The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

(a) E/2           

(b) 2E         

(c) E/√2         

(d) E

An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in the this medium is_____× 10⁷ m/s.

The displacement current of 4.425 $\mu A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs^-1. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x × 10^-3. The value of x is,

(Permittivity of free space, $E_0=8.85\times 10^{-12}{C}^2{N}^{-1}{m}^{-2})\_\_\_\_\_\_\_\_\_\_$  

The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light – wave is__________ × 10^-9 T.

(Given : Permittivity of vacuum ${\in }_0=8.85\times 10^{-12}{C}^2{N}^{-1}-m^{-2},$  speed of light in vacuum c = 3 × 10⁸ ms^-1)

An electromagnetic wave is represented by the electric field
E⃗ = E₀n̂[ωt + (y − z)]
The direction of propagation of the wave, ŝ is:

A point source surrounded by vacuum emits an electromagnetic wave of frequency of 900 kHz. If the power emitted by the source is 15 W, the amplitude of the electric field of the wave (in 10^-3V/m) at a distance 15 km from the source is :
( (1 / 4πε₀) = 9×10⁹ Nm²/C² and speed of light in vacuum = 3 × 10⁸ ms^-1)