Electromagnetic Waves
What to remember
- Displacement current Id = ε₀ dΦE/dt is produced by a changing electric flux. It makes the total current continuous, so Ampere's law becomes the Ampere–Maxwell law.
- An electromagnetic wave is transverse. The electric field E and the magnetic field B are perpendicular to each other and to the direction of travel, and they are in phase.
- In vacuum the speed is c = 1/√(μ₀ε₀) = 3 × 10⁸ m s⁻¹, and E₀/B₀ = c. All electromagnetic waves travel at this speed in vacuum, whatever their frequency.
- The spectrum in increasing frequency: radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. Wavelength falls as frequency rises, because c = fλ.
Displacement current
Ampere's circuital law says that the line integral of B around a closed loop equals μ₀ times the conduction current i that passes through any surface bounded by the loop. This works for a steady current in a wire.
Now take a parallel plate capacitor that is being charged. A conduction current i flows in the connecting wires, but no charge crosses the gap between the plates. Take a circular loop around the wire. Surface 1 is a flat disc that the wire pierces. Surface 2 is a bowl-shaped surface that passes between the plates and is not pierced by the wire.
- Through surface 1 the conduction current is i, so the law gives B around the loop = μ₀ i.
- Through surface 2 the conduction current is zero, so the law would give B = 0.
Both surfaces are bounded by the same loop, so the two results cannot both be right. Maxwell removed this contradiction by adding a new term called the displacement current.
Between the plates the electric field E = q/(ε₀A), where A is the plate area. The electric flux is ΦE = EA = q/ε₀. As the capacitor charges, q changes, so ΦE changes with time:
Id = ε₀ dΦE/dt = ε₀ A dE/dt = dq/dt = i
Thus the displacement current between the plates is equal to the conduction current in the wires. The total current i + Id is continuous everywhere.
The Ampere–Maxwell law is: ∮ B·dl = μ₀ (i + Id) = μ₀ i + μ₀ ε₀ dΦE/dt.
Key points about displacement current:
- It is not a flow of charge. It is a changing electric field that has the same magnetic effect as a current.
- It exists in vacuum as well as in matter.
- In a steady (dc) state the electric field is constant, so Id = 0.
- For a circular plate capacitor with uniform field between the plates, Id through a circle of radius r (r smaller than the plate radius) is Id × (r²/R²), where R is the plate radius.
- A changing electric field produces a magnetic field, and (Faraday's law) a changing magnetic field produces an electric field. This mutual production is the origin of electromagnetic waves.
Sources and nature of electromagnetic waves
An electric charge at rest produces only an electric field. A charge moving with constant velocity produces a steady magnetic field as well. An accelerated charge produces time-varying electric and magnetic fields that radiate away as electromagnetic waves. An oscillating charge (for example, charge oscillating in an LC circuit or in an antenna) is therefore the basic source. The frequency of the wave equals the frequency of oscillation of the charge.
Maxwell showed that the equations of electricity and magnetism predict waves whose speed is
c = 1/√(μ₀ε₀)
With μ₀ = 4π × 10⁻⁷ T m A⁻¹ and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², this gives c ≈ 3 × 10⁸ m s⁻¹, the measured speed of light. This showed that light is an electromagnetic wave. Hertz produced and detected such waves in the laboratory in 1887.
Properties of electromagnetic waves:
| Property | Statement |
|---|---|
| Type | Transverse wave |
| Direction of E | Perpendicular to the direction of travel |
| Direction of B | Perpendicular to both E and the direction of travel |
| Direction of travel | Along the direction of E × B |
| Phase | E and B reach their maxima and zeros at the same time and place |
| Amplitude ratio | E₀/B₀ = c (also E/B = c at every instant) |
| Speed in vacuum | c = 1/√(μ₀ε₀) = 3 × 10⁸ m s⁻¹ |
| Medium needed | None; the waves travel in vacuum |
| Relation | c = fλ (f is frequency, λ is wavelength) |
| Charge deflection | Not deflected by electric or magnetic fields, because the waves carry no charge |
For a wave travelling along the x-axis with E along y, B is along z. The fields can be written as
Ey = E₀ sin(kx − ωt), Bz = B₀ sin(kx − ωt)
Here k = 2π/λ is the wave number (rad m⁻¹), ω = 2πf is the angular frequency (rad s⁻¹), and the wave speed is ω/k = c.
In a material medium of permittivity ε and permeability μ, the speed is v = 1/√(με), which is less than c. The refractive index of the medium is n = c/v = √(μᵣεᵣ), where μᵣ and εᵣ are the relative permeability and relative permittivity.
Energy and momentum carried by the waves
Electromagnetic waves carry energy. The energy density (energy per unit volume) of the electric field is uE = ½ ε₀ E², and that of the magnetic field is uB = B²/(2μ₀). Using E = cB and c² = 1/(μ₀ε₀), the two are equal at every point: uE = uB. The average total energy density is
u = ½ ε₀ E₀² (this equals B₀²/(2μ₀))
The intensity (power per unit area, in W m⁻²) is the energy density times the speed:
I = u c = ½ ε₀ c E₀²
The waves also carry momentum. If a wave transfers a total energy U to a surface that absorbs it completely, the momentum delivered is
p = U/c
Because the waves carry momentum, they exert a force (radiation pressure) on a surface. This is why the tails of comets point away from the Sun, and it shows that the waves behave as a physical entity with energy and momentum, not only as a mathematical idea.
The electromagnetic spectrum
The spectrum has no sharp boundaries. Neighbouring regions overlap. The ranges below are approximate.
| Region | Approximate wavelength | How it is produced | Main uses |
|---|---|---|---|
| Radio waves | greater than 0.1 m (frequency about 500 kHz to 1000 MHz) | Oscillating electric circuits, charges accelerated in aerials | Radio and television broadcast, mobile communication |
| Microwaves | 0.1 m to 1 mm (frequency about 1 GHz to 300 GHz) | Special vacuum tubes: klystron, magnetron; Gunn diode | Radar, aircraft navigation, microwave oven, speed guns |
| Infrared | 1 mm to 700 nm | Hot bodies; vibrations of atoms and molecules | Remote controls, night vision, physical therapy, greenhouse effect, warming of Earth |
| Visible light | 700 nm to 400 nm | Electrons changing energy levels in atoms; hot filaments | Vision, photography, optical instruments |
| Ultraviolet | 400 nm to 1 nm | Very hot bodies, the Sun, special lamps, electric discharge | Sterilising water, killing germs, eye surgery (LASIK), detecting forged notes |
| X-rays | 1 nm to 10⁻³ nm | High-energy electrons striking a metal target in an X-ray tube | Medical diagnosis (bones), treating cancer, checking baggage |
| Gamma rays | less than 10⁻³ nm | Radioactive decay of nuclei, nuclear reactions | Treating cancer, sterilising equipment, nuclear studies |
Further facts about each region:
- Radio waves: they are produced by accelerated charges in conducting wires. AM broadcast uses about 530 kHz to 1710 kHz. Television and FM use higher frequencies (tens to hundreds of MHz). Cellphones use the ultra-high frequency band, about 800 MHz to 950 MHz.
- Microwaves: their short wavelength suits radar. In a microwave oven the frequency is chosen to match the natural frequency of water molecules, so the molecules absorb energy by resonance and the food warms.
- Infrared: it is often called heat radiation because most substances absorb it and warm up. Infrared is absorbed by water molecules, so it heats objects. The greenhouse effect traps infrared radiation re-emitted from the Earth's surface and keeps the Earth warm.
- Visible light: it is the part that the human eye detects, about 400 nm to 700 nm. Violet has the shortest wavelength and red has the longest.
- Ultraviolet: the Sun is a strong source, but most of it is absorbed by the ozone layer. Ordinary glass absorbs UV, so welders wear special goggles for protection.
- X-rays: they penetrate soft tissue and are stopped more by bone, which makes images of bones.
- Gamma rays: they have the highest frequency and energy of the spectrum, and the greatest penetration.
Order of properties along the spectrum, from radio to gamma:
| Quantity | Trend from radio to gamma |
|---|---|
| Wavelength | Decreases |
| Frequency | Increases |
| Photon energy (proportional to frequency) | Increases |
| Speed in vacuum | Same for all (3 × 10⁸ m s⁻¹) |
Formula summary
| Quantity | Formula | Symbols and units |
|---|---|---|
| Displacement current | Id = ε₀ dΦE/dt | ΦE is electric flux (V m); Id in A |
| Ampere–Maxwell law | ∮ B·dl = μ₀(i + Id) | i is conduction current in A |
| Speed in vacuum | c = 1/√(μ₀ε₀) | μ₀ = 4π × 10⁻⁷ T m A⁻¹; ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² |
| Field ratio | E₀/B₀ = c | E₀ in V m⁻¹; B₀ in T |
| Wave relation | c = fλ, ω = 2πf, k = 2π/λ | f in Hz; λ in m |
| Speed in medium | v = 1/√(με) = c/n | n is refractive index |
| Energy density | u = ½ ε₀ E₀² = B₀²/(2μ₀) | J m⁻³ |
| Intensity | I = u c = ½ ε₀ c E₀² | W m⁻² |
| Momentum | p = U/c | U in J; p in kg m s⁻¹ |
Worked examples
Example 1. The electric field amplitude of an electromagnetic wave in vacuum is 45 V m⁻¹. Find the amplitude of the magnetic field.
B₀ = E₀/c = 45 / (3 × 10⁸) = 1.5 × 10⁻⁷ T.
Example 2. A parallel plate capacitor with circular plates is being charged by a current of 3 A. Find the displacement current through a circle of one-third of the plate radius, drawn between the plates and coaxial with them.
The displacement current through the full plate area equals the conduction current, 3 A. The field is uniform, so the current through a circle is proportional to its area, which is proportional to the square of the radius. For one-third of the radius the area is 1/9.
Id = 3 × (1/3)² = 3/9 = 0.33 A.
Example 3. A plane electromagnetic wave in vacuum has E₀ = 60 V m⁻¹. Find the average intensity. Use ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² and c = 3 × 10⁸ m s⁻¹.
I = ½ ε₀ c E₀² = 0.5 × 8.85 × 10⁻¹² × 3 × 10⁸ × 3600 = 4.78 W m⁻², about 4.8 W m⁻².
Example 4. A surface completely absorbs 1.5 × 10⁻⁶ J of radiation energy. Find the momentum delivered to it.
p = U/c = 1.5 × 10⁻⁶ / (3 × 10⁸) = 5 × 10⁻¹⁵ kg m s⁻¹.
Example 5. A radio station broadcasts at 1.2 MHz. Find the wavelength.
λ = c/f = 3 × 10⁸ / (1.2 × 10⁶) = 250 m.
NEET traps
- Displacement current is not a movement of charge. It is produced by a changing electric field, and it can exist in vacuum.
- The displacement current between the plates equals the conduction current in the wires, but it is zero when the capacitor is fully charged and the field is steady.
- E and B in an electromagnetic wave are in phase, not 90° apart. This is different from an LC circuit, where the energy shifts between the electric and magnetic forms.
- Electromagnetic waves are transverse, but they do not need a medium. Sound is longitudinal in air and needs a medium.
- The ratio E₀/B₀ equals c, not 1/c. Check the units: V m⁻¹ divided by T gives m s⁻¹.
- The speed of all electromagnetic waves in vacuum is the same. Gamma rays are not faster than radio waves.
- Electromagnetic waves carry no charge, so they are not deflected by electric or magnetic fields. Cathode rays and beta rays are deflected.
- Infrared is called heat radiation, but the heating effect is not exclusive to it. The Sun's UV also warms objects, and every part of the spectrum can.
- Microwave ovens use resonance with water molecules. The microwaves do not heat the container directly.
- Ultraviolet has a longer wavelength than X-rays. A common error is to place X-rays before UV in the order of increasing wavelength.
One-liners
- Id = ε₀ dΦE/dt.
- Total current i + Id is continuous in a charging capacitor.
- Maxwell added the displacement current to Ampere's law.
- Accelerated charges radiate electromagnetic waves.
- Hertz first produced radio waves in the laboratory.
- c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m s⁻¹.
- E₀/B₀ = c.
- The direction of travel is along E × B.
- Electromagnetic waves are transverse and need no medium.
- Energy density of E equals energy density of B in the wave.
- Intensity I = ½ ε₀ c E₀².
- Momentum of absorbed radiation p = U/c.
- Radio waves have the longest wavelength of the spectrum.
- Gamma rays have the shortest wavelength and the highest frequency.
- Visible light spans about 400 nm to 700 nm; red has the longest wavelength.
Practice questions
Match List I with List II. List I: a. Displacement current b. Speed of electromagnetic waves in vacuum c. Average intensity of a wave d. Momentum of absorbed radiation List II: i. U/c ii. ε₀ dΦE/dt iii. 1/√(μ₀ε₀) iv. ½ε₀cE₀²
- a-iv, b-iii, c-ii, d-i
- a-ii, b-iii, c-i, d-iv
- a-ii, b-iii, c-iv, d-i
- a-i, b-iii, c-ii, d-iv
Answer
C. a-ii, b-iii, c-iv, d-i
a-ii, b-iii, c-iv, d-i.
Match List I with List II. List I: a. Infrared b. Ultraviolet c. Microwaves d. X-rays List II: i. Radar and microwave ovens ii. Water purification iii. Remote controls and night vision iv. Medical diagnosis of bone fractures
- a-iii, b-ii, c-i, d-iv
- a-ii, b-iii, c-i, d-iv
- a-iii, b-ii, c-iv, d-i
- a-i, b-ii, c-iii, d-iv
Answer
A. a-iii, b-ii, c-i, d-iv
Infrared: remote controls (a-iii). UV: water purifier (b-ii). Microwaves: radar and ovens (c-i). X-rays: bone diagnosis (d-iv).
A parallel plate capacitor is being charged by a steady conduction current of 2 A in the connecting wires. The displacement current between the plates is
- 4 A
- 2 A
- Zero
- 0.5 A
Answer
B. 2 A
The displacement current between the plates equals the conduction current in the wires, so Id = 2 A.
Food containing water becomes hot in a microwave oven quickly mainly because
- microwaves heat the dry container first and the container passes heat to the food
- microwaves have a greater speed than other electromagnetic waves in water
- microwaves are reflected by water molecules and trapped in the food
- the microwave frequency matches the natural frequency of water molecules, which absorb the energy by resonance
Answer
D. the microwave frequency matches the natural frequency of water molecules, which absorb the energy by resonance
Water molecules absorb microwaves strongly by resonance, so the food warms from within.
A surface completely absorbs 9 J of electromagnetic radiation energy. The momentum delivered to the surface (c = 3 × 10⁸ m s⁻¹) is
- 6 × 10⁻⁸ kg m s⁻¹
- 3 × 10⁻⁸ kg m s⁻¹
- 3 × 10⁻⁷ kg m s⁻¹
- 2.7 × 10⁹ kg m s⁻¹
Answer
B. 3 × 10⁻⁸ kg m s⁻¹
p = U/c = 9 / (3 × 10⁸) = 3 × 10⁻⁸ kg m s⁻¹. Using 2U/c would be wrong, because that applies to perfect reflection.
Which of the following statements are correct? 1. The visible region extends from about 400 nm to 700 nm. 2. X-rays have a longer wavelength than ultraviolet rays. 3. Gamma rays have the highest frequency in the spectrum.
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
Answer
C. 1 and 3 only
X-rays (1 nm to 10⁻³ nm) have shorter wavelength than UV.
A capacitor with circular plates is charged by a current of 4 A. The displacement current through a coaxial circle of half the plate radius, lying between the plates, is
- 4 A
- 2 A
- 0.5 A
- 1 A
Answer
D. 1 A
The field is uniform, so the current through a circle is proportional to its area, which is proportional to r². Id = 4 × (1/2)² = 1 A.
In a medium of relative permeability 1 and relative permittivity 4, the speed of an electromagnetic wave (c = 3 × 10⁸ m s⁻¹) is
- 1.5 × 10⁸ m s⁻¹
- 3 × 10⁸ m s⁻¹
- 6 × 10⁸ m s⁻¹
- 7.5 × 10⁷ m s⁻¹
Answer
A. 1.5 × 10⁸ m s⁻¹
v = 1/√(μ₀μᵣε₀εᵣ) = c/√(μᵣεᵣ) = c/√4 = 1.5 × 10⁸ m s⁻¹.
The amplitude of the electric field in a plane electromagnetic wave in vacuum is 200 V m⁻¹. Taking ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻² and c = 3 × 10⁸ m s⁻¹, the average intensity is
- 108 W m⁻²
- 27 W m⁻²
- 5.4 W m⁻²
- 54 W m⁻²
Answer
D. 54 W m⁻²
I = ½ ε₀ c E₀² = 0.5 × 9 × 10⁻¹² × 3 × 10⁸ × (200)² = 54 W m⁻². Missing the factor ½ gives 108.
Assertion (A): In an electromagnetic wave in vacuum, the electric field leads the magnetic field by 90°. Reason (R): The ratio E₀/B₀ equals c.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
D. A is false but R is true
E and B are in phase. Statement R is true.
Match List I with List II. List I: a. 3 × 10⁸ Hz b. 3 × 10¹⁰ Hz c. 6 × 10¹⁴ Hz d. 3 × 10¹⁸ Hz List II: i. 10⁻¹⁰ m ii. 1 m iii. 5 × 10⁻⁷ m iv. 1 cm
- a-i, b-ii, c-iii, d-iv
- a-iv, b-ii, c-iii, d-i
- a-ii, b-iv, c-iii, d-i
- a-ii, b-i, c-iii, d-iv
Answer
C. a-ii, b-iv, c-iii, d-i
λ = c/f: 1 m (a-ii), 1 cm (b-iv), 5 × 10⁻⁷ m (c-iii), 10⁻¹⁰ m (d-i). Here c = 3 × 10⁸ m s⁻¹.
Match List I with List II. List I: a. Radio waves b. Microwaves c. Visible light d. Gamma rays List II: i. 400 nm to 700 nm ii. greater than 0.1 m iii. 0.1 m to 1 mm iv. less than 10⁻³ nm
- a-ii, b-i, c-iii, d-iv
- a-ii, b-iii, c-i, d-iv
- a-i, b-ii, c-iii, d-iv
- a-iii, b-ii, c-i, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
Radio: above 0.1 m (a-ii). Microwave: 0.1 m to 1 mm (b-iii). Visible: 400-700 nm (c-i). Gamma: below 10⁻³ nm (d-iv).
An electromagnetic wave in vacuum has wavelength 3 m. Its frequency (c = 3 × 10⁸ m s⁻¹) is
- 100 MHz
- 1 GHz
- 10 MHz
- 900 MHz
Answer
A. 100 MHz
f = c/λ = 3 × 10⁸ / 3 = 10⁸ Hz = 100 MHz.
A radio signal is sent from the ground to a geostationary satellite at a height of 3.6 × 10⁷ m and the signal is received back on the ground after reflection (c = 3 × 10⁸ m s⁻¹). The time taken, for the path up and down, is
- 0.24 s
- 2.4 s
- 0.48 s
- 0.12 s
Answer
A. 0.24 s
Distance = 2 × 3.6 × 10⁷ = 7.2 × 10⁷ m. t = 7.2 × 10⁷ / 3 × 10⁸ = 0.24 s.
Which radiation is used in LASIK eye surgery?
- Infrared rays
- Ultraviolet rays
- X-rays
- Microwaves
Answer
B. Ultraviolet rays
Ultraviolet radiation is used in LASIK surgery to correct the shape of the cornea.
The intensity of a plane electromagnetic wave in vacuum is 2 W m⁻². The amplitude of its electric field is made 3 times larger. The new intensity is
- 54 W m⁻²
- 9 W m⁻²
- 18 W m⁻²
- 6 W m⁻²
Answer
C. 18 W m⁻²
I = ½ε₀cE₀² ∝ E₀². I′ = 2 × 3² = 18 W m⁻².
A perfectly absorbing plate of area 2 m² is placed normal to a beam of intensity 300 W m⁻². The momentum delivered to it in 10 s (c = 3 × 10⁸ m s⁻¹) is
- 1.8 × 10¹² kg m s⁻¹
- 2 × 10⁻⁶ kg m s⁻¹
- 4 × 10⁻⁵ kg m s⁻¹
- 2 × 10⁻⁵ kg m s⁻¹
Answer
D. 2 × 10⁻⁵ kg m s⁻¹
Energy U = 300 × 2 × 10 = 6000 J. p = U/c = 6000 / 3 × 10⁸ = 2 × 10⁻⁵ kg m s⁻¹. For reflection it would be 4 × 10⁻⁵, but this plate absorbs.
In an electromagnetic wave in vacuum, the energy density associated with the electric field, compared with that of the magnetic field, is
- twice as large
- c times as large
- equal
- half as large
Answer
C. equal
Using E = cB and c² = 1/(μ₀ε₀), ½ε₀E² = B²/(2μ₀). The two energy densities are equal.
In an electromagnetic wave, the electric field is along +y and the magnetic field is along +z. The wave travels along
- +z
- +x
- +y
- −x
Answer
B. +x
The direction of travel is along E × B. ŷ × ẑ = +x̂.
The wavelength of an electromagnetic wave of frequency 6 × 10¹⁴ Hz in vacuum (c = 3 × 10⁸ m s⁻¹) is
- 5 × 10⁻⁷ m
- 5 × 10⁻⁶ m
- 2 × 10⁶ m
- 1.8 × 10²³ m
Answer
A. 5 × 10⁻⁷ m
λ = c/f = 3 × 10⁸ / 6 × 10¹⁴ = 5 × 10⁻⁷ m = 500 nm, which is green light.
Light from the Sun takes about how long to reach the Earth, if the distance is 1.5 × 10¹¹ m (c = 3 × 10⁸ m s⁻¹)?
- 1500 s
- 5000 s
- 50 s
- 500 s
Answer
D. 500 s
t = d/c = 1.5 × 10¹¹ / 3 × 10⁸ = 500 s, about 8 minutes 20 seconds.
Which part of the electromagnetic spectrum has the shortest wavelength?
- Gamma rays
- Microwaves
- Ultraviolet rays
- X-rays
Answer
A. Gamma rays
Wavelength falls as frequency rises. Gamma rays have the highest frequency and so the shortest wavelength.
Assertion (A): A displacement current exists between the plates of a charging capacitor although no charge crosses the gap. Reason (R): A changing electric flux produces a magnetic field in the same way as a conduction current.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Id = ε₀dΦE/dt gives the same magnetic effect as a current.
Given below are two statements: Statement I: In vacuum, the energy density of the electric field of an electromagnetic wave equals that of the magnetic field. Statement II: The average total energy density of the wave is ½ε₀E₀².
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
uE = uB follows from E = cB. Their sum averaged over time is ½ε₀E₀².
The magnetic field amplitude of an electromagnetic wave in vacuum is 1.5 × 10⁻⁸ T. The electric field amplitude (c = 3 × 10⁸ m s⁻¹) is
- 45 V m⁻¹
- 4.5 V m⁻¹
- 0.45 V m⁻¹
- 5 × 10⁻¹⁷ V m⁻¹
Answer
B. 4.5 V m⁻¹
E₀ = cB₀ = 3 × 10⁸ × 1.5 × 10⁻⁸ = 4.5 V m⁻¹.
Which of the following statements are correct? 1. Electromagnetic waves are deflected by a magnetic field. 2. Light is an electromagnetic wave. 3. Electromagnetic waves carry momentum.
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
Answer
B. 2 and 3 only
The waves carry no charge and are not deflected. Light is an electromagnetic wave and the momentum is p = U/c.
The electric field of a wave in vacuum is Ey = 90 sin(2πx − 6π × 10⁸ t) V m⁻¹ (x in metre, t in second; c = 3 × 10⁸ m s⁻¹). Its wavelength and the amplitude of its magnetic field are
- 0.5 m and 6 × 10⁻⁷ T
- 2π m and 3 × 10⁻⁷ T
- 1 m and 3 × 10⁻⁷ T
- 1 m and 2.7 × 10¹⁰ T
Answer
C. 1 m and 3 × 10⁻⁷ T
k = 2π gives λ = 2π/k = 1 m. B₀ = E₀/c = 90/(3 × 10⁸) = 3 × 10⁻⁷ T. Check: ω/k = 6π × 10⁸/2π = 3 × 10⁸ m s⁻¹ = c.
Given below are two statements: Statement I: Displacement current is produced by a changing electric field. Statement II: Displacement current exists only in conducting wires.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Id = ε₀dΦE/dt exists wherever the electric flux changes, including vacuum.
Given below are two statements: Statement I: Microwaves are used in radar systems. Statement II: Microwaves have shorter wavelength than visible light.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Microwaves (1 mm to 0.1 m) have longer wavelength than visible light (400–700 nm), so Statement II is incorrect.
Match List I with List II. List I: a. Radio waves b. Microwaves c. X-rays d. Gamma rays List II: i. Klystron or magnetron valve ii. Oscillating electric circuits iii. High-energy electrons striking a metal target iv. Radioactive nuclei
- a-ii, b-i, c-iv, d-iii
- a-ii, b-i, c-iii, d-iv
- a-iii, b-i, c-ii, d-iv
- a-i, b-ii, c-iii, d-iv
Answer
B. a-ii, b-i, c-iii, d-iv
Radio waves: oscillating circuits (a-ii). Microwaves: klystron/magnetron (b-i). X-rays: electrons on metal target (c-iii). Gamma: nuclei (d-iv).
Given below are two statements: Statement I: X-rays are produced when high-energy electrons strike a metal target. Statement II: Gamma rays are emitted by radioactive nuclei.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard production methods.
Match List I with List II. List I: a. Energy density of the electric field b. Ratio E₀/B₀ c. Direction of travel d. Phase difference between E and B List II: i. Along E × B ii. c iii. ½ε₀E² iv. Zero
- a-iii, b-i, c-ii, d-iv
- a-i, b-ii, c-iii, d-iv
- a-ii, b-iii, c-i, d-iv
- a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
a-iii, b-ii, c-i, d-iv.
A 2 μF capacitor is charged so that the voltage across it rises at a steady rate of 5 V s⁻¹. The displacement current between the plates is
- 2.5 μA
- 0.4 μA
- 10 mA
- 10 μA
Answer
D. 10 μA
Id = i = dq/dt = C dV/dt = 2 × 10⁻⁶ × 5 = 10 × 10⁻⁶ A = 10 μA.
Which of the following statements are correct? 1. Ultraviolet rays can be used to kill germs in water. 2. Infrared radiation is emitted by hot bodies. 3. Radio waves are produced when fast electrons strike a metal target.
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
Answer
A. 1 and 2 only
Radio waves come from oscillating circuits. X-rays are produced by electrons striking a metal target.
Assertion (A): Food containing water is heated in a microwave oven. Reason (R): Glass and ceramic containers absorb microwaves strongly and pass the heat to the food.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
C. A is true but R is false
The container usually absorbs little; the water molecules absorb by resonance.
Assertion (A): Electromagnetic waves are not deflected by electric or magnetic fields. Reason (R): Electromagnetic waves carry no electric charge.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
A force on the wave would need a charge. The waves have none.
The electric field amplitude of a wave in vacuum is 100 V m⁻¹. Taking ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻², the average total energy density of the wave is
- 2.25 × 10⁻⁸ J m⁻³
- 4.5 × 10⁻⁸ J m⁻³
- 9 × 10⁻⁸ J m⁻³
- 4.5 × 10⁻⁶ J m⁻³
Answer
B. 4.5 × 10⁻⁸ J m⁻³
u = ½ε₀E₀² = 0.5 × 9 × 10⁻¹² × 10⁴ = 4.5 × 10⁻⁸ J m⁻³. Half of it is in E and half in B.
Which of the following statements are correct? 1. An accelerated charge radiates electromagnetic waves. 2. The direction of travel of the wave is along E × B. 3. In vacuum c = 1/√(μ₀ε₀).
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
Answer
D. 1, 2 and 3
All three are standard results.
Given below are two statements: Statement I: The wavelength of ultraviolet radiation is longer than that of infrared radiation. Statement II: The ozone layer absorbs most of the infrared radiation from the Sun.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
UV has a shorter wavelength than IR. The ozone layer absorbs ultraviolet, not infrared.
The radiation that is mainly responsible for the greenhouse effect is
- X-rays
- ultraviolet radiation
- infrared radiation
- microwaves
Answer
C. infrared radiation
The Earth re-emits absorbed energy as infrared radiation. Greenhouse gases and the atmosphere trap it and keep the Earth warm.
Assertion (A): Electromagnetic waves can travel through vacuum. Reason (R): Electromagnetic waves are transverse.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
They travel in vacuum because changing E and B sustain each other. Being transverse does not explain this, since transverse waves on a string need a medium.
Assertion (A): Gamma rays are the most penetrating part of the electromagnetic spectrum. Reason (R): Gamma rays are emitted by radioactive nuclei.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Penetration depends on their very high frequency and energy, not on the fact that nuclei emit them.
Who first produced and detected electromagnetic waves in the laboratory?
- Maxwell
- Oersted
- Hertz
- Faraday
Answer
C. Hertz
Maxwell predicted the waves. Hertz produced and detected them in 1887.
A magnetic field amplitude of an electromagnetic wave in vacuum when the electric field amplitude is 60 V m⁻¹ (c = 3 × 10⁸ m s⁻¹) is
- 1.8 × 10¹⁰ T
- 2 × 10⁷ T
- 5 × 10⁶ T
- 2 × 10⁻⁷ T
Answer
D. 2 × 10⁻⁷ T
B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T. Check by units: V m⁻¹ divided by m s⁻¹ gives T.