How Is Waveguide Conductivity Calculated with Changing Frequencies?

In summary, the wavelength along a certain dielectric-filled conducting waveguide at frequency f1=9 GHz is Lambda1=3.456 cm, and at frequency f2=10GHz is Lambda2=2.345 cm, for a particular mode. The cutoff frequency for this mode can be found by solving for Vp using Lambdag=Vp/f, then plugging in the numerical value for Vp=Lambdag*f and solving for Omegac using the equation Vp=c/n*[1-omegac/omega^2]^-1/2. However, it is important to know the value of n in order to accurately solve for Omegac. Additionally, it may be helpful to consider the relationship between
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Homework Statement


At frequency f1=9 GHz, the guide wavelength along a certain dielectric-filled conducting waveguide is found to be Lambda1=3.456 cm, for a particular mode. At frequency f2=10GHz, the wavelength is Lambda2=2.345 cm, for the same mode.

(a) What is the cutoff frequency Fo of that mode?

(b) What is the relative permitttivity epsilon of the dielectric


Homework Equations



Lambdag=Vp/f where Lambdag is the wavelength along the waveguide.

Vp=omega/Beta=c/n*[1-(omegac/omega)^2]^-1/2 where Vp is the phase velocity.





The Attempt at a Solution



I first solved for Vp using Lambdag=Vp/f and then I plugged in the numerical value for Vp=Lambdag*f and then I solved for Omegac which is the cutoff frequency from the above equation Vp=c/n*[1-omegac/omega^2]^-1/2. However, I am not getting the correct answer. Any ideas on what I should do?
 
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  • #2
Is this problem for a physics course or an engineering course? If engineering, do you have Paul and Nasar, Introduction to Electromagnetic Fields?

Does the "g" in "Lambdag" denote "guide" or "group"? (It should denote "guide".)

Hint 1a: How do you know n?
Hint 1b: You are given two sets of f and lambda. Why do you suppose that is?
Hint 2: f is not the same thing as omega.
 

1. What is the definition of conductivity in the context of waveguides?

Conductivity in waveguides refers to the ability of a material to conduct electrical current or electromagnetic waves. It is a measure of the ease with which electric charges can move through a material in response to an applied electric field.

2. How does the conductivity of a waveguide affect its performance?

The higher the conductivity of a waveguide, the lower the loss of energy as the electromagnetic waves travel through it. This results in better signal transmission and reduces the need for frequent signal amplification.

3. What factors can affect the conductivity of a waveguide?

The conductivity of a waveguide can be influenced by various factors, such as the material used, the dimensions and shape of the waveguide, and the frequency of the electromagnetic waves being transmitted. Temperature and impurities can also affect conductivity.

4. How is the conductivity of a waveguide measured?

The conductivity of a waveguide is typically measured using a technique called the "transmission line method". This involves measuring the loss of energy as a signal travels through the waveguide and using this information to calculate the conductivity of the material.

5. What are some common applications of high-conductivity waveguides?

High-conductivity waveguides are commonly used in communication systems, such as satellite and radar systems, to transmit signals over long distances without significant loss of energy. They are also used in medical imaging equipment and in research for particle accelerators.

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