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Power being transported down the cable

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38
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##
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?
 

collinsmark

Homework Helper
Gold Member
2,754
1,091
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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890
38
## \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 ##

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.

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.
 

collinsmark

Homework Helper
Gold Member
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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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collinsmark

Homework Helper
Gold Member
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## \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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