JEE Main 2022 · 26 July · morning · Single correct · +4 / −1
The magnetic field of a plane electromagnetic wave is given by :
B⃗=2 × 10⁻⁸ sin (0.5 × 10³ x+1.5 × 10¹¹ t) ĵ T.
The amplitude of the electric field would be :
- (a) 6 Vm⁻¹ along x-axis
- (b) 3 Vm⁻¹ along z-axis
- (c) 6 Vm⁻¹ along z-axis
- (d) 2 × 10⁻⁸ Vm⁻¹ along z-axis
Answer and solution
Answer (c) 6 Vm⁻¹ along z-axis
Solution
Physics sir · 3 steps
In vacuum E₀ = cB₀, and E⃗ must be perpendicular to both B⃗ and the direction of travel.
- Read the wave. Compare with B₀ sin(kx + ω t). So B₀ = 2 × 10⁻⁸ T, k = 0.5 × 10³ m⁻¹ and ω = 1.5 × 10¹¹ rad s⁻¹. The speed is ω/k = 3 × 10⁸ m s⁻¹, which is c.
E₀ = cB₀ = (3 × 10⁸) × (2 × 10⁻⁸) = 6 V m⁻¹
- Find the direction. The phase kx + ω t means the wave travels along -x. B⃗ is along ĵ. E⃗ must be perpendicular to both, so it lies along the z-axis. As a check, k̂ × ĵ = -î, which matches the direction of travel.
- So. The electric field has amplitude 6 V m⁻¹ along the z-axis, which is option (c).
The idea to keep. For any EM wave: E₀ = cB₀, and E⃗, B⃗ and the direction of travel are mutually perpendicular, with E⃗ × B⃗ pointing along the direction of travel.
Answer from the NTA final answer key. Solution by PrepWiser.