Power dissipated in a cylindrical cavity due to a current carrying filament

In summary, the problem involves a cylindrical cavity filled with a lossy dielectric and a filament that carries a current of 10 A. The electric field on the cavity, which is given as E=-z(i+j), is to be used to calculate the power dissipated in the cavity. The frequency is 5 kHz and the Poynting vector is relevant to the solution. However, the lack of information on the permeability and permittivity of the lossy material makes it difficult to find the magnetic field and solve the problem.
  • #1
nutan123
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Homework Statement



A cylindrical cavity oriented along z axis with length of 2 m has a filament in it upon which a current of 10 A is impressed. Cavity is perfectly conducting whereas it is filled with lossy dielectric. Electric field on the cavity is given as: E=-z(i+j). One has to calculate the power dissipated in the cavity. BTW, the frequency is 5 kHz.

Homework Equations



[tex]\nabla[/tex][tex]\times[/tex][tex]\vec{E}[/tex]=-del(B)/del(t)
[tex]\nabla[/tex][tex]\times[/tex][tex]\vec{H}[/tex]=del(D)/del(t)+J
S=\vec{E}[/tex]\times[/tex][tex]\vec{H}[/tex]

The Attempt at a Solution



It has to do with the Poynting vector. I tried to find the mangetic field from the current but without any value for the permeability and permittivity for the lossy material how can I find the power loss. Is the fact that question provides the electric field is in some way helpful.
 
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  • #2
nutan123 said:
A cylindrical cavity oriented along z axis with length of 2 m has a filament in it upon which a current of 10 A is impressed. Cavity is perfectly conducting whereas it is filled with lossy dielectric.

I assume you mean that the wall of the cavity is perfectly conducting while the cavity itself is filled with a dielectric?

Electric field on the cavity is given as: E=-z(i+j). One has to calculate the power dissipated in the cavity. BTW, the frequency is 5 kHz.

Is [itex]\textbf{E}=-z(\textbf{i}+\textbf{j})[/itex] the field on the cavity's wall, or throughout the cavity's interior?
 

Related to Power dissipated in a cylindrical cavity due to a current carrying filament

1. What is power dissipation?

Power dissipation is the amount of energy lost or converted into heat within a system. In the context of a cylindrical cavity, it refers to the energy lost as heat due to the presence of a current carrying filament.

2. How is power dissipation calculated?

The power dissipated in a cylindrical cavity can be calculated using the formula P = I^2*R, where P is power dissipation, I is the current flowing through the filament, and R is the resistance of the cavity.

3. What factors affect power dissipation in a cylindrical cavity?

The power dissipation in a cylindrical cavity is affected by the current flowing through the filament, the resistance of the cavity, and the material properties of the filament and cavity walls. The size and shape of the cavity can also influence power dissipation.

4. How does power dissipation affect the performance of a cylindrical cavity?

The power dissipation in a cylindrical cavity can cause a rise in temperature, which can affect the performance of the cavity. This rise in temperature can lead to changes in the electrical and magnetic properties of the cavity, which can impact its efficiency and stability.

5. How can power dissipation be reduced in a cylindrical cavity?

To reduce power dissipation in a cylindrical cavity, one can use a material with lower resistance for the filament or the cavity walls. Additionally, adjusting the current flowing through the filament can also help reduce power dissipation. Proper ventilation and cooling systems can also help dissipate heat and reduce power dissipation.

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