JEE Main 2023 · 11 April · evening · Single correct · +4 / −1
A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular space and time, E⃗=6.6 ĵ V / m. What is B⃗ at this point?
- (a) -2.2 × 10⁻⁸ î T
- (b) 2.2 × 10⁻⁸ î T
- (c) 2.2 × 10⁻⁸ k̂ T
- (d) -2.2 × 10⁻⁸ k̂ T
Answer and solution
Answer (c) 2.2 × 10⁻⁸ k̂ T
Solution
Physics sir · 4 steps
Get the size from B = E/c, then let E⃗ × B⃗ point along the wave to fix the direction.
- Magnitude. In free space the field magnitudes are linked by B = E/c. The frequency of 20 MHz does not enter this at all.
B = E/c = 6.6/(3 × 10⁸) = 2.2 × 10⁻⁸ T
- Direction. An EM wave is transverse, so B⃗ must be perpendicular to both E⃗ (along ĵ) and the propagation direction (along î). That leaves only ±k̂.
- Fix the sign. The wave travels along E⃗ × B⃗. Since ĵ × k̂ = î, choosing +k̂ makes the wave travel along +x, as it should.
- So. B⃗ = 2.2 × 10⁻⁸ k̂ T, which is option (c).
The idea to keep. For any EM wave, E⃗, B⃗ and the direction of travel form a right-handed set: E⃗ × B⃗ points along propagation. Magnitudes always follow E = cB.
Answer from the NTA final answer key. Solution by PrepWiser.