JEE Main 2024 · 9 April · evening · Single correct · +4 / −1
The magnetic field in a plane electromagnetic wave is B_y=(3.5 × 10⁻⁷) sin (1.5 × 10³ x+0.5 × 10¹¹ t) T. The corresponding electric field will be :
- (a) E_z=105 sin (1.5 × 10³ x+0.5 × 10¹¹ t) Vm⁻¹
- (b) E_y=10.5 sin (1.5 × 10³ x+0.5 × 10¹¹ t) Vm⁻¹
- (c) E_y=1.17 sin (1.5 × 10³ x+0.5 × 10¹¹ t) Vm⁻¹
- (d) E_z=1.17 sin (1.5 × 10³ x+0.5 × 10¹¹ t) Vm⁻¹
Answer and solution
Answer (a) E_z=105 sin (1.5 × 10³ x+0.5 × 10¹¹ t) Vm⁻¹
Solution
Physics sir · 4 steps
Size from E₀ = cB₀, direction perpendicular to both B⃗ and the direction of travel.
- Direction of E. In an EM wave, E⃗, B⃗ and the direction of travel are all perpendicular to each other. The wave varies with x, so it travels along the x-axis, and B⃗ is along y. So E⃗ must be along z, which rules out E_y.
- Amplitude of E. The two field amplitudes are linked by the wave speed. The options are built on c = 3 × 10⁸ m s⁻¹, so use that.
E₀ = cB₀ = (3 × 10⁸)(3.5 × 10⁻⁷) = 105 V m⁻¹
- Same phase. E⃗ and B⃗ oscillate in phase, so E_z has the same sine term as B_y. The phase kx + ω t means the wave moves along -x. Since z × y = -x, an E_z field fits this direction of travel.
- So. E_z = 105 sin(1.5 × 10³ x + 0.5 × 10¹¹ t) V m⁻¹, which is option (a).
The idea to keep. In every EM wave question, fix two things. The direction of E⃗ is perpendicular to both B⃗ and the direction of travel, and its size is E₀ = cB₀.
Answer from the NTA final answer key. Solution by PrepWiser.