JEE Main 2020 · 3 September · morning · Single correct · +4 / −1
The magnetic field of a plane electromagnetic wave is
B⃗ = 3 × 10⁻⁸sin [ 200π ( y + ct ) ]î T
where c = 3 × 10⁸ ms –1 is the speed of light. The corresponding electric field is :
- (a) E⃗ = - 10⁻⁶sin [ 200π ( y + ct ) ]k̂ V/m
- (b) E⃗ = - 9 sin [ 200π ( y + ct ) ]k̂ V/m
- (c) E⃗ = 9 sin [ 200π ( y + ct ) ]k̂ V/m
- (d) E⃗ = 3 × 10⁻⁸sin [ 200π ( y + ct ) ]k̂
Answer and solution
Answer (b) E⃗ = - 9 sin [ 200π ( y + ct ) ]k̂ V/m
Solution
Physics sir · 4 steps
Get the size from E₀ = cB₀, then pick the sign so E⃗ × B⃗ points along the travel.
- Direction of travel. The phase is 200π(y + ct). A phase of the form y + ct means the wave moves along -y, so it travels along -ĵ.
- Amplitude of E. E₀ = cB₀ = 3 × 10⁸ × 3 × 10⁻⁸ = 9 V m⁻¹. This rules out (a) and (d).
- Sign of E. E⃗ must be perpendicular to B⃗ (along î) and to the travel direction (along ĵ), so it lies along ±k̂. The rule is that E⃗ × B⃗ points along the direction of travel.
(-k̂) × î = -ĵ (direction of travel)
- So. Taking +k̂ would give k̂ × î = +ĵ, which is the wrong way. So E⃗ = -9 sin[200π(y + ct)]k̂ V m⁻¹, which is option (b).
The idea to keep. For any EM wave, E₀ = cB₀ gives the size. Read the travel direction from the phase, then choose the sign of E⃗ so that E⃗ × B⃗ points that way.
Answer from the NTA final answer key. Solution by PrepWiser.