JEE Main202126 Feb 2021Morning ShiftPhysicsElectromagnetic WavesActual
A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P , placed at a distance of 2 m . The efficiency of the bulb is 1 . 25 % . The value of peak electric field at P is x × 10 - 1 V m - 1 . Value of x is (Rounded-off to the nearest integer) [Take ε 0 = 8 . 85 × 10 - 12 C 2 N - 1 m - 2 , c = 3 × 10 6 m s - 1 ]
Correct answer
0
Step-by-step solution
I avg = 1 2 ε 0 E 0 2 c 1 . 25 100 × 1000 4 π 2 2 = 1 2 × 8 . 85 × 10 - 12 × 3 × 10 8 × E 0 2 E 0 2 = 187 . 4 ∴   E 0 = 13 . 689   V   m - 1 = 136 . 89 × 10 - 1   V   m - 1 ∴   x = 136 . 89 Rounding off to nearest integer x = 137