Multiple Choice
What is the average magnitude of the Poynting vector for an electromagnetic wave with magnetic field amplitude equal to ?
32. Electromagnetic Waves
Intensity of EM Waves
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A light bulb emits light uniformly in all directions. What is the electric field amplitude away from the light bulb?
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Verified step by step guidance1
Understand that the power of the light bulb is given as 50 W, and it emits light uniformly in all directions, which means it can be treated as a point source radiating spherically.
Recall that the intensity (I) of the light at a distance (r) from a point source is given by the formula: I = P / (4πr²), where P is the power of the source.
Substitute the given values into the intensity formula: P = 50 W and r = 0.1 m (since 10 cm = 0.1 m). Calculate the intensity at this distance.
The intensity is related to the electric field amplitude (E) by the formula: I = (1/2)ε₀cE², where ε₀ is the permittivity of free space and c is the speed of light. Rearrange this formula to solve for E.
Substitute the calculated intensity and known constants (ε₀ and c) into the rearranged formula to find the electric field amplitude E at 10 cm from the light bulb.
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