68 free MCQs on Electromagnetic Waves with worked answers and explanations. Maxwell's equations, EM spectrum, properties, and applications of each band.
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In an electromagnetic wave, the oscillating electric field (E) and magnetic field (B) are perpendicular to each other and to the direction the wave travels - a purely transverse wave that needs no medium, unlike sound.
Q1.
Electromagnetic waves are produced by:
Answer: A. Oscillating or accelerating electric charges
Why: Accelerating or oscillating electric charges create changing electric fields, which create changing magnetic fields, which in turn create changing electric fields - resulting in EM waves that propagate outward.
Q2.
The speed of electromagnetic waves in vacuum is:
Answer: A. 3 × 10⁸ m/s (same for all EM waves)
Why: All electromagnetic waves travel at c = 3 × 10⁸ m/s in vacuum, regardless of their frequency or wavelength. This is a fundamental constant of nature.
Q3.
In an electromagnetic wave, the electric field E and magnetic field B are:
Answer: A. Perpendicular to each other and to the direction of propagation
Why: EM waves are transverse. E, B, and the propagation direction are all mutually perpendicular. If E is along x-axis and B along y-axis, propagation is along z-axis.
Q4.
Which part of the electromagnetic spectrum has the highest frequency?
Answer: A. Gamma rays
Why: The EM spectrum from lowest to highest frequency: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Gamma rays have the highest frequency (and energy) and shortest wavelength.
Q5.
Microwaves are used in cooking because they:
Answer: A. Cause water molecules to vibrate (dielectric heating)
Why: Microwave ovens use frequencies (~2.45 GHz) that efficiently couple to water molecules, causing them to rotate and vibrate, generating heat throughout the food uniformly.
Q6.
X-rays are used in medical imaging because they:
Answer: A. Penetrate soft tissue but are absorbed by denser materials like bone
Why: X-rays pass easily through soft tissue (mostly water and carbon) but are absorbed by denser materials like bone and metal. This differential absorption creates the image on X-ray film.
Q7.
Gamma rays are produced by:
Answer: A. Nuclear transitions (radioactive decay)
Why: Gamma rays are emitted during nuclear reactions and radioactive decay when the nucleus transitions from an excited to a lower energy state. They have the highest energy EM radiation.
Q8.
Infrared radiation is associated with:
Answer: A. Heat (thermal radiation from warm objects)
Why: All objects above absolute zero emit infrared radiation. The hotter the object, the more IR it emits. This is why thermal cameras detect temperature differences using IR.
Q9.
The displacement current was introduced by Maxwell because:
Answer: A. Ampere's law was inconsistent for a changing electric field (capacitor plates)
Why: Maxwell noticed that Ampere's law gave contradictory results for the magnetic field around a charging capacitor. Adding displacement current I<sub>d</sub> = ε_0 dΦ_E/dt made the law consistent.
Q10.
The relationship between electric and magnetic field amplitudes in an EM wave is:
Answer: A. E<sub>0</sub>/B<sub>0</sub> = c (speed of light)
Why: For an EM wave: E<sub>0</sub>/B<sub>0</sub> = c. Since c = 3 × 10⁸ m/s, the magnetic field amplitude is much smaller (by factor c) than the electric field amplitude in SI units.
Q11.
Ultraviolet radiation is responsible for:
Answer: A. Sunburn and skin cancer (damages DNA), also synthesizes vitamin D
Why: UV radiation (especially UVB and UVC) can damage DNA, causing sunburn and increasing skin cancer risk. However, UVB also triggers vitamin D synthesis in the skin.
Q12.
Radio waves are used for communication because they:
Answer: A. Travel long distances, can be reflected by the ionosphere, and carry modulated information
Why: Radio waves (long wavelength) can travel around the Earth by ionospheric reflection, and carry information through amplitude (AM) or frequency (FM) modulation.
Q13.
The ozone layer in the atmosphere protects Earth by absorbing:
Answer: A. Ultraviolet radiation
Why: The ozone layer (stratosphere, 15-35 km altitude) absorbs harmful ultraviolet radiation, particularly UVC (most harmful) and much of UVB. Depletion of ozone (by CFCs) allows more UV to reach Earth.
Q14.
Which EM radiation is used in remote controls for TV?
Answer: A. Infrared
Why: TV remote controls use infrared LEDs (wavelength ~940 nm). The IR signal carries coded information (pulse patterns). The TV sensor detects these IR pulses.
Q15.
Which property does an electromagnetic wave NOT possess?
Answer: A. It requires a medium to propagate
Why: Unlike sound (mechanical wave), EM waves do NOT require a medium. They can travel through vacuum. This is why we can see sunlight and receive satellite signals through the vacuum of space.
Q16.
The frequency of visible light (wavelength 500 nm) is approximately:
Answer: A. 6 × 10¹⁴ Hz
Why: f = c/λ = (3×10⁸)/(500×10⁻⁹) = 3×10⁸/5×10⁻⁷ = 6×10¹⁴ Hz. Visible light is in the range 4-7.5 × 10¹⁴ Hz.
Q17.
The intensity of an EM wave is proportional to:
Answer: A. The square of the amplitude (E<sub>0</sub>² or B<sub>0</sub>²)
Why: Intensity I = average power/area. For an EM wave, I ∝ E<sub>0</sub>² (or B<sub>0</sub>²). Doubling the amplitude quadruples the intensity.
Q18.
Maxwell unified electricity, magnetism, and optics by showing that:
Answer: A. Light is an electromagnetic wave
Why: Maxwell derived from his equations that EM waves propagate at speed 1/√(μ_0 ε_0) = c = 3 × 10⁸ m/s, matching the known speed of light. This proved light is an EM wave.
Q19.
RADAR (Radio Detection and Ranging) uses:
Answer: A. Microwaves (cm-wavelength)
Why: RADAR uses microwave pulses. They are reflected by metal objects (aircraft, ships). The time delay between sending and receiving gives the distance. Weather RADAR detects water droplets.
Q20.
The unit of intensity of electromagnetic radiation is:
Answer: A. W/m²
Why: Intensity = power per unit area = W/m². The intensity of sunlight at Earth surface is approximately 1000 W/m² (solar constant ≈ 1361 W/m² above atmosphere).
Q21.
A microwave oven operates at 2.45 GHz. What is the wavelength of the microwaves used?
Answer: A. 12.2 cm
Why: λ = c/f = 3×10⁸/(2.45×10⁹) = 3/2.45 × 10⁻¹ ≈ 0.122 m = 12.2 cm.
Q22.
The Poynting vector S gives:
Answer: A. Energy flux (power per unit area) of an EM wave, S = E × B/μ_0
Why: Poynting vector S = (1/μ_0)(E × B). Its magnitude is the intensity (W/m²) and its direction is the propagation direction of the EM wave.
Q23.
Radiation pressure of an EM wave on a perfectly absorbing surface is:
Answer: A. P = I/c (intensity divided by speed of light)
Why: Radiation pressure P = I/c (for perfect absorption). For perfect reflection, P = 2I/c (momentum change is doubled as direction reverses). This pressure is extremely small but detectable.
Q24.
In which region of the EM spectrum does VIBGYOR (visible light) lie?
Answer: A. Between infrared and ultraviolet (wavelengths 400-700 nm)
Why: Visible light occupies a narrow band: violet ≈400 nm to red ≈700 nm. Below violet is UV; above red is infrared. The visible spectrum is a tiny part of the EM spectrum.
Q25.
The energy density of an EM wave has equal contributions from E and B fields:
Answer: A. u = ε_0 E² = B²/μ_0 (each term equal at any instant)
Why: Total energy density u = ½ε_0 E² + ½B²/μ_0. At any instant, ½ε_0 E² = ½B²/μ_0 (equal electric and magnetic energy densities). Total u = ε_0 E² = B²/μ_0.
Q26.
Which EM radiation is used in PET (Positron Emission Tomography) scans?
Answer: A. Gamma rays (from positron-electron annihilation)
Why: In PET, radioactive tracers emit positrons. When a positron meets an electron, they annihilate producing two gamma-ray photons in opposite directions. Detectors locate the annihilation events to create 3D images.
Q27.
The ionosphere reflects radio waves of frequency below a critical frequency. This is used in:
Answer: A. AM radio broadcasting (long distance, ground and sky waves)
Why: AM radio uses frequencies (535-1705 kHz) that are reflected by the ionosphere. This allows AM signals to travel beyond the horizon. FM radio (88-108 MHz) passes through the ionosphere and is used for shorter ranges.
Q28.
Displacement current in a capacitor charging is:
Answer: A. I<sub>d</sub> = ε_0 dΦ_E/dt (linked to changing electric flux)
Why: Between capacitor plates, no real current flows but the electric field changes. The displacement current I<sub>d</sub> = ε_0 dΦ_E/dt ensures continuity of current in Ampere-Maxwell law.
Q29.
Which characteristic of an EM wave changes when it passes from vacuum to glass?
Answer: A. Frequency remains same; wavelength and speed decrease
Why: When an EM wave enters a medium, its speed decreases (v = c/n), and wavelength decreases (λ' = λ/n), but the frequency remains unchanged (f = c/λ = v/λ').
Q30.
The wavelength of gamma rays is typically:
Answer: A. Less than 0.01 nm (< 10⁻² nm)
Why: Gamma rays: λ < 0.01 nm (10 pm). X-rays: 0.01-10 nm. UV: 10-400 nm. Visible: 400-700 nm. IR: 700 nm-1 mm. Microwaves: 1 mm-0.1 m. Radio: > 0.1 m.
Q31.
EM waves carry linear momentum. The momentum of a photon of frequency f is:
Answer: A. p = hf/c = h/λ
Why: Photon momentum p = E/c = hf/c = h/λ. This momentum causes radiation pressure and is used in laser cooling of atoms and solar sail propulsion concepts.
Q32.
Infrared waves are used in:
Answer: A. Night vision cameras, TV remotes, and detecting heat sources
Why: IR applications: remote controls, night vision, thermal imaging, IR spectroscopy, heating, fiber optic communications (near IR), and detecting warm objects in the dark.
Q33.
According to Maxwell, a changing magnetic field creates:
Answer: A. An electric field (Faraday-Maxwell law)
Why: Maxwell extended Faraday's law: a changing magnetic field creates an electric field. Symmetrically, Ampere-Maxwell law: a changing electric field creates a magnetic field. Together, these sustain EM waves.
Q34.
If the electric field amplitude of an EM wave is E<sub>0</sub> = 100 V/m, what is the magnetic field amplitude?
Answer: A. B<sub>0</sub> = 3.33 × 10⁻⁷ T
Why: B<sub>0</sub> = E<sub>0</sub>/c = 100/(3×10⁸) = 3.33×10⁻⁷ T. The magnetic field amplitude is much smaller than the electric field amplitude in SI units.
Q35.
Long wavelength radio waves can diffract around mountains and follow Earth's curvature because:
Answer: A. Their wavelength is comparable to or larger than obstacle dimensions
Why: Diffraction is significant when wavelength ≳ obstacle size. Long radio waves (km wavelengths) diffract around mountains and curve with Earth. Short waves (MHz range) require line-of-sight or reflection.
Q36.
The greenhouse effect works because:
Answer: A. CO<sub>2</sub> and H<sub>2</sub>O absorb outgoing infrared radiation from Earth but allow incoming visible light
Why: Greenhouse gases are transparent to incoming solar visible light but absorb the longer-wavelength IR re-emitted by Earth. This traps heat, raising the temperature.
Q37.
Frequency modulation (FM) radio is better quality than amplitude modulation (AM) because:
Answer: A. Noise creates amplitude variations (easily filtered in FM); FM has wider bandwidth for more audio detail
Why: Most noise/interference affects signal amplitude. In FM, information is in frequency (constant amplitude), so amplitude noise is stripped out, giving cleaner reception.
Q38.
The electric field of an EM wave is E = E<sub>0</sub> sin(kx - ωt). The wave is traveling in the:
Answer: A. Positive x direction
Why: E = E<sub>0</sub> sin(kx - ωt): the wave travels in the positive x direction (kx and ωt terms: positive k coefficient, negative ωt coefficient means positive x travel).
Q39.
The intensity of electromagnetic radiation from a point source varies with distance r as:
Answer: A. I ∝ 1/r² (inverse square law)
Why: Power spreads over a sphere: I = P/(4πr²). As r doubles, area quadruples, I decreases 4 times. This inverse square law applies to all EM radiation from point sources.
Q40.
In the EM spectrum, as frequency increases:
Answer: A. Wavelength decreases and photon energy increases
Why: c = fλ, so λ = c/f decreases as f increases. Photon energy E = hf increases with frequency. Higher frequency waves are more penetrating and more dangerous (UV, X-rays, gamma).
Q41.
Displacement current was introduced by Maxwell to account for:
Answer: B. Current through capacitor (changing E field)
Why: Maxwell added displacement current (epsilon<sub>0</sub> x dE/dt) to Ampere's law to account for changing electric field in capacitor gaps.
Q42.
In electromagnetic waves, E and B fields are:
Answer: B. Perpendicular to each other and both perpendicular to propagation
Why: EM waves: E perpendicular to B, both perpendicular to direction of propagation. All three are mutually perpendicular.
Q43.
Derive the speed of light from Maxwell's equations: c = 1/√(μ_0ε_0). Given μ_0 = 4π×10⁻⁷ H/m and ε_0 = 8.85×10⁻¹² F/m. Calculate c.
Answer: A. 3 × 10⁸ m/s
Why: c = 1/√(μ_0ε_0) = 1/√(4π×10⁻⁷ × 8.85×10⁻¹²) = 1/√(1.11×10⁻¹⁷) = 1/(3.33×10⁻⁹) ≈ 3×10⁸ m/s. This is how Maxwell first derived the speed of light.
Q44.
The amplitude of the E field in sunlight (I = 1000 W/m²) is:
Answer: A. 868 V/m
Why: I = ½cε_0 E<sub>0</sub>². E<sub>0</sub> = √(2I/(cε_0)) = √(2×1000/(3×10⁸×8.85×10⁻¹²)) = √(2000/2.65×10⁻³) = √(7.54×10⁵) ≈ 868 V/m.
Q45.
An EM wave has electric field E = 6 sin(ωt - kx) V/m. The average intensity is:
Answer: A. 0.0477 W/m²
Why: I = E<sub>0</sub>²/(2cμ_0) = 36/(2×3×10⁸×4π×10⁻⁷) = 36/(2×120π) = 36/(240π) ≈ 36/754 ≈ 0.0477 W/m².
Q46.
A laser beam (λ = 600 nm, power = 1 mW, beam area = 1 mm²) exerts radiation pressure on a perfect mirror. The force is:
Answer: A. 6.67 × 10⁻⁹ N
Why: Radiation pressure on mirror = 2I/c. I = P/A = 10⁻³/10⁻⁶ = 10³ W/m². Pressure = 2×10³/(3×10⁸) = 6.67×10⁻⁶ Pa. Force = P×A = 6.67×10⁻⁶ × 10⁻⁶ = 6.67×10⁻¹² N.
Q47.
The wave equation for the electric field component of an EM wave is:
Answer: A. ∇²E = μ_0ε_0 ∂²E/∂t² (wave equation with speed c)
Why: Combining Maxwell's equations gives the wave equation: ∇²E = μ_0ε_0 ∂²E/∂t². Comparing with the general wave equation ∇²E = (1/v²)∂²E/∂t² gives v = 1/√(μ_0ε_0) = c.
Q48.
In Compton scattering, an X-ray photon scatters off an electron. The wavelength shift is:
Answer: A. Δλ = (h/m<sub>e</sub> c)(1 - cosθ) (Compton formula)
Why: Compton shift: Δλ = (h/m<sub>e</sub> c)(1 - cosθ) where θ is the scattering angle and h/m<sub>e</sub> c = 2.43 pm is the Compton wavelength. This confirms the particle nature of X-rays.
Q49.
The skin depth (depth at which EM wave amplitude falls to 1/e) in a conductor is:
Answer: A. δ = √(2/μσω) where σ is conductivity
Why: Skin depth δ = √(2ρ/μω) = √(2/μσω). For copper at 60 Hz, δ ≈ 8.5 mm. At 1 MHz, δ ≈ 0.07 mm. This limits AC current to the conductor surface at high frequency.
Q50.
A gamma-ray photon of wavelength 1 pm has energy approximately:
Answer: A. 1.24 MeV
Why: E = hf = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(10⁻¹²) = 1.99×10⁻¹³ J = 1.99×10⁻¹³/1.6×10⁻¹⁹ eV ≈ 1.24×10⁶ eV = 1.24 MeV.
Q51.
The critical frequency (MUF - Maximum Usable Frequency) for ionospheric reflection depends on:
Answer: A. Electron density in the ionosphere (f<sub>c</sub> = 9√N where N = electron density)
Why: Plasma frequency f<sub>p</sub> = (1/2π)√(Ne²/ε_0 m<sub>e</sub>) = 9√N (in SI). Waves above this frequency pass through the ionosphere; waves below are reflected. This determines which frequencies can be used for long-range HF communications.
Q52.
Synchrotron radiation is emitted when:
Answer: A. Relativistic electrons travel in circular paths in magnetic fields
Why: Accelerating charges emit EM radiation. Relativistic electrons in synchrotrons are centripetally accelerated, emitting intense, highly collimated EM radiation from radio waves to X-rays.
Q53.
If the amplitude of an EM wave in vacuum is E<sub>0</sub>, in a medium of refractive index n (transparent), the E field amplitude:
Answer: D. E<sub>0</sub>/√n
Why: In a dielectric: intensity I = ½nε_0 c E<sub>0</sub>². For same intensity: E<sub>medium</sub> = E<sub>vacuum</sub>/√n. The electric field amplitude decreases in a denser medium.
Q54.
The Cherenkov radiation is produced when:
Answer: A. A charged particle moves faster than light in that medium
Why: Cherenkov radiation is emitted when a charged particle travels faster than the phase velocity of light in that medium (c/n). It appears as a blue glow in nuclear reactors (water acts as the medium).
Q55.
In the Sun, energy reaches Earth as visible light but is re-emitted as infrared. The peak wavelength (λ_max) for a blackbody at 5800 K (Sun) by Wien's law is:
Answer: A. 500 nm (visible)
Why: Wien's displacement law: λ_max T = 2.898 × 10⁻³ m·K. For T = 5800 K: λ_max = 2.898×10⁻³/5800 ≈ 500 nm (visible light). Earth at 300 K emits at λ_max ≈ 10 μm (infrared).
Q56.
Which phenomenon demonstrates the vector nature of the electric field in EM waves?
Answer: A. Polarization (wave can be polarized only for transverse waves with vector displacement)
Why: Polarization shows that E has a specific direction perpendicular to propagation (vector nature of transverse EM wave). Unpolarized light has E randomly oriented; polarized light has a defined E direction.
Q57.
The average energy density of an EM wave is related to the electric field by:
Answer: A. <u> = ε_0 E<sub>0</sub>²/2 = ε_0 E<sub>rms</sub>²
Why: Instantaneous u = ε_0 E² (total, including both E and B contributions). Time-averaged: <u> = ε_0 E<sub>rms</sub>² = ε_0 E<sub>0</sub>²/2. Intensity = <u> × c = ε_0 c E<sub>0</sub>²/2.
Q58.
In medical MRI (Magnetic Resonance Imaging), the EM waves used are:
Answer: A. Radio waves (RF pulses, ~64 MHz for 1.5 T MRI)
Why: MRI uses radio frequency pulses that match the Larmor precession frequency of hydrogen nuclei in the magnetic field. At 1.5 T, this is about 64 MHz (FM radio range). MRI is non-ionizing and safer than X-rays.
Q59.
The concept of displacement current ensures conservation of charge (continuity equation). In a capacitor being charged with current I, the displacement current between plates is:
Answer: A. Exactly equal to I (continuous current through circuit)
Why: Displacement current I<sub>d</sub> = ε_0 dΦ_E/dt. As the capacitor charges, Φ_E increases. Maxwell showed I<sub>d</sub> = I (conduction current in wires), ensuring charge conservation across the capacitor gap.
Q60.
The photon energy of a UV photon (λ = 200 nm) is:
Answer: A. 6.2 eV
Why: E = hc/λ = (4.136×10⁻¹⁵ eV·s × 3×10⁸ m/s)/(200×10⁻⁹ m) = 1.24×10⁻⁶/200×10⁻⁹ = 6.2 eV. UV photons have enough energy to break chemical bonds (≥ 3.5 eV).
Q61.
The speed of electromagnetic waves in vacuum, 1/√(μ₀ε₀), is:
Answer: C. 3×10⁸ m/s
Why: c = 1/√(μ₀ε₀) = 3×10⁸ m/s.
Q62.
In an electromagnetic wave in vacuum, the ratio of the amplitudes of the electric field to the magnetic field (E₀/B₀) equals:
Answer: A. c
Why: For an EM wave in vacuum, E₀/B₀ = c.
Q63.
Among radio waves, X-rays, ultraviolet, and gamma rays, the one with the highest frequency is:
Answer: D. gamma rays
Why: Gamma rays have the highest frequency (and shortest wavelength) of these.
Q64.
An electromagnetic wave carries energy 3×10⁸ J and is completely absorbed. The momentum delivered is:
Answer: B. 1 kg·m/s
Why: p = U/c = 3×10⁸/(3×10⁸) = 1 kg·m/s.
Q65.
The concept added by Maxwell to Ampere law to make it consistent is the:
Answer: A. displacement current
Why: Maxwell introduced the displacement current term ε₀ dΦ_E/dt to complete Ampere law.
Q66.
A radio station broadcasts at 100 MHz. The wavelength of its EM waves is:
Answer: B. 3 m
Why: λ = c/f = 3×10⁸/(100×10⁶) = 3 m.
Q67.
The average intensity of an electromagnetic wave in vacuum in terms of the peak electric field E₀ is:
Answer: A. (1/2)cε₀E₀²
Why: Average intensity I = (1/2)cε₀E₀².
Q68.
The type of electromagnetic radiation commonly used in TV remote controls is:
Answer: A. infrared
Why: Remote controls use infrared radiation to transmit signals.