EM: Evaluating a Poynting vector

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Homework Statement



I reference problem 9.10 Purcell's Electricity and Magnetism (3rd ed).

A very thin straight wire carries current ##I## from infinity radially inward onto a conducting shell with radius R. Show that the total flux of the Poynting vector away from an imaginary tube of radius ##b## surrounding the wire equals the rate of change of the electric field.

I'm having trouble with several key ideas necessary to solve this problem, any assistance is greatly appreciated. I will write my understanding of the problem below to illustrate the issues.

Homework Equations




The Attempt at a Solution



There are 2 fields coming from the wire. There is a ##\vec{B} _w## looping around the wire and an ##\vec{E} _w## emanating radially from the wire, as if it were from a static line of charge.

There are also 2 fields associated with the sphere. There is an ##\vec{E} _s (t)## pointing radially out from the sphere, which induces a ##\vec{B} _s##, which has a radially -pointing (wrt sphere) curl as given by
$$\nabla \times \vec{B} _s = \mu _0 \epsilon _0 \ \partial _t \vec{E} _s (t)$$

To me, the Poynting vector evaluated at the surface of the tube should be in the form
$$\vec{S} = \frac{1}{\mu _0} \big ( [\vec{E} _w + \vec{E} _s (t)] \times [\vec{B} _w + \vec{B} _s ] \big )$$
However, the solutions manual states that
$$\vec{S} = \frac{1}{\mu _0} \big ( \vec{E} _s (t) \times \vec{B} _w \big )$$
I can sort of understand why ##\vec{E} _w## was omitted; the resulting Poynting vector term points along the wire, that is to say energy (associated with the current) travels towards the sphere instead of going "outwards" from the imaginary tube. However, I can't understand why the ##\vec{B} _s ## term was omitted as well. Should we not be evaluating the total fields at the points of interest?

Also, let's say I find that the ##\nabla \times \vec{B} _s## is independent of time, am I then allowed to say that ## \vec{B} _s## is independent of time as well, and so the flux of the Poynting vector does not go into increasing the energy density of the ##\vec{B} _s## field?

I apologize in advance for my slightly confused phrasing. Many thanks in advance for any assistance.
 

Answers and Replies

  • #2
TSny
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I reference problem 9.10 Purcell's Electricity and Magnetism (3rd ed).

A very thin straight wire carries current ##I## from infinity radially inward onto a conducting shell with radius R. Show that the total flux of the Poynting vector away from an imaginary tube of radius ##b## surrounding the wire equals the rate of change of the electric field.
Have you stated the problem exactly as given in the textbook (word for word)? The dimensions of "the flux of the Poynting vector" are not the same as the dimensions of "the rate of change of electric field".
 
  • #3
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Have you stated the problem exactly as given in the textbook (word for word)? The dimensions of "the flux of the Poynting vector" are not the same as the dimensions of "the rate of change of electric field".
Sincerest apologies, I meant to type "rate of change of energy stored in the electric field"!
 
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TSny
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Sincerest apologies, I meant to type "rate of change of energy stored in the electric field"!
OK. But the wording of the problem as you've given still seems incomplete to me. Is any information given about the size of ##b##? Is it assumed to be small compared to ##R##?

So, it would be nice to have the exact wording of the question. I have the first two editions of Purcell, but this problem is not in either of those.
 
  • #5
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OK. But the wording of the problem as you've given still seems incomplete to me. Is any information given about the size of ##b##? Is it assumed to be small compared to ##R##?

So, it would be nice to have the exact wording of the question. I have the first two editions of Purcell, but this problem is not in either of those.

Apologies, I was writing rather quickly and missed out these details. I think its best I upload a few images. Thank you for your patience.

Screen Shot 2018-05-03 at 11.29.00 PM.png
Screen Shot 2018-05-03 at 11.29.06 PM.png
 

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  • #6
TSny
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OK. Thanks for posting the exact statement of the problem.

So you are asking about the choice of E and B fields used to construct the Poynting vector for the outward flow of energy through the surface of the tube of small radius ##b##. In particular, I think you are asking the following questions

(i) Why is E chosen to be the field due to the sphere alone? What about any E field associated with the induced charges on the current carrying wire?

(ii) Why is B chosen to be just the field due to the current in the wire? What about any B field due to currents on the sphere; or, what about any B field which is induced by the time-changing E field?

There are 2 fields coming from the wire. There is a ##\vec{B} _w## looping around the wire and an ##\vec{E} _w## emanating radially from the wire, as if it were from a static line of charge.
##\vec{E} _w## would not be purely radial. There would also be a component of ##\vec{E} _w## that is parallel to the wire and in the direction of the current. As you noted, the radial component would not contribute to any outward or inward flux of energy through the tube of radius ##b##. The parallel component of ##\vec{E} _w## along with the B field due to the current yields a Poynting vector that is radially inward toward the wire. This inward flow of energy is related to the Joule heating of the wire. The question is not interested in this. So you only need to worry about the outward flux of energy due to ##\vec{E} _s## in the Poynting vector.


There are also 2 fields associated with the sphere. There is an ##\vec{E} _s (t)## pointing radially out from the sphere, which induces a ##\vec{B} _s##, which has a radially -pointing (wrt sphere) curl as given by
$$\nabla \times \vec{B} _s = \mu _0 \epsilon _0 \ \partial _t \vec{E} _s (t)$$
You might try applying the Maxwell equation ##\oint \vec B \cdot d \vec l = \mu_0 I_{\rm enc} + \mu_0 \epsilon_0 \frac{d}{dt} \Phi_E##.

For the path of integration on the left, choose a circumference of the tube of radius ##b##. Try to construct an argument for why the second term on the right is negligible. Thus, the B field at the surface of the tube can be approximated by ##\vec{B} _w##.
 
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  • #7
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Thank you for your response,
OK. Thanks for posting the exact statement of the problem.

You might try applying the Maxwell equation ##\oint \vec B \cdot d \vec l = \mu_0 I_{\rm enc} + \mu_0 \epsilon_0 \frac{d}{dt} \Phi_E##.

For the path of integration on the left, choose a circumference of the tube of radius ##b##. Try to construct an argument for why the second term on the right is negligible. Thus, the B field at the surface of the tube can be approximated by ##\vec{B} _w##.
I have tried but what I thought out seemed a little "hand-wavy" to me, could you see my argument is alright?

If I apply the Maxwell equation and integrate ##\vec{B}## along a path of radius b, the right hand side disappears as the area ##\pi b^2## is much smaller than the overall size of the sphere. The looping integral path can therefore be treated almost as a point, and the magnitude of ##\vec{B}## can be approximated as uniform (and I can pull it out of the path integral) , thus ##B \approx 0##. Does any of this make sense?

The Poynting Vector from this ##\vec{B} _s## term seems to point into the wire, does this also contribute to Joule heating?

Thank you for your assistance!
 
  • #8
TSny
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If I apply the Maxwell equation and integrate ##\vec{B}## along a path of radius b, the right hand side disappears as the area ##\pi b^2## is much smaller than the overall size of the sphere.
Yes, that's the reason for neglecting the electric flux term (displacement current). This term is small and of order ##b^2##.

So, we are left with
##\oint \vec B \cdot d \vec l = \mu_0 I_{\rm enc} ##.

Let ##B_t## be the component of B that is tangent to the path of integration. Since the system is rotationally invariant when rotating about the axis of the wire, ##B_t## is constant around the path of integration. So, ##\oint \vec B \cdot d \vec l = B_t \cdot 2 \pi b##.

Therefore, ##B_t \cdot 2 \pi b = \mu_0 I_{\rm enc} = \mu_0 I ##. Here, ##I## is the current of the wire.

Thus, ##B_t = \frac{\mu_0 I}{2 \pi b}##.

Note this field is not small. ##B_t## is seen to be the same as what you called ##B_w##.

Consider the direction of ##\vec B_t## along with the direction of ##\vec E_s## for points on the surface of the tube of radius ##b## and deduce the direction of the Poynting vector associated with these fields.
 
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