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Question

How can we obtain the answer here? How can we deal with the thicknesses here?

Thank you!

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  • $\begingroup$ Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better $\endgroup$ – pentane Jun 14 '18 at 0:01
  • $\begingroup$ Thank you! I tried out with the question but I do not know how to deal with the thicknesses here $\endgroup$ – Ravindu Ranganath Jun 14 '18 at 0:04
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The problem is easier than you think, and you don't have to worry about the thickness because your interested in the field OUTSIDE the conductor. I'm sure by now you know that the easiest way to solve this is by using Ampere's law. Our closed loop will be a circumference of radius r (distance of point P to the center) and Ampere's law tells us that: $$\oint _C {\bf B}\cdot d{\bf l}=\mu _0 I_{enclosed}$$

As for the first integral, since B is uniform through our curve, it will simply become $$B2\pi r$$

Now for $I_{enclosed}$ what is it? It's the current the traverses our closed loop. Now if we have a current I going on one direction in one conductor, and the same current I going in the opposite direction in outer conductor the currents will cancel out (notice that they give you the total current I not the density of current (meaning that the dimensions don't matter).

So that will lead up to a B of zero. The remarkable conclusion is that it doesn't matter how thick the outer or the inner conductor are, if they are traversed by the same current but in opposite directions the B field will always be zero. I hope this helped.

Inside the cable the scenario will be different if you want to we can work on that too.

EDIT1: OOPS I forgot to add the permeability of vacuum on Ampere's law. My apologies

EDIT2: So let's see what's happening inside the cylinder. For that we take the same considerations as before: circular closed loop centered in the axis of the cable now with an arbitrary radius r.

For all of the cases we will consider, the left side of Ampere laws always results on

$$B2\pi r$$

So the difference will lie on our $I_{enclosed}$

For $r<a$ you will need the current density that goes through that current. That will be given by the total current divided by the section on the conductor (think of it as water through a pipe, the flux of water in that quantity divided by the section of the pipe). That is $\frac{I}{\pi a^2}$. Now that you have your density you multiply it by the section of your current loop (you now think of the loop as a disk). You will then have $$I_{enclosed}=\frac{I \pi r^2}{\pi a^2}$$

Putting everything together ´

$$B2\pi r=\frac{\mu_0 I \pi r^2}{\pi a^2}$$

so

$$B=\frac{\mu_0 I r}{2 \pi a^2}$$

The direction of B is given by the right hand rule, so it's clockwise.

Now the situation of a

For b

And putting everything together leaves you with $$B=\frac{\mu_o I}{2 \pi r}(1-\frac{r^2-b^2}{c^2 - b^2})$$

The direction of B I will also leave up to you now :)

Tell me if there were any questions left, did you understand everything?

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  • $\begingroup$ Thank you very much! It would be great to know the scenario inside the cable too.. $\endgroup$ – Ravindu Ranganath Jun 14 '18 at 13:38
  • $\begingroup$ You're very welcome. Check my edits then :) $\endgroup$ – Granger Obliviate Jun 14 '18 at 13:58
  • $\begingroup$ I totally understood what you said. Thank you very much for your effort.. The problem I had is that we do not have Ampere's Law in our syllabi due to lack of Integral knowledge of Biology students.. We do only have derived equations given in the Syllabus.. $\endgroup$ – Ravindu Ranganath Jun 14 '18 at 15:03
  • $\begingroup$ Man that really sucks, I'm studying engineering and we can't take Electromagnetism without first succeeding in Mathematical Analysis 3 (so multiple integrals, stokes theorem, divergence theorem, etc.). It must be even harder for you guys. Did you understand how I integrated the B field above? Glad that I helped! $\endgroup$ – Granger Obliviate Jun 14 '18 at 15:26
  • $\begingroup$ Also I'm curious about your answers to my questions: what did you get for B between a and b and for the direction of B between b and c? $\endgroup$ – Granger Obliviate Jun 14 '18 at 15:31

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