Skip to content

Question

An electromagnetic wave in vacuum has a magnetic field amplitude of \( B_0 = 3 \times 10^{-8} \,
\text{T} \). What is the electric field amplitude? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

Options

Choose one · Correct answer highlighted

Explanation

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Using E₀ = B₀ c , we have E₀ = (3 × 10⁻⁸) × (3 × 10⁸) = 9 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 9 V/m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.