A 1000 W bulb with efficiency 1.25% emits radiation that sets up electric and magnetic fields at point P located 2 m away. The peak electric field at P is x * 10⁻¹ V/m. Find x (nearest integer). [e0 = 8.85e-12 C² N⁻¹ m⁻², c = 3e8 m/s]

Question

Accepted Answer

x = 137. Radiated power P = 0.0125 * 1000 = 12.5 W. Intensity at 2 m: I = P/(4*pi*r²) = 12.5/(4*pi*4) = 12.5/50.27 = 0.2487 W/m². Average intensity I = (1/2)*e0*c*E0², so E0 = sqrt(2I/(e0*c)) = sqrt(2*0.2487/(8.85e-12*3e8)) = sqrt(0.4974/2.655e-3) = sqrt(187.3) = 13.7 V/m. Then x*10⁻¹ = 13.7 means x = 137. Note: this gives x = 137; the simple reading is x = 137 V/m / 0.1. The nearest integer for x is 137.

A 1000 W bulb with efficiency 1.25% emits radiation that sets up electric and magnetic fields at point P located 2 m away. The peak electric field at P is x * 10⁻¹ V/m. Find x (nearest integer). [e0 = 8.85e-12 C² N⁻¹ m⁻², c = 3e8 m/s]

Solution

Related JEE Advanced Physics questions