JEE Main 2025 · 28 January · morning · Single correct · +4 / −1
Due to presence of an em-wave whose electric component is given by E=100 sin (ω t-k x) NC⁻¹ a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as
- (a) 50 sin (ω t-kx) NC⁻¹
- (b) 400 sin (ω t-kx) NC⁻¹
- (c) 200 sin (ω t-k x) NC⁻¹
- (d) 25 sin (ω t-kx) NC⁻¹
Answer and solution
Answer (c) 200 sin (ω t-k x) NC⁻¹
Solution
Physics sir · 3 steps
Energy is energy density times volume: shrink the volume four times, so the field only needs to double.
- Energy in a cylinder. The average energy density of an EM wave is u = 1/2ε₀ E₀², counting both the electric and magnetic parts. The energy held in a cylinder is U = u × π r² L, so U ∝ E₀² r² when L is fixed.
- Volume shrinks. Halving the diameter halves the radius, so π r² L becomes 1/4 of the first volume. To hold the same energy, the energy density must become 4 times larger.
E₀'² (r/2)² = E₀² r² ⇒ E₀' = 2E₀ = 2 × 100 = 200 N C⁻¹
- So. The field must be 200 sin(ω t - kx) N C⁻¹, which is option (c).
The idea to keep. EM energy density goes as E₀², not E₀. When the volume changes, scale the energy density, then take a square root to get the change in field amplitude.
Answer from the NTA final answer key. Solution by PrepWiser.