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#hertz

3 public questions tagged with this topic.

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

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EM Wave Characteristics - Transverse Nature and E/B Ratio

An electromagnetic wave has an angular frequency \( \omega = 6 \times 10^{11} \, \text{rad/s} \). What is its frequency

**Relationship E and B** in EM wave E₀ = c B₀, B₀ = E₀/c, for vacuum. Fields sustain each other via Maxwell's equations ∇×E = -∂B/∂t, ∇×B = μ₀ ε₀ ∂E/∂t, time-varying E produces B and vice versa, self-sustaining propagation without medium, speed c. Frequency v = (ω/2π) . Given ω = 6 × 10¹¹ rad/s , v = (6 × 10¹¹/2 π) ≈ 9.55 × 10¹⁰ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 9.55 × 10¹⁰ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

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Displacement Current and Ampere-Maxwell Law

An electromagnetic wave has an angular frequency \( \omega = 8 \times 10^{11} \, \text{rad/s} \). What is its frequency

**Displacement current** I_d = ε₀ dΦ_E/dt, Φ_E = ∫ E·dA electric flux (V·m), ε₀=8.85×10⁻¹² F/m, ensures continuity of current in charging capacitor where conduction current stops between plates, I_d equals conduction current in wires, 3 A conduction ⇒ 3 A displacement, maintaining Ampere's law ∮ B·dl = μ₀(I_c+I_d). Frequency v = (ω/2π) . Given ω = 8 × 10¹¹ rad/s , v = (8 × 10¹¹/2 π) ≈ 1.27 × 10¹¹ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.27 × 10¹¹ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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