Solving a Flux Density Problem: Explaining E and B Field Amplitudes

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SUMMARY

This discussion addresses the calculation of flux density and the amplitudes of electric (E) and magnetic (B) fields from an isotropic quasimonochromatic point source radiating at 100 W. The flux density at a distance of 1 m is calculated using the formula S = P/4πr², resulting in approximately 7.9577 W/m². The amplitude of the electric field (E0) is determined to be approximately 5.6719 x 10^-3 V/m, while the magnetic field amplitude (B0) is approximately 3.1834 x 10^-9 T, utilizing the relationships involving the permittivity (ε0) and permeability (μ0) of free space.

PREREQUISITES
  • Understanding of the Poynting vector in electromagnetic theory
  • Familiarity with the concepts of electric field (E) and magnetic field (B) amplitudes
  • Knowledge of the permittivity (ε0) and permeability (μ0) of free space
  • Basic proficiency in solving equations involving power and flux density
NEXT STEPS
  • Study the derivation and applications of the Poynting vector in electromagnetic radiation
  • Learn about the relationship between intensity and field amplitudes in electromagnetic waves
  • Explore the implications of isotropic radiation in practical scenarios
  • Investigate the effects of distance on flux density and field amplitudes in various media
USEFUL FOR

Students and professionals in physics, electrical engineering, and telecommunications who are involved in electromagnetic theory and applications, particularly those focusing on wave propagation and field analysis.

jlmac2001
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I have no ideal how to do this problem. Can someone explain it to me? What formula would i use?

An isotropic quasimonochromatic point source radiates at a rate of 100 W. What is the flux density at a distance of 1 m? What are the amplitudes of the E and B fields at that point?
 
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"Isotropic" means that the emitted power is distributed uniformly over a circular shell. In this case, of radius 1m.
"Quasimonochromatic" means, you may use a harmonic (sinus) wave. Just look up how E and B in such a wave depend on intensity. Key word: Poynting vector.
 


To solve this problem, we can use the formula for flux density, which is given by the equation:

S = P/4πr^2

Where S is the flux density, P is the power radiated by the source, and r is the distance from the source.

In this case, we are given that the power radiated by the source is 100 W and the distance from the source is 1 m. Plugging in these values into the formula, we get:

S = 100/4π(1)^2 = 25/π W/m^2

This means that the flux density at a distance of 1 m from the source is approximately 7.9577 W/m^2.

To find the amplitudes of the E and B fields at this point, we can use the formula:

S = (1/2)√(ε0/μ0)E0^2

Where S is the flux density, ε0 and μ0 are the permittivity and permeability of free space respectively, and E0 is the amplitude of the electric field.

To solve for E0, we rearrange the formula to get:

E0 = √(2Sμ0/ε0)

Plugging in the values for S, μ0, and ε0, we get:

E0 = √(2(7.9577)(4π x 10^-7)/(8.854 x 10^-12)) = 5.6719 x 10^-3 V/m

Similarly, we can find the amplitude of the magnetic field using the formula:

S = (1/2)√(ε0/μ0)B0^2

Rearranging the formula and plugging in the values, we get:

B0 = √(2Sε0/μ0) = √(2(7.9577)(8.854 x 10^-12)/(4π x 10^-7)) = 3.1834 x 10^-9 T

Therefore, the amplitudes of the E and B fields at a distance of 1 m from the source are approximately 5.6719 x 10^-3 V/m and 3.1834 x 10^-9 T, respectively.

In summary, to solve this flux density problem, we used the formula for flux density and the formulas for the amplit
 

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