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

In summary, the conversation discusses the concept of an isotropic quasimonochromatic point source and its properties, such as emitting power at a rate of 100 W and having a uniform distribution of power over a circular shell. The conversation also mentions the use of a harmonic wave and the relationship between intensity and the E and B fields, with the Poynting vector being a key factor.
  • #1
jlmac2001
75
0
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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  • #2
"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.
 
  • #3


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
 

What is flux density?

Flux density refers to the amount of magnetic field passing through a specific area. It is represented by the symbols B or D and is measured in Teslas (T) or Webers per square meter (Wb/m²).

What are E and B field amplitudes?

E and B field amplitudes refer to the strength or magnitude of the electric and magnetic fields, respectively. They are represented by the symbols E and B and are measured in volts per meter (V/m) and Teslas (T), respectively.

How is flux density calculated?

Flux density can be calculated by dividing the magnetic flux (Φ) by the area (A) it passes through. This can be expressed as B = Φ/A. The electric field amplitude can be calculated using the equation E = cB, where c is the speed of light.

Why is solving a flux density problem important?

Solving a flux density problem is important because it allows scientists to understand and predict the behavior of electromagnetic fields. This is crucial in many applications, such as designing electronic devices, studying the Earth's magnetic field, and developing medical imaging technologies.

What factors can affect the E and B field amplitudes?

The E and B field amplitudes can be affected by the strength and orientation of the current or source creating the fields, the distance from the source, and the medium through which the fields are passing. Additionally, the E and B fields are interrelated and can influence each other's amplitudes.

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