Power being transported down the cable

In summary: VI\frac {aL}{\ln \frac b a } ##L.H.S. is the power spent on charges due to the electromagnetic force acting on the charges of the inner cylinder, while R.H.S. is the power radiated by the charges of the inner cylinder. In short, the power radiated by the charges is the power given to the charges by the electromagnetic force.
  • #1
Pushoam
962
51
Homework Statement
Problem 8.1 Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.62, assuming the two conductors are held at potential difference V, and carry current I (down one and back up the other).

Example 7.13. A long coaxial cable carries current I (the current flows down the
surface of the inner cylinder, radius a, and back along the outer cylinder, radius
b) as shown in Fig. 7.40. Find the magnetic energy stored in a section of length l.
Relevant Equations
## \frac {dW}{ dt} = - \frac d {dt} \int_{\upsilon}~~ \frac 1 2~ \left( \epsilon_{\omicron} E^2 + \frac {B^2} \mu_o~\right) d \tau ~~ -\frac 1 \mu_o \oint_{S_{in}} \vec E \times \vec B \cdot d\vec a##
244417

Relevant equation:
244418

Since the potential difference V is independent of time and current I is steady, both the electric field and magnetic field are independent of time. Hence, the first term of R.H.S. of (8.9) is 0.

If V >0, then ## V_{in} - V_{out} ## = V,
## \Rightarrow V_{in} \gt V_{out} \Rightarrow ## There is a positive charge distribution ## \sigma ## on the inner cylinder.
Using Gauss's law,
## \vec E_{in} = \frac V {\ln \left (\frac b a \right) } \frac {\hat s } s~~~~~~ ~a\lt s\lt b ## ...(1)
Using Ampere's law,
## \vec B_{in} = \frac {\mu_0 I} {2\pi s} \hat \phi ~~~ ~~~~a\lt s\lt b## ...(2)

Assuming that (1) and (2) is applicable on the surface of the inner cylinder,
on the inner cylinder,
## \frac {dW} {dt} = - \frac 1 {\mu_0} \oint_{S_{in} }\vec E \times \vec B \cdot d \vec a##
## = - VI\frac {aL}{\ln \frac b a } ##

L.H.S. is the power spent on charges due to the electromagnetic force acting on the charges of the inner cylinder, while R.H.S. is the power radiated by the charges of the inner cylinder. In short, the power radiated by the charges is the power given to the charges by the electromagnetic force.
Is this correct?
The question says this is the power transported down the cables. How?
 
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  • #2
Pushoam said:
Problem Statement: Problem 8.1 Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.62, assuming the two conductors are held at potential difference V, and carry current I (down one and back up the other).

Example 7.13. A long coaxial cable carries current I (the current flows down the
surface of the inner cylinder, radius a, and back along the outer cylinder, radius
b) as shown in Fig. 7.40. Find the magnetic energy stored in a section of length l.
Relevant Equations: ## \frac {dW}{ dt} = - \frac d {dt} \int_{\upsilon}~~ \frac 1 2~ \left( \epsilon_{\omicron} E^2 + \frac {B^2} \mu_o~\right) d \tau ~~ -\frac 1 \mu_o \oint_{S_{in}} \vec E \times \vec B \cdot d\vec a##

View attachment 244417
Relevant equation:
View attachment 244418
Since the potential difference V is independent of time and current I is steady, both the electric field and magnetic field are independent of time. Hence, the first term of R.H.S. of (8.9) is 0.

If V >0, then ## V_{in} - V_{out} ## = V,
## \Rightarrow V_{in} \gt V_{out} \Rightarrow ## There is a positive charge distribution ## \sigma ## on the inner cylinder.
Using Gauss's law,
## \vec E_{in} = \frac V {\ln \left (\frac b a \right) } \frac {\hat s } s~~~~~~ ~a\lt s\lt b ## ...(1)
Using Ampere's law,
## \vec B_{in} = \frac {\mu_0 I} {2\pi s} \hat \phi ~~~ ~~~~a\lt s\lt b## ...(2)

Assuming that (1) and (2) is applicable on the surface of the inner cylinder,
on the inner cylinder,
## \frac {dW} {dt} = - \frac 1 {\mu_0} \oint_{S_{in} }\vec E \times \vec B \cdot d \vec a##
## = - VI\frac {aL}{\ln \frac b a } ##

L.H.S. is the power spent on charges due to the electromagnetic force acting on the charges of the inner cylinder, while R.H.S. is the power radiated by the charges of the inner cylinder. In short, the power radiated by the charges is the power given to the charges by the electromagnetic force.
Is this correct?
The question says this is the power transported down the cables. How?
Of course you know that if the result is anything other than [itex] \frac{dW}{dt} = VI [/itex], then something went wrong.

So far so good with your [itex] \vec E = \frac{V}{s \ln{\frac{a}{b}}} \hat s [/itex] and [itex] \vec B = \frac{\mu_0 I}{2 \pi s} \hat \phi[/itex], where [itex] \hat s [/itex] denotes the radial direction (I would have used [itex] \hat \rho [/itex] or [itex] \hat r [/itex] for the radial direction, but use whatever variable you want. We'll just continue to use [itex] \hat s [/itex] here). These look correct to me.

Now find

[tex] \frac{dW}{dt} = -\frac{1}{\mu_0} \int_S \left( \vec E \times \vec B \right) \cdot d\vec {A} [/tex]

Notice the above integral is not under a closed surface anymore. That's actually intentional. It's an open surface, and thus an open integral, because we are determining the power flow in only a single direction ("down the cables.")

Also, it's an area integral. And the area we're integrating over is the cross sectional area of the cable, and only the area between [itex] a [/itex] and [itex] b [/itex]. And we're only integrating a single section corresponding to our single direction.

But integrating over area isn't very convenient. Wouldn't it be nicer to integrate over [itex] s [/itex]? What is the relationship between [itex] ds [/itex] and [itex] d \vec A [/itex]?
 
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  • #3
## \frac { dW} {dt} = - \frac 1 {\mu_0}\, \int_S\, \vec E \times \vec B \cdot \, d\vec a##
## d\vec a = s \,ds \,d\phi ~\hat z ##
## \frac { dW} {dt} = - \frac 1 {\mu_0} \,\int_a^b \, \int_0 ^{2\pi}\,
\frac{V}{s \ln{\frac{b}{a}}} \,
\frac{\mu_0 I}{2 \pi s}
\, s \, ds \, d \phi = - VI ##

collinsmark said:
Notice the above integral is not under a closed surface anymore. That's actually intentional. It's an open surface, and thus an open integral, because we are determining the power flow in only a single direction ("down the cables.")

Also, it's an area integral. And the area we're integrating over is the cross sectional area of the cable, and only the area between a and b . And we're only integrating a single section corresponding to our single direction.

I did not get this.

Pushoam said:
1559388001728-png.png

According to the Poynting's theorem - equation (8.9), ## \frac { dW} {dt} ## is the work done by the electromagnetic force on the charges in a volume ## \upsilon ##. Here the charges are on the surface of inner cylinder and so a cylindrical volume ## a \leq s \leq a + \epsilon ## where ## \epsilon ## is a small positive quantity. I did not get why I should take s from a to b as there is no charge in region between a and b. So, taking s from a to a + ## \epsilon ## should be fine. ## \frac 1 {\mu_0}\, \int_S\, \vec E \times \vec B \cdot \, d\vec a## is the power radiated by the charges through the surface of the cylinderical volume. Since the curl part of the cylindrical surface area is perpendicular to Poynting vector, the integration over this area is 0. So, power flow in ## \hat s ## direction is 0. So, we have to integrate over the two cross - sectional area. This should be the power flow in ## \hat z ## direction.

I did not get why you are telling to integrate over only one cross-sectional area.
 
  • #4
If you wish, think of the power source (maybe a battery with its terminals being connected to one end of the coaxial cable) as being somewhere inside the closed surface, with the coaxial cable being the only thing that exits the surface. In this case, the only place where [itex] \vec E \times \vec B [/itex] is not zero is the single cross section of the coaxial cable where it passes through the closed surface. And even then it's still zero inside the center conductor. So the only place that it's not zero is in this single cross section and only in the region between [itex] a [/itex] and [itex] b [/itex]. That's why you only need that single integral. Note that the surface vector of this area (and thus the surface vector for [itex] d \vec A [/itex]) is along the [itex] \hat z [/itex] direction. Also note that [itex] \vec E \times \vec B [/itex] also points along the [itex] \hat z [/itex] direction. This makes sense because you're taking the dot product of of [itex] \left( \vec E \times \vec B \right) [/itex] and [itex] d \vec A [/itex].

You need to integrate over the surface area of this cross sectional region of coaxial cable. Note the [itex] d \vec A [/itex] in
[tex] - \frac{1}{\mu_0} \int_S \left( \vec E \times \vec B \right) \cdot d \vec A [/tex].

But your [itex] \vec E [/itex] and [itex] \vec B [/itex] equations are all in terms of [itex] s [/itex].

[tex] \vec E = \frac{V}{s \ln \frac{a}{b}} \hat s [/tex]

[tex] \vec B = \frac{\mu_0 I}{2 \pi s} \hat \phi [/tex].

If you can somehow express [itex] d \vec A [/itex] in terms of [itex] ds [/itex] you should be able to integrate with respect to [itex] s [/itex].

Hint: If the area of the cross section is [itex] \vec A = (\pi s^2 - \pi a^2) \hat z [/itex], and where [itex] \pi a^2 [/itex] is a constant, what is [itex] d \vec A [/itex] in terms of [itex] ds [/itex]? [Edit: I see you did this up at the very top of your last post. See my post below for additional comments on that.]
 
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  • #5
Pushoam said:
## \frac { dW} {dt} = - \frac 1 {\mu_0} \,\int_a^b \, \int_0 ^{2\pi}\,
\frac{V}{s \ln{\frac{b}{a}}} \,
\frac{\mu_0 I}{2 \pi s}
\, s \, ds \, d \phi = - VI ##

Also, I just noticed something. In your electric field equation, you had
[tex] \vec E = \frac{V}{s \ln \frac{b}{a}} \hat s [/tex]

You might want to double check that. I think you have the [itex] b [/itex] and the [itex] a [/itex] swapped.

Related to that, I'm pretty sure the final answer, [itex] \frac{dW}{dt} = VI, [/itex] should be positive.
 
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1. How is power transported down a cable?

Power is transported down a cable through the flow of electrons. When a power source, such as a battery or generator, is connected to one end of a cable, it creates a potential difference that causes the electrons to flow through the cable towards the other end.

2. What factors affect the transportation of power down a cable?

The transportation of power down a cable can be affected by various factors, such as the type and quality of the cable, the distance the power needs to travel, the amount of power being transmitted, and external factors like temperature and interference.

3. Can power be transported down any type of cable?

No, not all types of cables are suitable for transporting power. The cable must have the appropriate conducting material, insulation, and size to handle the amount of power being transmitted without overheating or causing a voltage drop.

4. Is there a limit to how far power can be transported down a cable?

Yes, there is a limit to how far power can be transported down a cable. The longer the cable, the more resistance it will have, which can result in a voltage drop. This can affect the efficiency and reliability of power transmission. Therefore, the distance of power transportation is limited by the cable's properties and the amount of power being transmitted.

5. How is power loss minimized during transportation down a cable?

Power loss during transportation down a cable can be minimized by using high-quality cables with low resistance, reducing the distance the power needs to travel, and ensuring proper maintenance and insulation of the cable. Additionally, using higher voltages can also help minimize power loss as it reduces the amount of current needed to transmit the same amount of power.

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