JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
The oscillating magnetic field in a plane electromagnetic wave is given by
B_y=5 × 10⁻⁶ sin 1000 π(5 x-4 × 10⁸ t) T. The amplitude of electric field will be :
- (a) 15 × 10² Vm⁻¹
- (b) 5 × 10⁻⁶ Vm⁻¹
- (c) 16 × 10¹² Vm⁻¹
- (d) 4 × 10² Vm⁻¹
Answer and solution
Answer (d) 4 × 10² Vm⁻¹
Solution
Physics sir · 3 steps
Do not assume the speed is c here; read the wave speed straight from the equation, then use E₀ = vB₀.
- Read k and ω. Open the bracket: the phase is 5000π x - 4 × 10¹¹π t. So k = 5000π m⁻¹ and ω = 4 × 10¹¹π rad s⁻¹.
- Wave speed. v = ω/k = (4 × 10¹¹π)/5000π = 8 × 10⁷ m s⁻¹. This is less than c, so the wave is travelling through a medium, not vacuum.
E₀ = vB₀ = (8 × 10⁷)(5 × 10⁻⁶) = 400 V m⁻¹
- So. The electric field amplitude is 4 × 10² V m⁻¹, option (d).
The idea to keep. In any EM wave the amplitudes are linked by E₀ = vB₀. Use v = c only in vacuum; if the equation is given, find v = ω/k from it first.
Answer from the NTA final answer key. Solution by PrepWiser.