JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is 6.0 × 10⁻⁷ T. The maximum value of electric field in the electromagnetic wave is
- (a) 6.0 × 10⁻⁷ Vm⁻¹
- (b) 5 × 10¹⁴ Vm⁻¹
- (c) 180 Vm⁻¹
- (d) 2 × 10¹⁵ Vm⁻¹
Answer and solution
Answer (c) 180 Vm⁻¹
Solution
Physics sir · 3 steps
In an EM wave the fields are tied by E₀ = cB₀: multiply the magnetic amplitude by the speed of light.
- Field relation. In an electromagnetic wave in vacuum, the electric and magnetic field amplitudes are linked by the speed of light. The direction of travel (here the y-axis) does not change this relation.
E₀ = c B₀
- Put in numbers. E₀ = (3 × 10⁸ m s⁻¹) × (6.0 × 10⁻⁷ T) = 18 × 10¹ = 180 V m⁻¹.
- So. The maximum electric field is 180 V m⁻¹, which is option (c).
The idea to keep. For any EM wave in vacuum, E₀/B₀ = c. The electric field number is always about 3 × 10⁸ times the magnetic field number in SI units.
Answer from the NTA final answer key. Solution by PrepWiser.