Power Loss Of An EM Wave Hitting A Cu Wire

In summary, when an EM wave hits a copper wire, it creates an alternating electric field that induces a current, causing the wire to heat up and dissipate energy as heat. The power loss of an EM wave hitting a copper wire depends on factors such as frequency, intensity, length, diameter, and resistance. Thicker wires have lower resistance and therefore experience less power loss. Power loss can be reduced by using thicker wires, shorter lengths, and materials with lower resistance. Understanding power loss in a copper wire is important in various industries and can lead to more efficient systems, reduced energy waste, and the development of new technologies and materials.
  • #1
H_man
145
0
Hi all, :uhh:

I am trying to calculate the loss in power of a wave which is partially

reflected and partially transmitted from a wire.


The skin depth of the wire is many times the thickness of the wire.

My first attempt to solve this problem was to consider seperately the

energy lost due to currents induced by the electric field and then the

magnetic field.


I started by calculating the skin depth of the Cu wire in order to

calculate the effective resistance at the frequency of the wave.

I then thought to calculate the current I, by first taking J0=sigma.E,

using the electric field derived from the Poynting vector and the

conductivity of Cu. Then integrating with respect to area bearing in

mind that J falls off exponentially from the surface.


Finally to use P=I2R to determine power loss from the electric field.

The answer I got was non-sensical.


When then trying to calculate the losses from the changing magnetic

field I found the helpful expression:
I/l = H (where I is the current, l the length of the field in the

conductor and H the field strength).


The book didn't show any calculation relating to the electric field

which makes me think I've been barking up the wrong tree and in fact the

only losses which occur are entirely due to the magnetic field. However,

I find this very very confusing, why should one field do work and not

the other?

I think I am possibly missing some basic physical knowledge
 
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  • #2
here.



Hi there,

Calculating the loss in power of a wave that is partially reflected and transmitted from a wire is indeed a complex problem. However, your approach of considering the electric and magnetic fields separately is a good starting point.

Firstly, it is important to note that both the electric and magnetic fields contribute to the power loss in a wave. The electric field induces currents in the wire, which then dissipate energy as heat due to the resistance of the wire. On the other hand, the changing magnetic field also induces currents in the wire, leading to the same dissipation of energy as heat.

In your approach, you correctly calculated the skin depth of the wire to determine the effective resistance at the frequency of the wave. However, when calculating the current using J0=sigma.E, it is important to consider the direction of the electric field and the orientation of the wire. This will affect the direction and magnitude of the induced current, which in turn affects the power loss.

Similarly, when calculating the losses from the changing magnetic field, it is important to consider the orientation of the wire with respect to the magnetic field. The expression I/l = H is a good starting point, but it may need to be modified depending on the geometry of the wire and the direction of the magnetic field.

In summary, both the electric and magnetic fields contribute to the power loss in a wave that is partially reflected and transmitted from a wire. It is important to consider the direction and orientation of both fields when calculating the power loss. I hope this helps in your calculations. Good luck!
 

1. How does an EM wave cause power loss in a copper wire?

When an EM wave hits a copper wire, it creates an alternating electric field around the wire. This electric field induces a current in the wire, which causes the wire to heat up and dissipate energy as heat. This process is known as Joule heating and is the main cause of power loss in a copper wire.

2. What factors affect the power loss of an EM wave hitting a copper wire?

The power loss of an EM wave hitting a copper wire depends on several factors, including the frequency and intensity of the EM wave, the length and diameter of the wire, and the resistance of the wire. Higher frequencies and intensities, longer and thicker wires, and higher resistance all result in greater power loss.

3. How does the thickness of a copper wire affect its power loss?

The thickness of a copper wire plays a significant role in determining its power loss when hit by an EM wave. Thicker wires have a lower resistance, which means they dissipate less energy as heat. This results in lower power loss compared to thinner wires.

4. Can power loss be reduced in a copper wire hit by an EM wave?

Yes, power loss in a copper wire can be reduced through various methods such as using thicker wires, reducing the wire's length, and using materials with lower resistance. Additionally, using special coatings or insulating materials on the wire can also help reduce power loss.

5. What are some practical applications of understanding power loss in a copper wire?

Understanding power loss in a copper wire is crucial in various industries, including electrical engineering, telecommunications, and power transmission. It helps engineers design more efficient systems and equipment, reduces energy waste, and improves overall performance. Additionally, understanding power loss can also aid in the development of new technologies and materials for more efficient energy transfer.

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