻⁶ N = 2.0 µN.
Painted matt black the surface absorbs instead of reflecting, so Prad = I/c and both answers are halved: 2.0×10⁻⁶ Pa and 1.0×10⁻⁶ N. A mirror gets twice the push because the light’s momentum is reversed rather than merely stopped — the same reason a bouncing ball hits a wall harder than a lump of clay of equal momentum.
Q6. An electromagnetic wave travels through a medium of relative permittivity 2.25 and relative permeability 1.0. Find (a) the speed of the wave, (b) the refractive index, and (c) the wavelength in the medium of light whose wavelength in vacuum is 600 nm. Does the frequency change? (4 marks)
(b) n = c/v = 1.5 (equally, n = √(μrεr) = 1.5).
(c) λ′ = λ/n = 600/1.5 = 400 nm.
The frequency does not change. It is fixed by the source: f = c/λ = (3.0×10⁸)/(600×10⁻⁹) = 5.0×10¹⁴ Hz in vacuum, and in the medium f = v/λ′ = (2.0×10⁸)/(400×10⁻⁹) = 5.0×10¹⁴ Hz. Identical.
Q7. Name the band and give the frequency for each wavelength: 3.0 m, 3.0 cm, 3.0 µm, 30 nm, 30 pm. Arrange them in order of increasing photon energy. (5 marks)
| 3.0 m | f = 1.0×10⁸ Hz | radio |
| 3.0 cm = 3.0×10⁻² m | f = 1.0×10¹⁰ Hz | microwave |
| 3.0 µm = 3.0×10⁻⁶ m | f = 1.0×10¹⁴ Hz | infrared |
| 30 nm = 3.0×10⁻⁸ m | f = 1.0×10¹⁶ Hz | ultraviolet |
| 30 pm = 3.0×10⁻¹¹ m | f = 1.0×10¹⁹ Hz | X-ray |
radio (3.0 m) < microwave (3.0 cm) < infrared (3.0 µm) < ultraviolet (30 nm) < X-ray (30 pm). Using E ≈ 1240 eV/λ(nm) the energies are about 4×10⁻⁷ eV, 4×10⁻⁵ eV, 0.41 eV, 41 eV and 4.1×10⁴ eV.
Q8. A plane electromagnetic wave in vacuum has a peak electric field of 30 V m⁻¹. Find (a) the peak magnetic field, (b) the peak electric and magnetic energy densities and comment on them, (c) the average total energy density, and (d) the intensity. (5 marks)
(b) uE = ½ε₀E₀² = ½(8.854×10⁻¹²)(900) = 3.98×10⁻⁹ J m⁻³;
uB = B₀²/2μ₀ = (1.0×10⁻¹⁴)/(2.513×10⁻⁶) = 3.98×10⁻⁹ J m⁻³.
They are equal — the electric and magnetic halves of an electromagnetic wave always carry the same energy. (Any difference in the last digit is only the rounding of c to 3.0×10⁸.)
(c) ⟨u⟩ = ½ε₀E₀² = 3.98×10⁻⁹ J m⁻³. (Each field averages half its peak density, and there are two fields, so the average total equals the peak density of one of them.)
(d) I = ⟨u⟩c = (3.98×10⁻⁹)(3.0×10⁸) = 1.2 W m⁻².
Q9. (a) Explain why a beam of light passing between the poles of a strong magnet is not deflected, although a beam of electrons would be. (b) Explain why displacement current is called a “current” even though no charge moves. (4 marks)
(b) Because of the role it plays, not the mechanism behind it. Maxwell found that a changing electric flux produces a magnetic field in exactly the same way a conduction current does, and that the quantity ε₀ dΦE/dt has to be added to the conduction current in Ampere’s law for the law to be consistent. It has the units of current (amperes), it enters the equation in the current slot, and it makes the total current continuous through a capacitor gap. So it is named a current by its function. No charge crosses the gap.
Q10. Compare the photon energies at the two edges of the visible band, 400 nm and 700 nm, in eV, and use the result to explain why ultraviolet light can cause sunburn but red light cannot. (h = 6.63×10⁻³⁴ J s, 1 eV = 1.60×10⁻¹⁹ J.) (4 marks)
For 700 nm: E = (6.63×10⁻³⁴)(3.0×10⁸)/(7.0×10⁻⁷) = 2.84×10⁻¹⁹ J = 1.8 eV.
The ratio is 700/400 = 1.75, so violet photons carry 1.75 times the energy of red ones.
Photochemical damage is a per-photon effect: a single photon must carry enough energy to break a chemical bond. Red photons at 1.8 eV cannot, however many of them arrive — a bright red lamp just warms you. Ultraviolet photons lie beyond 3.1 eV and keep climbing, and somewhere past the violet edge they carry enough to damage the molecules in skin cells. That is the whole reason the danger begins just outside the band we can see, and it is why sunscreen is rated for UV and not for brightness.
Kaizen — one small improvement at a time. You do not need a heroic study session to own this chapter. Tonight, learn the seven bands in order and say them out loud. Tomorrow, add one use to each. The day after, do three c = fλ conversions in your head before breakfast. Small, honest, repeated steps beat any all-nighter — and this is the one chapter in Class 12 Physics where a week of five-minute habits genuinely finishes the job.
Continue the Physics sequence: connect field ideas through Electric Charges and Fields, revise circuit foundations in Current Electricity, then compare wave behaviour in Ray Optics and Optical Instruments and applications in Semiconductor Electronics.