Exams › JEE Main › Physics

A source of monochromatic green light radiates equally in every direction. If it converts only 3% of the 100 W electrical power it uses into electromagnetic radiation, then the electric-field amplitude of the emitted wave at a point 5 m away from the source is closest to:

1. 1.34 V/m
2. 2.68 V/m
3. 4.02 V/m
4. 5.36 V/m

Correct answer: 2.68 V/m

Solution

Radiated power = 3% of 100 W = 3 W. Intensity at 5 m: I = 3/(4*pi*5^2) = 9.55e-3 W/m^2. From I = (1/2)*e0*c*E0^2, E0 = sqrt(2I/(e0 c)) = 2.68 V/m.

Related JEE Main Physics questions

• A plane electromagnetic wave falls on a material surface. If it transfers momentum p and energy E, which statement is true?
• For an electromagnetic wave moving in vacuum, the electric field is expressed as E = E₀ sin(kx − ωt). Which of the following quantities does not depend on the wavelength?
• When an electromagnetic wave enters the ionised region of the ionosphere, the moving electron cloud gives rise to a space current, while the electric field also produces a capacitive displacement current. In this situation, what is the phase relation between the space current and the displacement current?
• The rms electric field associated with sunlight is 720 N/C. What is the mean total energy density of this electromagnetic wave?
• The electric field component of an electromagnetic wave in a medium is given by Eₓ = 0, E_y = 2.5 N/C cos[(2π × 10⁶ rad/m)x - (π × 10⁻² rad/s)t] ŷ, and E_z = 0. The wave is:
• For an electromagnetic wave traveling through a material medium, which statement about the electric and magnetic energy densities is correct?

⚔️ Practice JEE Main Physics free + battle 1v1 →