JEE Main 2025 · 23 January · evening · Single correct · +4 / −1
A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is E_y=9.3 Vm⁻¹. Then, the magnetic field vector of the wave at that point is
- (a) B_z=1.55 × 10⁻⁸ T
- (b) B_z=6.2 × 10⁻⁸ T
- (c) B_z=3.1 × 10⁻⁸ T
- (d) B_z=9.3 × 10⁻⁸ T
Answer and solution
Answer (c) B_z=3.1 × 10⁻⁸ T
Solution
Physics sir · 3 steps
In free space B is just E divided by c, and E × B must point along the direction of travel.
- Size of B. In a plane EM wave in free space, the field magnitudes at any point and instant are linked by E = cB. The frequency does not enter this relation, so the 20 MHz is not needed.
B = E/c = 9.3/(3 × 10⁸) = 3.1 × 10⁻⁸ T
- Direction of B. The wave travels along E⃗ × B⃗. With E⃗ along +y and travel along +x, B⃗ must lie along +z, because ĵ × k̂ = î.
- So. The magnetic field is B_z = 3.1 × 10⁻⁸ T, which is option (c).
The idea to keep. For any EM wave in vacuum, B = E/c at every point and instant, whatever the frequency. Fix the direction with E⃗ × B⃗ along the direction of travel.
Answer from the NTA final answer key. Solution by PrepWiser.