JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the x-direction. At a particular point in space and time, B⃗ = 2.0 × 10⁻⁸k̂T. (where, k̂ is unit vector along z-direction) What is E⃗ at this point?
- (a) 0.6 ĵ V/m
- (b) 6.0 k̂ V/m
- (c) 6.0 ĵ V/m
- (d) 0.6 k̂ V/m
Answer and solution
Answer (c) 6.0 ĵ V/m
Solution
Physics sir · 3 steps
Get the size from E = cB, then use E × B along x to fix the direction.
- Magnitude. In vacuum the field amplitudes are linked by the speed of light. The frequency of 100 MHz does not enter this relation.
E = cB = (3 × 10⁸ m s⁻¹)(2.0 × 10⁻⁸ T) = 6.0 V m⁻¹
- Direction. E⃗ and B⃗ are perpendicular to each other and to the direction of travel. E⃗ × B⃗ points along the propagation direction, +î. With B⃗ along k̂, E⃗ must be along ĵ, because ĵ × k̂ = î.
- So. E⃗ = 6.0 ĵ V/m, which is option (c).
The idea to keep. For any EM wave in vacuum, E = cB gives the size. E⃗, B⃗ and the direction of travel form a right-handed set, so E⃗ × B⃗ points where the wave goes.
Answer from the NTA final answer key. Solution by PrepWiser.