JEE Main 2024 · 29 January · evening · Single correct · +4 / −1
A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E⃗=9.6 ĵ V / m. The value of magnetic field at this point is :
- (a) 9.6 ĵ T
- (b) 3.2 × 10⁻⁸ î T
- (c) 9.6 × 10⁻⁸ k̂ T
- (d) 3.2 × 10⁻⁸ k̂ T
Answer and solution
Answer (d) 3.2 × 10⁻⁸ k̂ T
Solution
Physics sir · 3 steps
Get the size from B = E/c, then let E⃗ × B⃗ point along the travel direction.
- Magnitude. In free space the field amplitudes at any point are linked by E = cB. The frequency of 35 MHz plays no part here.
B = E/c = 9.6/(3 × 10⁸) = 3.2 × 10⁻⁸ T
- Direction. The wave travels along the direction of E⃗ × B⃗, which here is î. With E⃗ along ĵ, we need ĵ × B = î. Since ĵ × k̂ = î, B⃗ must point along k̂.
- So. B⃗ = 3.2 × 10⁻⁸ k̂ T, which is option (d).
The idea to keep. For any EM wave, E⃗, B⃗ and the travel direction are mutually perpendicular, with E⃗ × B⃗ along the travel direction. Divide E by c for the size of B.
Answer from the NTA final answer key. Solution by PrepWiser.