JEE Main 2025 · 23 January · morning · Single correct · +4 / −1
The electric field of an electromagnetic wave in free space is E⃗=57 cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](4 î-3 ĵ) N / C. The associated magnetic field in Tesla is
- (a) B⃗=57/(3 × 10⁸) cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](k̂)
- (b) B⃗=57/(3 × 10⁸) cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](5 k̂)
- (c) B⃗=-57/(3 × 10⁸) cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](k̂)
- (d) B⃗=-57/(3 × 10⁸) cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](5 k̂)
Answer and solution
Answer (d) B⃗=-57/(3 × 10⁸) cos [7.5 × 10⁶ t-5 × 10⁻³(3 x+4 y)](5 k̂)
Solution
Physics sir · 4 steps
Read the travel direction from the phase, then cross it with E and divide by c to get B.
- Direction of travel. The phase is ω t - k⃗·r⃗ with k⃗ ∝ 3î + 4ĵ. So the wave moves along n = (3î + 4ĵ)/5, and B has the same phase as E.
B⃗ = 1/c n × E⃗
- Cross product. 1/5(3î + 4ĵ) × (4î - 3ĵ) = 1/5(-9k̂ - 16k̂) = -5k̂. So B⃗ = -57/(3 × 10⁸)cos[…](5k̂).
- Check. E⃗ has amplitude 57 × |4î - 3ĵ| = 285 N/C, and |B⃗| = 285/c matches E₀ = cB₀. Also (4î - 3ĵ) × (-5k̂) = 15î + 20ĵ, which points along n, as E⃗ × B⃗ must.
- So. The magnetic field is -57/(3 × 10⁸)cos[7.5 × 10⁶ t - 5 × 10⁻³(3x + 4y)](5k̂) T, which is option (d).
The idea to keep. For any EM wave, B⃗ = 1/cn × E⃗, with n taken from the phase. Fix the sign with E⃗ × B⃗ along n, and include the length of E's vector in its amplitude.
Answer from the NTA final answer key. Solution by PrepWiser.