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Energy, Intensity and Momentum of EM Waves

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29 questions

What is the typical wavelength range of radio waves according to the document?

**Momentum of EM wave** p = U/c, U energy, radiation pressure exerts force F = I A/c, small but measurable, comet tail pushed by sunlight, solar sail concept. For E₀=45 V/m, B₀=1.5×10⁻⁷ T, intensity I =0.5×3×10⁸×8.85×10⁻¹²×45²≈2.69 W/m². The document mentions that radio waves have wavelengths greater than 0.1 m , with long radio waves extending up to 10⁶ m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields > 0.1 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

What is the primary source of a magnetic field as per Maxwell's generalization?

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Maxwell generalized Ampere's circuital law by introducing displacement current. According to this, the source of a magnetic field is not only the conduction current but also the time-varying electric field (displacement current). Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ =

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

Why are visible light waves critical for human vision?

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. Visible light waves have wavelengths that the human eye’s photoreceptors can detect, triggering neural responses that allow perception of color and detail in the environment. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Detectable wavelengths, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

What is the significance of the equation \( \oint \mathbf{E} \cdot \mathrm{d} \mathbf{l} = -\frac{d \Phi_B}{dt} \) in Ma

**Momentum of EM wave** p = U/c, U energy, radiation pressure exerts force F = I A/c, small but measurable, comet tail pushed by sunlight, solar sail concept. For E₀=45 V/m, B₀=1.5×10⁻⁷ T, intensity I =0.5×3×10⁸×8.85×10⁻¹²×45²≈2.69 W/m². This is Faraday's Law, which states that a changing magnetic flux induces an electric field, a key mechanism for electromagnetic wave propagation. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields It describes the induction of an electric field by changing magnetic flux, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has an electric field amplitude of \( 90 \, \text{V/m} \). What is the magnetic field

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Using B₀ = (E₀/c) , we have B₀ = (90/3 × 10⁸) = 3 × 10⁻⁷ T . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3 × 10⁻⁷ T, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

Which of the following statements about the electromagnetic spectrum is true?

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. The document explains that the boundaries between different regions of the electromagnetic spectrum are not sharply defined, and there are overlaps. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Boundaries are not sharply defined, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

What is the approximate wavelength range of infrared waves as per the document?

**Momentum of EM wave** p = U/c, U energy, radiation pressure exerts force F = I A/c, small but measurable, comet tail pushed by sunlight, solar sail concept. For E₀=45 V/m, B₀=1.5×10⁻⁷ T, intensity I =0.5×3×10⁸×8.85×10⁻¹²×45²≈2.69 W/m². The document states that infrared waves range from 1 mm to 700 nm . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1 mm to 700 nm, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

In electromagnetic theory, what explains the constant speed of light in vacuum across all frequencies?

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. The speed of light in vacuum is determined by the fundamental constants of permittivity ( ε₀ ) and permeability ( μ₀ ), independent of frequency, as c = (1/√(μ₀ ε₀)) . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f,

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has a magnetic field amplitude of \( B_0 = 2 \times 10^{-8} \, \text{T} \). What is th

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Using E₀ = B₀ c , we have E₀ = (2 × 10⁻⁸) × (3 × 10⁸) = 6 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6 V/m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has an angular frequency \( \omega = 4 \times 10^{11} \, \text{rad/s} \). What is its

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. Frequency v = (ω/2π) . Given ω = 4 × 10¹¹ rad/s , v = (4 × 10¹¹/2 π) ≈ 6.37 × 10¹⁰ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6.37 × 10¹⁰ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

Why are ultraviolet rays used in sterilization processes?

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Ultraviolet rays have high energy due to their short wavelengths, allowing them to disrupt molecular bonds in microorganisms, effectively killing them for sterilization purposes. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields High energy disruption, illustrating EM

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

Why do electromagnetic waves not require a medium for propagation?

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. Electromagnetic waves are self-sustaining oscillations of electric and magnetic fields, where a changing electric field generates a magnetic field and vice versa, requiring no material medium. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields They are self-sustaining field oscillations, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves