JEE Main 2024 · 9 April · morning · Single correct · +4 / −1
A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60 Vm⁻¹, the equation for magnetic field is :
- (a) B_z=2 × 10⁻⁷ sin [π/2(x-3 × 10⁸ t)] k̂ T
- (b) B_z=2 × 10⁻⁷ sin [π/2 × 10³(x-3 × 10⁸ t)] k̂ T
- (c) B_z=60 sin [π/2(x-3 × 10⁸ t)] k̂ T
- (d) Bₓ=60 sin [π/2(x-3 × 10⁸ t)] î T
Answer and solution
Answer (b) B_z=2 × 10⁻⁷ sin [π/2 × 10³(x-3 × 10⁸ t)] k̂ T
Solution
Physics sir · 4 steps
Get three things right for B: its direction, its amplitude E₀/c, and k = 2π/λ with λ in metres.
- Direction of B. In an EM wave, E⃗, B⃗ and the direction of travel are mutually perpendicular, with E⃗ × B⃗ along the motion. Since ĵ × k̂ = î, with E⃗ along y and travel along x, B⃗ must be along z.
- Amplitude of B. The field amplitudes are linked by E₀ = cB₀, so B₀ = 60/(3 × 10⁸) = 2 × 10⁻⁷ T.
k = 2π/λ = 2π/(4 × 10⁻³ m) = 500π = π/2 × 10³ m⁻¹
- Build the equation. Write B_z = B₀ sin[k(x - ct)] with c = 3 × 10⁸ m s⁻¹. This gives B_z = 2 × 10⁻⁷ sin[π/2 × 10³ (x - 3 × 10⁸ t)]k̂ T.
- So. Only option (b) has the right direction, the right amplitude and the right wave number.
The idea to keep. For B from E in an EM wave: B is perpendicular to both E and the direction of travel, B₀ = E₀/c, and k = 2π/λ with λ in metres.
Answer from the NTA final answer key. Solution by PrepWiser.