JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm⁻¹. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :
- (a) E_y = 60 sin [ π/4 × 10³(x - 3 × 10⁸t) ]ĵ Vm⁻¹
B_z = 2 sin [ π/4 × 10³(x - 3 × 10⁸t) ]k̂ T - (b) E_y = 60 sin [ π/4 × 10³(x - 3 × 10⁸t) ]ĵ Vm⁻¹
B_z = 2 × 10⁻⁷sin [ π/4 × 10³(x - 3 × 10⁸t) ]k̂ T - (c) E_y = 2 × 10⁻⁷sin [ π/4 × 10³(x - 3 × 10⁸t) ]ĵ Vm⁻¹
B_z = 60 sin [ π/4 × 10³(x - 3 × 10⁸t) ]k̂ T - (d) E_y = 2 × 10⁻⁷sin [ π/4 × 10⁴(x - 4 × 10⁸t) ]ĵ Vm⁻¹
B_z = 60 sin [ π/4 × 10⁴(x - 4 × 10⁸t) ]k̂ T
Answer and solution
Answer (b) E_y = 60 sin [ π/4 × 10³(x - 3 × 10⁸t) ]ĵ Vm⁻¹ B_z = 2 × 10⁻⁷sin [ π/4 × 10³(x…
Solution
Physics sir · 3 steps
Get k from the wavelength and the field amplitude B₀ from E₀/c; everything else is already fixed.
- Wave number. The wavelength is λ = 8 mm = 8 × 10⁻³ m, so k = 2π/λ = 2π/(8 × 10⁻³) = π/4 × 10³ m⁻¹. In vacuum the wave moves at c = 3 × 10⁸ m s⁻¹, so the phase is k(x - ct).
B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T
- Directions. E⃗ is along ĵ and the wave travels along î. Since ĵ × k̂ = î, the magnetic field must be along k̂, in phase with E⃗.
- So. E_y = 60 sin[π/4 × 10³(x - 3 × 10⁸ t)] and B_z = 2 × 10⁻⁷ sin[π/4 × 10³(x - 3 × 10⁸ t)], which is option (b).
The idea to keep. In an EM wave, E₀ = cB₀, so the magnetic amplitude is tiny compared to the electric one. Both fields share the same phase k(x - ct), with k = 2π/λ.
Answer from the NTA final answer key. Solution by PrepWiser.