Electromagnetic Waves
Displacement current, nature of EM waves & the electromagnetic spectrum
Nature of Electromagnetic Waves
Displacement Current and Maxwell's EquationsTopic 1
Ampere's law in its original form related the magnetic field around a loop to the conduction current threading it. Maxwell spotted a gap: during the charging of a capacitor, current flows in the wires but stops at the plates, yet a magnetic field still exists between them. To fix this inconsistency he introduced the displacement current, a current-like term produced by a changing electric field.
The displacement current is $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux. Between capacitor plates, as charge builds up, the changing electric flux produces exactly the displacement current needed to keep the magnetic field continuous. The key NEET idea is that a changing electric field produces a magnetic field, just as a changing magnetic field produces an electric field (Faraday's law).
With this addition, the four Maxwell's equations unify all of electricity and magnetism. In words: electric field lines begin and end on charges (Gauss's law); magnetic field lines form closed loops with no monopoles; a changing magnetic flux induces an electric field (Faraday's law); and a current or a changing electric flux produces a magnetic field (the modified Ampere–Maxwell law).
The profound consequence is that a changing electric field makes a changing magnetic field, which in turn makes a changing electric field, and so on — a self-sustaining disturbance that travels through space as an electromagnetic wave. Maxwell's theory predicted these waves and even their speed, before they were ever observed, a landmark NEET likes to highlight.
| Idea | Statement |
|---|---|
| Displacement current | $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}$ |
| Changing $E$ | produces a magnetic field |
| Changing $B$ | produces an electric field |
What is the source of the displacement current between the plates of a charging capacitor?
Show solution
The changing electric field (and hence changing electric flux) between the plates: $I_d = \varepsilon_0\,d\Phi_E/dt$. No charge actually crosses the gap, yet this term produces a magnetic field there.
Which of Maxwell's equations was modified by adding the displacement current?
Show solution
Ampere's circuital law. The modified Ampere–Maxwell law includes both the conduction current and the displacement current $\varepsilon_0\,d\Phi_E/dt$.
The displacement current arises from a changing:
Maxwell modified which law by adding displacement current?
A changing electric field produces a:
Magnetic field lines, according to Maxwell's equations:
Maxwell's equations predicted the existence of:
NEET tip: Displacement current $I_d = \varepsilon_0\,d\Phi_E/dt$ comes from a changing electric field and completes Ampere's law. Remember the symmetry: changing $E$ makes $B$, changing $B$ makes $E$.
Properties of Electromagnetic WavesTopic 2
In an electromagnetic wave, the electric field $E$ and magnetic field $B$ oscillate perpendicular to each other and to the direction of propagation — so EM waves are transverse. The two fields are in phase, reaching their maxima and zeros together, and the wave travels in the direction given by $E \times B$. This transverse, mutually perpendicular geometry is the single most-tested fact about EM waves.
EM waves need no medium and travel through vacuum, which is how sunlight crosses empty space. Their speed in vacuum is fixed by two fundamental constants: $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^{8}\ \text{m/s}$. The fact that this matched the known speed of light convinced Maxwell that light itself is an electromagnetic wave.
The magnitudes of the two fields are linked by $c = \dfrac{E_0}{B_0}$, so the electric field is numerically much larger than the magnetic field. In a material medium the speed drops to $v = \dfrac{1}{\sqrt{\mu\varepsilon}}$, always less than $c$, with the refractive index $n = c/v$. The frequency stays fixed while the wavelength shortens in a denser medium.
EM waves carry both energy and momentum, shared equally between the electric and magnetic fields. Because they carry momentum, they exert a small radiation pressure on surfaces they strike — the principle behind solar sails. The average energy density and the intensity both depend on the square of the field amplitude, and these waves are produced by accelerating charges, such as electrons oscillating in an antenna.
| Property | Detail |
|---|---|
| Nature | transverse; $E \perp B \perp$ propagation |
| Speed in vacuum | $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$ |
| Field ratio | $c = E_0/B_0$ |
| Carry | energy & momentum (radiation pressure) |
The electric field amplitude of an EM wave is $E_0 = 6\ \text{V/m}$. Find the magnetic field amplitude.
Show solution
$B_0 = \dfrac{E_0}{c} = \dfrac{6}{3\times10^{8}} = 2\times10^{-8}\ \text{T}$.
In which directions do $E$, $B$ and the propagation of an EM wave point relative to one another?
Show solution
All three are mutually perpendicular. $E$ and $B$ oscillate at right angles to each other, and both are perpendicular to the direction of propagation (the direction of $E \times B$).
Electromagnetic waves are:
The speed of EM waves in vacuum is:
In an EM wave, $E$ and $B$ are:
The ratio $E_0/B_0$ for an EM wave in vacuum equals:
EM waves exert radiation pressure because they carry:
NEET tip: EM waves are transverse, need no medium, travel at $c = 1/\sqrt{\mu_0\varepsilon_0}$ with $E_0/B_0 = c$, and carry energy and momentum (radiation pressure). $E$, $B$ and propagation are mutually perpendicular.
The Electromagnetic Spectrum
The Electromagnetic Spectrum and Its OrderTopic 3
All electromagnetic waves are the same kind of disturbance, differing only in frequency and wavelength. Arranged from longest wavelength (lowest frequency) to shortest wavelength (highest frequency), the electromagnetic spectrum runs: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. Remembering this order is essential, because NEET frequently asks you to rank waves by wavelength, frequency or energy.
Since all EM waves travel at $c$ in vacuum, frequency and wavelength are tied by $c = f\lambda$. As you move up the spectrum the wavelength decreases, the frequency increases, and the photon energy $E = hf$ increases. So gamma rays have the shortest wavelength, highest frequency and highest energy, while radio waves are at the opposite extreme — a chain of three inverse/direct relations worth fixing firmly.
Visible light occupies only a tiny band, roughly $400$ to $700\ \text{nm}$, between ultraviolet and infrared. Within it, violet has the shortest wavelength and red the longest — the order VIBGYOR runs from short to long wavelength. The narrowness of the visible band, sandwiched in the middle of a vast spectrum, is itself a point NEET likes to make.
Each band is produced differently and used differently. The higher-energy waves (UV, X-rays, gamma) can ionise atoms and damage tissue, while lower-energy waves are generally safer. Understanding that wavelength, frequency and energy march together — and knowing roughly where each band sits — lets you answer most spectrum questions without memorising exact numbers.
| Spectrum (long to short $\lambda$) | Trend |
|---|---|
| Radio to gamma | $\lambda$ decreases |
| Radio to gamma | $f$ and energy increase |
| Wave relation | $c = f\lambda$, $E = hf$ |
| Visible light | $\sim 400$–$700\ \text{nm}$ (VIBGYOR) |
Arrange in increasing order of frequency: X-rays, radio waves, visible light.
Show solution
Frequency increases as wavelength decreases: radio waves $<$ visible light $<$ X-rays.
Find the frequency of light of wavelength $600\ \text{nm}$.
Show solution
$f = \dfrac{c}{\lambda} = \dfrac{3\times10^{8}}{600\times10^{-9}} = 5\times10^{14}\ \text{Hz}$.
Which has the longest wavelength?
Which has the highest photon energy?
Visible light lies between:
As frequency increases across the spectrum, wavelength:
In the visible band, the longest wavelength colour is:
NEET tip: Memorise the order radio → micro → IR → visible → UV → X-ray → gamma. Across it, $\lambda$ falls while $f$ and energy ($E = hf$) rise. Use $c = f\lambda$ for numericals.
Sources and Uses of Each BandTopic 4
Each region of the spectrum is produced by a characteristic process and put to characteristic uses, and NEET very often tests these source–use pairings directly. Radio waves are produced by oscillating currents in antennas and are used in radio, television and mobile communication. Microwaves come from special vacuum tubes (klystrons, magnetrons) and are used in microwave ovens, radar and satellite communication — their absorption by water molecules is what heats food.
Infrared (IR) radiation is emitted by hot bodies and molecules; it is felt as heat and used in night-vision devices, remote controls and thermal imaging, and it is also responsible for the greenhouse effect that warms the Earth. Visible light, emitted by atoms and hot objects like the Sun and lamps, is the only band our eyes detect and the basis of photography and illumination.
Ultraviolet (UV) is produced by very hot bodies (such as the Sun) and special lamps; it causes sunburn and is used for sterilisation and detecting forgeries, but is largely absorbed by the ozone layer, which protects life on Earth. X-rays are produced when fast electrons strike a metal target and are used in medical imaging and security scanning because they penetrate soft tissue but not bone.
Gamma rays, the most energetic band, come from radioactive nuclei and nuclear reactions; they are used in cancer radiotherapy and to sterilise medical equipment. The general trend NEET expects you to grasp is that the higher-energy bands (UV, X-ray, gamma) are penetrating and potentially harmful, used where their energy is an asset, while the lower-energy bands dominate everyday communication and heating.
| Band | Typical use |
|---|---|
| Radio waves | radio, TV, mobile communication |
| Microwaves | radar, ovens, satellite links |
| Infrared | remote controls, thermal imaging |
| Ultraviolet | sterilisation, detecting forgery |
| X-rays | medical imaging, security |
| Gamma rays | cancer therapy, sterilisation |
Which electromagnetic waves are used in a microwave oven, and why do they heat food?
Show solution
Microwaves. They are strongly absorbed by water molecules in the food, setting them into vigorous motion (rotation), which raises the food's temperature.
Which band is used in medical imaging of bones, and which in cancer radiotherapy?
Show solution
X-rays are used for imaging bones (they penetrate soft tissue but not bone). Gamma rays, being highly energetic, are used to destroy cancerous tissue in radiotherapy.
Microwaves are commonly used in:
Infrared radiation is used in:
X-rays are produced when:
Gamma rays are emitted by:
The ozone layer protects life mainly by absorbing:
NEET tip: Learn one source and one use per band. High-energy bands (UV, X-ray, gamma) are penetrating/harmful; lower-energy bands handle communication and heating. These source–use pairings are direct one-mark questions.
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