Electromagnetic Waves | ISC Class 12 Physics Notes

This note covers displacement current, the production and transverse nature of electromagnetic waves, their speed and field relationships, the electromagnetic spectrum, the sources and detection of its different regions, their properties and uses, and simple calculations involving wavelength, frequency and field strength.

Why is displacement current needed?

An electric field describes the electric force per unit positive test charge. A magnetic field describes magnetic influence, including the force on moving charges. An electric current produces a magnetic field, and a changing magnetic field produces an electric field.

Maxwell recognised the converse connection: a changing electric field also produces a magnetic field. This connection is essential when considering a capacitor, a device that stores charge on separated conductors, while it is being charged.

What happens in a charging capacitor?

Conduction current is current associated with the flow of charges through a conductor. In an ideal parallel-plate capacitor circuit, charges flow through the connecting wires and accumulate on the plates. They do not flow across the insulating gap between the plates.

Nevertheless, the charge accumulating on the plates changes the electric field between them. The associated contribution to the magnetic field is described by displacement current. It is not a stream of charges crossing that gap.

Definition: Displacement current is the current term associated with a changing electric flux. It acts as a source of magnetic field, just as conduction current does.

Ampere's circuital law originally related the magnetic field around a closed loop to the conduction current through a surface bounded by that loop. A surface cutting the wire includes current; a surface passing between the plates includes no conduction current.

Those choices cannot give different magnetic fields around the same loop. Maxwell's correction counts displacement current as well. The Ampere-Maxwell law therefore uses total current: conduction current plus displacement current.

Three drawings show a parallel-plate capacitor and the same circular loop. The second uses a pot-shaped surface; the third uses a surface with a flat circular bottom between the plates. Arrows show the electric field across the plate gap.

See Fig. 8.1 in your NCERT textbook

How does the current remain consistent?

Let i be the total current, i꜀ the conduction current and iᵈ the displacement current. Then i = i꜀ + iᵈ. These are currents, measured in amperes, symbol A. SI means International System of Units. The SI unit of electric current is the ampere.

In the ideal charging arrangement, the wire carries conduction current and the gap carries displacement current of equal magnitude. This does not mean that conduction and displacement currents must always occupy separate regions.

In most cases, both may be present in the same region because no medium is perfectly conducting or perfectly insulating. A steady electric field has no associated displacement current because it does not change with time.

How is displacement current related to electric flux?

Electric flux, written Φₑ, measures the electric field passing through a surface, accounting for its area and orientation. For a uniform field perpendicular to a flat surface, flux is the product of field magnitude and surface area.

Let E denote electric field magnitude, A plate area, Q charge magnitude on a plate and ε₀ the permittivity of free space, the constant relating electric field to its electric sources in vacuum, a region without matter. Here A is an area symbol, not the ampere unit.

Derivation: Displacement current in an ideal capacitor

Consider parallel plates in vacuum, with a uniform field between them and the field outside neglected. These are the assumptions of the ideal capacitor model.

1. The field between the plates is E = Q/(ε₀A). Multiplication by the plate area gives the flux through a surface across the gap.
2. Thus Φₑ = EA = Q/ε₀. For fixed plate area and constant ε₀, a changing charge produces a changing electric flux.
3. Let t denote time. The notation dQ/dt means the rate of change of charge with time, so the charging current is i = dQ/dt. Differentiating the flux relation gives ε₀ dΦₑ/dt = dQ/dt.

iᵈ = ε₀ dΦₑ/dt = i for this charging capacitor. The equality follows from the changing field; it does not require charge to cross the insulating space.

Which quantities and units are involved?

The SI unit of charge is the coulomb, symbol C. The SI unit of electric field strength is the newton per coulomb, written N/C, equivalently the volt per metre, written V/m. A newton measures force and a volt measures potential difference.

The SI unit of magnetic field is the tesla, symbol T. Metre, symbol m, and second, symbol s, are the units of length and time used here. Distinguishing field strength from current prevents confusion about what actually exists in the capacitor gap.

The side view marks electric and magnetic fields near the plates. The circular cross-sectional view shows electric-field crosses and magnetic-field arrows around circular paths. The two field directions are perpendicular.

See Fig. 8.2 in your NCERT textbook

The changing electric field supplies the missing magnetic effect. This complements electromagnetic induction, in which a changing magnetic field gives rise to an electric field. Together, the two connections make the propagation of coupled fields possible.

How are electromagnetic waves produced, and why are they transverse?

Definition: An electromagnetic wave consists of coupled, time-varying electric and magnetic fields that propagate through space and carry energy.

Accelerated charges, whose velocity changes with time, radiate electromagnetic waves. An oscillating charge repeatedly moves about an equilibrium position, so its velocity changes. It provides a simple way to understand the production of radiation.

How does an oscillating charge generate a wave?

1. The charge oscillates and produces an electric field that changes with time.
2. The changing electric field produces a changing magnetic field.
3. The changing magnetic field is associated with a changing electric field.
4. These coupled fields propagate away from the source, carrying energy supplied by the accelerated charge.

The wave has the same frequency, or number of oscillations per second, as the oscillating charge. Stationary charges and steady currents are not sources of electromagnetic radiation: the required time-varying fields are absent.

The propagating wave takes energy from its source. The repeated association of changing electric and magnetic fields does not create energy without a supply. Radio and television signals carry energy, as does sunlight travelling from the Sun to Earth.

What does transverse mean here?

A transverse wave has its oscillations perpendicular to its direction of propagation, meaning the direction in which it travels. In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and both are perpendicular to that direction.

For a wave travelling along the z-axis, the electric field can lie along the x-axis and the magnetic field along the y-axis. These are three mutually perpendicular coordinate directions. The wave travels along z; its field oscillations do not point along z.

The wave is drawn travelling along the z-axis. Electric-field arrows lie along the x-direction and magnetic-field arrows along the y-direction. The two sinusoidal patterns are drawn in perpendicular planes and share the same positions of their zero values and peaks.

See Fig. 8.3 in your NCERT textbook

The fields are in phase: corresponding zero values and maxima occur together. Their amplitudes, meaning maximum magnitudes, are related but are not numerically equal in SI units. Direction, phase and amplitude are distinct aspects of the wave.

Which relations describe electromagnetic waves in vacuum?

Electromagnetic waves require no material medium. Electric and magnetic fields can oscillate and propagate in vacuum. This distinguishes them from waves whose propagation depends on motion of the particles of a material medium.

Let c be their speed in vacuum. For the calculations here, c = 3 × 10⁸ m/s. All regions of the electromagnetic spectrum have this same vacuum speed; a higher frequency does not mean a higher speed.

How are wavelength and frequency connected?

Wavelength, symbol λ, is the spatial separation of successive corresponding points of a wave. Frequency is written ν. The SI unit of frequency is the hertz, symbol Hz, meaning one oscillation per second. The relation is c = νλ.

For a sinusoidal wave, let k be the angular wave number, the spatial rate of change of phase, and ω its angular frequency, the rate of change of phase with time. Phase specifies the stage of an oscillation.

Derivation: The frequency-wavelength relation

1. The angular wave number is k = 2π/λ, where π is the circle constant.
2. The angular frequency is ω = 2πν. Angular quantities describe a complete oscillation using an angle of 2π radians; a radian is a unit of angle.
3. For a vacuum electromagnetic wave, ω = ck. Substitution gives 2πν = c(2π/λ), from which the common factor cancels.

νλ = c. Therefore, wavelength decreases as frequency increases, provided the wave remains in vacuum and the propagation speed remains constant.

How are the field amplitudes related?

Write the electric and magnetic field amplitudes as E₀ and B₀. They satisfy E₀ = cB₀, or B₀ = E₀/c. Use V/m for electric field and tesla for magnetic field when using c in metres per second.

The vacuum speed also satisfies c = 1/√(μ₀ε₀), where μ₀ is the permeability of free space, the magnetic constant in the field equations, and √ denotes a square root. This connects light propagation with electric and magnetic properties.

In a material medium, v = 1/√(με), where v is the wave speed and μ and ε are the medium's permeability and permittivity. These describe its magnetic and electric response. The statement about a common speed c applies specifically to vacuum.

Note: Equal vacuum speed does not imply equal wavelength, frequency or interaction with matter. Those differences distinguish regions of the spectrum even though their electromagnetic nature is shared.

How is the electromagnetic spectrum arranged?

The electromagnetic spectrum is the classification of electromagnetic waves by frequency. In order of increasing frequency, its main regions are radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.

The same sequence runs from longer to shorter wavelengths. Reverse it for increasing wavelength. Because the vacuum speed is fixed, the frequency and wavelength orderings must be opposite. Gamma rays occupy the high-frequency end; long radio waves occupy the long-wavelength end.

Why are the boundaries approximate?

There is no sharp division between one region and the next. The classification depends roughly on how radiation is produced or detected. Different regions overlap, so a boundary should not be treated as an exact physical barrier.

The following summary keeps the conventional wavelength bands as listed. Prefixes used here are: millimetre, mm, equal to 10⁻³ m; nanometre, nm, equal to 10⁻⁹ m. The sign > means greater than, and < means less than.

An electron is a negatively charged atomic particle; the nucleus is the central part of an atom. An energy level is an allowed energy of an atomic electron. Inner-shell electrons belong to shells closer to the nucleus. Radioactive decay is the spontaneous transformation of an unstable nucleus, which can be accompanied by radiation.

Frequency and wavelength scales stand beside labelled spectrum regions. Gamma rays are at the high-frequency end, followed by X-rays, ultraviolet, visible, infrared and radio regions. A separate enlarged visible strip runs from violet near 400 nm to red near 700 nm.

See Fig. 8.4 in your NCERT textbook

Do not infer a new kind of wave whenever a label changes. These regions share transverse electric and magnetic fields. Their different wavelengths and frequencies lead to different interactions with the atoms and molecules of matter.

How are radio waves and microwaves produced, detected and used?

Radio waves arise from accelerated charges in conducting wires. A transmitting aerial contains accelerating and decelerating electrons. A receiving aerial detects the radiation. Radio and television communication are major applications.

Which radio frequency bands are commonly identified?

Radio waves are generally in the range from 500 kHz to about 1000 MHz. The prefixes are kilohertz, kHz, equal to 10³ Hz; megahertz, MHz, equal to 10⁶ Hz; and gigahertz, GHz, equal to 10⁹ Hz.

Modulation means varying a wave property to carry a signal. Amplitude modulation (AM) varies amplitude and frequency modulation (FM) varies frequency. Cellular phones use radio waves in the ultrahigh-frequency band, abbreviated UHF, for voice communication.

What distinguishes microwaves?

Microwaves are short-wavelength radio waves with frequencies in the gigahertz range. Sources include klystrons and magnetrons. Point-contact diodes, electronic devices with a small contact region used to detect microwave signals, are detectors.

Their short wavelengths make them suitable for radar, a system using reflected radio waves to locate objects or measure their motion. Radar is used for aircraft navigation and provides the basis of speed guns for fast balls, tennis serves and automobiles.

Microwave ovens transfer energy from microwaves to water-containing food, increasing molecular motion and temperature. This is an application of the interaction of the radiation with matter, rather than evidence of a different vacuum speed.

Radio waves and microwaves illustrate why classification includes production, detection and use. A region's name alone is not a complete description: connect the accelerating charges or generating device to the detector and then to the practical application.

Why are infrared and visible radiation important?

Infrared radiation lies beside the low-frequency, long-wavelength end of visible light. Hot bodies and molecules produce it. Its conventional band extends from 1 mm to 700 nm, placing it between microwaves and visible radiation.

Why is infrared often called heat radiation?

Water molecules in most materials readily absorb infrared radiation. Carbon dioxide and ammonia molecules also absorb it. Absorption increases thermal motion, so the absorbing substance heats and can heat its surroundings.

Infrared is sometimes referred to as heat waves; this name reflects its heating effect on absorption. Infrared can increase the vibration of entire atoms or molecules, raising internal energy and temperature. It does not follow that other electromagnetic waves cannot transfer energy.

Infrared lamps are used in physical therapy. Detectors in Earth satellites are used for military purposes and for observing crop growth. Infrared-emitting diodes are also used in remote switches for household electronic systems, including television sets.

A light-emitting diode is a semiconductor device that emits radiation when operated electrically. A semiconductor is a material whose electrical conduction can be controlled. The radiation emitted by such a diode need not lie in the visible range.

How does infrared contribute to Earth's warmth?

1. Incoming visible radiation passes relatively easily through the atmosphere.
2. The Earth's surface absorbs part of that incoming radiation.
3. The surface emits radiation at longer, infrared wavelengths.
4. Greenhouse gases, such as carbon dioxide and water vapour, trap this infrared radiation, helping maintain Earth's warmth.

This is the greenhouse effect, the warming associated with the trapping of outgoing infrared radiation by atmospheric gases. It connects absorption and re-emission, rather than requiring infrared radiation to travel more slowly in vacuum.

What makes visible light different?

Visible light is the part of the spectrum detected by the human eye. Its frequency extends from about 4 × 10¹⁴ Hz to about 7 × 10¹⁴ Hz, corresponding to an approximate wavelength range of 700 nm to 400 nm.

Light emitted or reflected by objects provides visual information. Human sensitivity does not define the sensitivity of every animal: snakes can detect infrared, while the visible range of many insects extends well into ultraviolet.

How do ultraviolet, X-rays and gamma rays differ?

These regions lie towards the higher-frequency side of visible light. Their sources and interactions help distinguish them, but their wavelength ranges are not sharply separated. Use approximate ranges with the correct ordering rather than rigid boundaries.

What produces ultraviolet radiation, and what does it do?

Ultraviolet radiation, abbreviated UV, is produced by special lamps and very hot bodies. The Sun is an important source. Most solar UV is absorbed by the atmospheric ozone layer, a region rich in ozone, a form of oxygen.

Large quantities of UV are harmful to humans. Exposure can stimulate production of melanin, the pigment responsible for skin tanning. Welding arcs produce large amounts of UV, so welders use special glass goggles or face masks to protect their eyes.

UV lamps are used to kill germs in water purifiers. Short wavelengths allow UV to be focused into narrow beams for high-precision applications such as LASIK, laser-assisted in situ keratomileusis, a form of eye surgery.

How are X-rays produced and used?

X-rays can be generated by bombarding a metal target with high-energy electrons. This is one common production method. They are used for medical diagnosis and treatment of certain forms of cancer.

Because X-rays can damage or destroy living tissues and organisms, unnecessary or excessive exposure must be avoided. Their usefulness in medicine does not remove their capacity to damage tissue. The intended application depends on how the radiation interacts with matter.

What is distinctive about gamma rays?

Gamma rays are high-frequency radiation produced in nuclear reactions and emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. The nucleus is the central part of an atom.

Broader descriptive ranges are about 400 nm to 0.6 nm for UV, about 10 nm to 10⁻⁴ nm for X-rays, and about 10⁻¹⁰ m to less than 10⁻¹⁴ m for gamma rays. These overlap and supplement the conventional bands listed earlier.

The overlap reinforces the importance of the production process. For gamma rays, think of nuclear processes; for common X-ray generation, think of high-energy electrons striking a metal target. Both remain transverse electromagnetic radiation with the same vacuum speed.

How are the different regions of the spectrum detected?

A detector responds to incident radiation and makes its presence observable. The appropriate detector depends on the radiation's interaction with matter. There is no single human sensory response that detects the complete electromagnetic spectrum.

A thermopile uses connected thermocouples to detect heating; a thermocouple produces a voltage from a temperature difference. A bolometer detects absorbed radiation through the resulting temperature-dependent change in electrical resistance, the opposition to current flow.

Photographic film records radiation through a light-sensitive or radiation-sensitive material. A photocell produces an electrical response to incident light. Infrared photographic film is used for the infrared region rather than assuming all film responds identically to every wavelength.

Ionisation means formation of electrically charged atoms or molecules through removal or addition of electrons. A Geiger tube detects radiation through gas ionisation and electrical pulses. An ionisation chamber detects radiation by collecting charge produced through ionisation of a gas.

Which detector belongs with which region?

What can be inferred from shared detectors?

Visible light and UV can both be detected using photocells or photographic film. X-rays and gamma rays also share detection methods. Thus, sharing a detector does not mean two named regions have identical production processes.

Similarly, the human eye's response identifies visible light but does not set an absolute limit on electromagnetic radiation. Regions beyond human vision remain detectable through other interactions, including heating and ionisation.

A useful comparison therefore links four features: position in the spectrum, source, detector and application. Keeping these categories separate avoids confusing an instrument that produces radiation with an instrument that responds to it.

How are simple electromagnetic-wave calculations solved?

Begin by identifying whether the question concerns frequency, wavelength or field strength. Write the relevant relation and convert prefixes before substitution. All six examples below use vacuum propagation, with c = 3 × 10⁸ m/s.

Additional prefixes are centimetre, cm, equal to 10⁻² m, and nanotesla, nT, equal to 10⁻⁹ T. A larger frequency gives a shorter wavelength. Electric and magnetic amplitudes require different units even though their ratio is fixed.

How are wavelength and frequency converted?

Worked example 1. A wave travelling in vacuum has frequency ν = 30 MHz. Find its wavelength, taking c = 3 × 10⁸ m/s.

Formula: λ = c/ν. Substitute: ν = 30 × 10⁶ Hz, so λ = (3 × 10⁸)/(30 × 10⁶). Answer: 10 m. The frequency must first be expressed in hertz.

Worked example 2. A radio tunes from 7.5 MHz to 12 MHz. Find the corresponding vacuum wavelength band using c = 3 × 10⁸ m/s.

Formula: λ = c/ν. Substitute: (3 × 10⁸)/(7.5 × 10⁶) = 40 m, and (3 × 10⁸)/(12 × 10⁶) = 25 m. Answer: 25 m to 40 m. The higher-frequency endpoint corresponds to the shorter wavelength.

For a band, calculate both endpoints separately. Do not preserve the original endpoint order when writing wavelengths in increasing order: the inverse relationship reverses it. The two answers describe the same set of waves.

How are electric and magnetic fields related?

Worked example 3. A plane wave of frequency 25 MHz travels in vacuum along the positive x-direction. At an instant its electric field is 6.3 V/m along positive y. Find the magnetic field using c = 3 × 10⁸ m/s.

Formula: B = E/c, where B is magnetic field magnitude at that instant. Substitute: B = 6.3/(3 × 10⁸). Answer: 2.1 × 10⁻⁸ T along positive z. Checking the magnitude, multiplication by c returns 6.3 V/m. The electric field, magnetic field and propagation direction have the required mutually perpendicular orientation.

Worked example 4. The magnetic amplitude of a vacuum wave is B₀ = 510 nT. Find E₀ using c = 3 × 10⁸ m/s.

Formula: E₀ = cB₀. Substitute: E₀ = (3 × 10⁸)(510 × 10⁻⁹). Answer: 153 V/m. Convert nanotesla into tesla before using the field-amplitude relation.

For direction, mere perpendicularity leaves two opposite possibilities. Use the vector product of electric field with magnetic field, written E × B: its right-hand-rule direction must be the direction of propagation. In that rule, turning from E towards B gives the thumb's perpendicular direction.

How are two relations combined?

Worked example 5. A vacuum wave has electric amplitude E₀ = 120 N/C and frequency ν = 50.0 MHz. Find magnetic amplitude and wavelength using c = 3 × 10⁸ m/s and the equivalence of N/C and V/m.

Formula: c = E₀/B₀; c = νλ. Rearranging gives B₀ = E₀/c and λ = c/ν. Substitute: B₀ = 120/(3 × 10⁸); λ = (3 × 10⁸)/(50.0 × 10⁶). Answer: B₀ = 4 × 10⁻⁷ T and λ = 6 m.

Worked example 6. A vacuum wave has frequency ν = 2.0 × 10¹⁰ Hz and electric amplitude E₀ = 48 V/m. Find wavelength and magnetic amplitude, taking c = 3 × 10⁸ m/s.

Formula: c = νλ; c = E₀/B₀. Rearranging gives λ = c/ν and B₀ = E₀/c. Substitute: λ = (3 × 10⁸)/(2.0 × 10¹⁰); B₀ = 48/(3 × 10⁸). Answer: λ = 0.015 m and B₀ = 1.6 × 10⁻⁷ T.

The two-part examples use independent relationships. Frequency fixes wavelength when speed is known, while electric amplitude fixes magnetic amplitude. Frequency alone does not supply an electric or magnetic amplitude.

Glossary

• Displacement current — Current term associated with changing electric flux, producing a magnetic field without requiring conduction across the region.
• Conduction current — Electric current arising from the actual flow of charged particles through a material.
• Electric flux — Measure of the electric field passing through a surface, including the effect of area and orientation.
• Electromagnetic wave — Coupled electric and magnetic fields that vary with time and propagate through space carrying energy.
• Transverse wave — Wave whose oscillations are perpendicular to the direction in which the wave propagates.
• Frequency — Number of complete oscillations per second, measured in the SI unit hertz.
• Wavelength — Distance between successive corresponding points, such as adjacent peaks, in a wave pattern.
• Amplitude — Maximum magnitude of an oscillating quantity, such as electric or magnetic field strength.
• Electromagnetic spectrum — Classification of electromagnetic radiation by frequency, with approximate and overlapping boundaries between named regions.
• Infrared radiation — Electromagnetic radiation adjacent to the long-wavelength end of visible light, emitted by hot bodies and molecules.
• Ultraviolet radiation — Electromagnetic radiation beyond the violet end of visible light, produced by special lamps and very hot bodies.
• Gamma rays — High-frequency electromagnetic radiation produced in nuclear reactions and emitted by radioactive nuclei.

Common errors and misconceptions

• Misconception: Displacement current means charges cross an ideal capacitor's insulating gap. Correct: It is associated with changing electric flux; conduction current does not cross that gap.
• Misconception: A steady electric field produces displacement current. Correct: Displacement current requires electric flux to change with time.
• Misconception: Electromagnetic waves need air or another material medium. Correct: They can propagate through vacuum as coupled electric and magnetic fields.
• Misconception: The fields point along the propagation direction. Correct: Both are perpendicular to that direction and to one another.
• Misconception: Gamma rays travel faster than radio waves in vacuum. Correct: Both have speed c; their frequencies and wavelengths differ.
• Misconception: Spectrum boundaries are sharp and exact. Correct: Regions overlap, and their classification depends roughly on production and detection.
• Misconception: The electric and magnetic amplitudes are equal in SI units. Correct: In vacuum their ratio is c, with E₀ measured in V/m and B₀ in tesla.

Exam-style questions with model answers

Q1. Define displacement current and state its cause in the gap of a charging ideal capacitor. [2 marks]

1. Displacement current is the current term associated with a changing electric flux and acts as a source of magnetic field.
2. During charging, increasing plate charge changes the electric field and flux across the capacitor gap, although conduction charge does not cross that gap.

Q2. Explain the transverse nature of an electromagnetic wave using a wave travelling along the z-axis. State the phase relation of its fields. [3 marks]

1. The electric field can oscillate along the x-axis, perpendicular to the z-axis along which the electromagnetic wave travels.
2. The magnetic field then oscillates along the y-axis, perpendicular to both the electric field and propagation direction. These perpendicular field oscillations establish the transverse nature.
3. The two fields are in phase, so corresponding zero values and maxima occur together at a given position.

Q3. Give the seven principal spectrum regions in increasing frequency. Explain the wavelength ordering and the nature of their boundaries, and state their common vacuum-speed property. [4 marks]

1. Increasing frequency order is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.
2. This is decreasing wavelength order because frequency multiplied by wavelength equals the constant vacuum speed.
3. Boundaries are approximate rather than sharp, and regions overlap. Classification depends roughly on production or detection methods.
4. All these electromagnetic waves travel with the same speed in vacuum, despite their different frequencies and wavelengths.

Q4. Compare infrared and ultraviolet radiation in five respects: position relative to visible light, sources, detectors, uses and effects on matter. [5 marks]

1. Infrared lies on the lower-frequency, longer-wavelength side of visible light; ultraviolet lies on its higher-frequency, shorter-wavelength side. Both belong to the electromagnetic spectrum.
2. Hot bodies and molecules produce infrared. Special lamps and very hot bodies produce ultraviolet, with the Sun being an important ultraviolet source.
3. Infrared detectors include thermopiles, bolometers and infrared photographic film. Ultraviolet can be detected using photocells and photographic film.
4. Infrared is used in physical therapy and household remote switches. Ultraviolet lamps are used to kill germs in water purifiers.
5. Absorbed infrared increases thermal motion and heats matter. Large quantities of ultraviolet can harm humans, and ultraviolet exposure stimulates melanin production and skin tanning.

Q5. A plane electromagnetic wave travels in vacuum with frequency 2.0 × 10¹⁰ Hz and electric field amplitude 48 V/m. Taking c = 3 × 10⁸ m/s, calculate its wavelength and magnetic field amplitude. Show a relation and substitution for each. [4 marks]

1. Use λ = c/ν, where λ is wavelength, c is vacuum speed and ν is frequency.
2. Substitution gives λ = (3 × 10⁸)/(2.0 × 10¹⁰) = 0.015 m. The supplied frequency is already in hertz.
3. Use B₀ = E₀/c, where B₀ is magnetic amplitude and E₀ is electric amplitude.
4. Substitution gives B₀ = 48/(3 × 10⁸) = 1.6 × 10⁻⁷ T, with the magnetic field perpendicular to the electric field.

Q6. Give one production method and one use each for X-rays and gamma rays. [4 marks]

1. One common method of producing X-rays is to bombard a metal target with high-energy electrons, as in an X-ray tube.
2. X-rays are used as a diagnostic tool in medicine, allowing medical investigation through their interaction with matter.
3. Gamma rays are produced in nuclear reactions and can also be emitted by radioactive nuclei during decay.
4. Gamma rays are used in medicine to destroy cancer cells. Their nuclear origin distinguishes this production process from electron bombardment of a target.

Q7. A radio receives frequencies from 7.5 MHz to 12 MHz. Taking c = 3 × 10⁸ m/s and 1 MHz = 10⁶ Hz, calculate the wavelength at each endpoint and state the corresponding wavelength band in increasing order. [3 marks]

1. Using λ = c/ν, the lower frequency 7.5 × 10⁶ Hz gives λ = (3 × 10⁸)/(7.5 × 10⁶) = 40 m.
2. The higher frequency 12 × 10⁶ Hz gives λ = (3 × 10⁸)/(12 × 10⁶) = 25 m.
3. The wavelength band in increasing order is 25 m to 40 m. The ordering reverses because wavelength decreases as frequency increases at fixed vacuum speed.

Key takeaways

• Displacement current accounts for the magnetic effect of changing electric flux, including the field between a charging capacitor's plates.
• Accelerated charges radiate electromagnetic waves; an oscillating charge produces radiation at the same frequency as its own oscillation.
• Electric and magnetic fields are mutually perpendicular and transverse to propagation, with corresponding oscillations in phase.
• All electromagnetic waves have the same vacuum speed, while frequency and wavelength obey the inverse relationship expressed by c = νλ.
• The spectrum runs from radio waves through microwaves, infrared, visible, ultraviolet and X-rays to gamma rays in increasing frequency.
• Spectrum boundaries are approximate and overlapping, so production and detection methods help distinguish the named regions.
• Learn each region through its source, detector and use; sharing a detector does not imply identical sources.
• Convert frequency and field prefixes before substitution, and keep electric-field and magnetic-field units distinct throughout numerical work.

Test yourself

What causes displacement current between the plates of a charging capacitor?

The accumulating plate charge changes the electric field and hence the electric flux through the gap.

What is the displacement current if the electric flux remains constant?

It is zero because displacement current depends on the rate of change of electric flux.

A charge oscillates at 10⁹ Hz. What frequency does its electromagnetic radiation have?

The radiation has frequency 10⁹ Hz, the same as the oscillating charge that produces it.

Which regions lie immediately beside visible light?

Infrared lies on the longer-wavelength side, and ultraviolet lies on the shorter-wavelength side of visible light.

What happens to vacuum wavelength when frequency increases?

Wavelength decreases because frequency multiplied by wavelength must remain equal to the constant vacuum speed.

Name a microwave source and a microwave detector.

A magnetron is a microwave source, and a point-contact diode can serve as a microwave detector.

Why are infrared waves often associated with heating?

Absorption increases thermal motion of atoms and molecules, raising the internal energy and temperature of the absorbing substance.

Why should spectrum boundaries not be memorised as exact divisions?

Different regions overlap, and their classification depends roughly on how the radiation is produced or detected.