JEE Main 2021 · 16 March · morning · Single correct · +4 / −1
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time,
B⃗ = 8.0 × 10⁻⁸ zT. The value of electric field at this point is :
(speed of light = 3 × 10⁸ ms⁻¹)
x, y, z are unit vectors along x, y and z directions.
- (a) 2.6 x V/m
- (b) -24 x V/m
- (c) 24 x V/m
- (d) -2.6 y V/m
Answer and solution
Answer (b) -24 x V/m
Solution
Physics sir · 4 steps
Size from E = cB, sign from making E⃗ × B⃗ point along the travel direction.
- Magnitude. In vacuum the field magnitudes are tied by E = cB. The frequency does not enter this at all.
E = cB = (3 × 10⁸)(8.0 × 10⁻⁸) = 24 V m⁻¹
- Direction. E⃗ and B⃗ are perpendicular to each other and to the travel direction, so E⃗ lies along ±x. The wave travels along E⃗ × B⃗, which must be +y.
- Check the sign. x × z = -y, which is the wrong way. (-x) × z = +y, which is correct, so E⃗ points along -x.
- So. E⃗ = -24 x V/m, which is option (b).
The idea to keep. For any EM wave, E = cB gives the size, and E⃗ × B⃗ must point along the direction of travel. Use that cross product to pick the sign.
Answer from the NTA final answer key. Solution by PrepWiser.