Question

Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 25% and it is a point source.

Answer 1

Answer:

Explanation:

As per the question,

Given:

Power = 100 W

Efficiency = 2.5 % = 0.025

Radius = 3 m

Now

The surface area of surrounding of sphere is given by

A = 4 πr²

⇒ A = 4 × 3.14 × 3² = 113 m²

And the intensity at this radius is given by

I = 0.002 W/m²

As it is given that total intensity is half provided by the electric field and half by the magnetic field.

∴

c = speed of light = 3 × 10⁸ ms⁻¹

That is,

As the electric field in a light beam is of the form of sinusoidal, then the peak electric field is given by

Now,

we know that,

∴

   

Also,

As the magnetic field in a light beam is of the form of sinusoidal, then the peak magnetic field is given by

Answer 2

The value of electric field is 1 / 2 ( 0.0.22 W/m^2) and the value of magnetic field is 4.07 V/m

Explanation:

The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding sphere is

A = 4πr^2 = 4π(3)^2 = 113 m^2

The intensity l at this distance is

l = Power / Area = 100 W × 2.5% / 113 m^2

I = 0.022 W/m^2

Half of this intensity is provided by the electric field and half by the magnetic field.

1 / 2 l = 1 / 2(ε0 E^2 rms C)

= 1 / 2 ( 0.0.22 W/m^2)

E (rms) = √0.022 / (8.85 × 10^−12) ( 3×10^−8) V/m

The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E0 is

E0 = √2 E(rms) = √2 × 2.9 V/m

= 4.07 V/m

Thus the value of electric field is 1 / 2 ( 0.0.22 W/m^2) and the value of magnetic field is 4.07 V/m