Electromagnetic wave, intensity, electric field problem

In summary, the problem involves finding the rms value of the electric field for beaming energy via an electromagnetic wave with uniform intensity within a cross-sectional area of 80 m2. The speed of light is given as 3×10^8 m/s and the permeability of free space as 4×10^-7 T · m/A. The relevant equations include P=IV, poynting vector = S = EB/mu0, Power = I/Area, and S=(E^2)/mu0*c. The value of <S> is calculated to be 11687.5 watts/m2, which can be used to solve for the electric field, Eo.
  • #1
ashleyd
4
0

Homework Statement


Instead of sending power by a 850 kV, 1100 A transmission line, one desires to beam this energy via an electromagnetic wave. The beam has uniform intensity within a cross-sectional area of 80 m2. The speed of light is 3×10^8 m/s and the permeability of free space is 4×10^-7 T · m/A.
Find the rms value of the electric field.
Answer in units of kV/m

Homework Equations


P=IV
poynting vector = S = EB/mu0
Not sure if the following equations are relevant:
Power = I/Area
S=(E^2)/mu0*c

The Attempt at a Solution


I have tried using the formulas above, along with changing the kv to v
the answers i have gotten through this that are incorrect include:
18774.62904
593705.9
2099.06734
66378.33756
 
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  • #2
[tex] <S> = \frac{E_o^2}{2 \mu_0 c} [/tex]

Where <S> is the time averaged magnitude of the Poynting vector for a plane wave.

What value did you calculate for <S>?
 
  • #3
The value I got for S, without converting kV to V was 11687.5
 
  • #4
Okay, convert to watts/m2 and use the given formula to solve for Eo.
 
  • #5


A possible response could be:

I would approach this problem by first understanding the concept of intensity of an electromagnetic wave. Intensity is defined as the amount of energy passing through a unit area per unit time. In this case, the beam has a uniform intensity within a cross-sectional area of 80 m^2.

To find the rms value of the electric field, we can use the formula for power, P=IV, where P is the power, I is the current, and V is the voltage. However, since we are dealing with an electromagnetic wave, we need to use the Poynting vector, S=EB/μ0, where E is the electric field, B is the magnetic field, and μ0 is the permeability of free space.

Substituting the given values, we get S=IV/μ0. Since we are given the power (P), we can rearrange this equation to solve for the electric field (E). This gives us E=Sμ0/V.

To find V, we can use the speed of light (c) and the given values for the voltage and current. We know that P=IV, and since the power is the same for both the transmission line and the electromagnetic wave, we can equate the two equations, giving us I=PV. Substituting this into the equation for V=IR, we get V=PR/I.

Finally, we can plug in all the values and solve for the electric field (E). The answer will be in units of volts per meter (V/m). To convert this to kilovolts per meter (kV/m), we can divide the answer by 1000.

Therefore, the rms value of the electric field is approximately 2.1 kV/m.
 

1. What is an electromagnetic wave?

An electromagnetic wave is a type of energy that is created by the oscillation of electric and magnetic fields. It is a form of radiation that can travel through space and is responsible for the transmission of energy in the form of light, radio waves, microwaves, and other forms of electromagnetic radiation.

2. How is the intensity of an electromagnetic wave measured?

The intensity of an electromagnetic wave is measured by the amount of energy it carries per unit area per unit time. This can be measured using instruments such as a power meter, which measures the power of the electromagnetic wave at a specific point in space.

3. What factors affect the intensity of an electromagnetic wave?

The intensity of an electromagnetic wave is affected by the amplitude of the electric and magnetic fields, the distance from the source of the wave, and any obstructions in the path of the wave. Additionally, the type of material the wave is traveling through can also affect its intensity.

4. How does the electric field of an electromagnetic wave interact with matter?

The electric field of an electromagnetic wave can interact with matter in several ways. It can cause charged particles within the material to vibrate, creating heat, or it can cause electrons to move, creating an electric current. The strength and frequency of the electric field determine the type of interaction that will occur.

5. What are some applications of electromagnetic waves?

Electromagnetic waves have many practical applications, including communication (such as radio waves and microwaves), medical imaging (such as x-rays and MRI), and energy production (such as solar panels). They are also used in technologies like radar, GPS, and satellite communication.

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