A monochromatic point like light source emits 1.50 W of electromagnetic power uniformly in all directions.The Poynting vector in a point situated at 0.750 m from the source is

Question

Quizplus · Accepted Answer

The intensity of the electromagnetic wave is given by the Poynting vector, which is defined as the cross product of the electric field vector and the magnetic field vector. In this case, the electric field vector and the magnetic field vector are perpendicular to each other and have the same magnitude. Therefore, the magnitude of the Poynting vector is given by the product of the electric field strength and the magnetic field strength. The electric field strength is given by the following equation:$$E = \sqrt{\frac{2P}{4\pi r^2}}$$where P is the power of the light source and r is the distance from the light source. The magnetic field strength is given by the following equation:$$B = \frac{E}{c}$$where c is the speed of light. Substituting these equations into the equation for the Poynting vector, we get the following:$$S = \frac{P}{4\pi r^2}$$Substituting the given values into this equation, we get the following:$$S = \frac{1.50 W}{4\pi (0.750 m)^2} = 0.212 W/m^2$$Therefore, the correct answer is A.

A Monochromatic Point Like Light Source Emits 1