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I'm doing some self studying of introductory physics, and was working a question from a textbook that has several parts. The first few parts asked me to show that the magnetic field of an infinite current carrying cylinder has the form $\vec{B} = f(r)\begin{bmatrix}0 & z & -y\end{bmatrix}$ if we choose the x axis to be the central axis of the pipe. where $r = \sqrt{y^2 + z^2}$. I was then asked to show that in the empty space interior that $f(r) = \frac{a}{r^2}$, where $a$ is a constant of integration. I succeeded at that as well, following the outline in the textbook, I used the definition of curl as a derivative, computed curl and performed some integration. In summary, I have shown that inside the cylinder $\vec{B} = \frac{a}{r^2}\begin{bmatrix}0 & z & -y\end{bmatrix}$.

However, I'm stuck on what should be a simple corollary of my result, which is to show that the magnetic field inside the cylinder in empty space is always $0$. The hint is that the field at the central axis is $0$ by symmetry, which I understand. But, then I should use my result above for the field inside and the fact that the field is $0$ at the central axis to show it is $0$ everywhere inside. This is supposed to be a simple conclusion to all of the work I did above, but I'm just not seeing it.

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  • $\begingroup$ There seem to be some mistakes in your writeup. In your second paragraph you mention the electric field, which I'm guessing was a typo? More importantly, in the first paragraph you say you showed that $\vec{B} = \frac{a}{r^2}[\cdots]$ inside the cylinder, but did you mean to say you showed that for outside the cylinder? $\endgroup$
    – David Z
    Commented Aug 23, 2015 at 0:57
  • $\begingroup$ Thanks, fixed. I showed generally, that the field form holds in empty space which would include outside or inside the cylinder as the charge is located on the cylinder's surface. This is a freshman physics question so it shouldn't be that hard. I just can't think of it, and it's bothering me. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 1:00
  • $\begingroup$ It's not fixed though. The contradiction is that, in the first paragraph, you say you've showed that $\vec B\neq 0$ inside the cylinder, but then in the second paragraph you are trying to show that $\vec B=0$ inside the cylinder. They can't both be true. $\endgroup$
    – David Z
    Commented Aug 23, 2015 at 1:05
  • $\begingroup$ @DavidZ It's not a contradiction if a = 0, which is precisely what I'm trying to show. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 1:06
  • $\begingroup$ Ah, gotcha. I'm a little distracted I suppose. $\endgroup$
    – David Z
    Commented Aug 23, 2015 at 1:07

2 Answers 2

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So you have that $\vec{B} = \frac{a}{r^2}\begin{bmatrix}0 & z & -y\end{bmatrix}$ thus the magnitude is $B = \frac{a}{r^2}\sqrt{z^2+(-y)^2},$ where $a$ is unknown.

Can you write that as a function of $r?$

Can you investigate what happens as $r$ goes to zero?

Are magnetic fields continuous in empty space (a vacuum)?

If so, try the next five:

What magnetic field do you expect at the origin?

What is its magnitude?

Remember that $a$ is an unknown constant. Is there any choice of $a$ that allows $B$ to approach the magnitude it needs as you approach the origin?

Is it the only choice? Is it what you wanted?

If not, can you compute the line integral of the magnetic field in a circle about the origin and compare that to some thing?

For educational purposes I can share some actual requirements of continuity. Across any surface bounded by a triangle the normal component of the $\vec B$ field must give the same average flux just on one side of the triangle as on the other (or else you can't take the divergence and hence you can't say the divergence is zero).

There is a similar rule for tangential components but they can jump depending one whether you have surface currents or if something super extreme is happening to an electric field in a particular instant.

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  • $\begingroup$ I've been thinking along these lines for hours. To answer your question, B = $\frac{a}{r}$ of course. I've been thinking about taking the limit as r goes to 0, but there is absolutely no reason why lim(B) as r --> 0 must be 0. It's just not true that the limit of a magnetic field as it approaches a point has to equal the field at that point. For example, there is a discontinuity in the electric field of a sphere as you move from inside to outside. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 4:57
  • $\begingroup$ I shouldn't have to compute a line integral. The book hasn't covered it yet. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 5:04
  • $\begingroup$ If magnetic fields are continuous in empty space it's trivial. We require Lim(B) as r --> 0 = B(0) = 0 therefore a = 0. But, the author never mentioned magnetic fields are continuous in empty space, and I can't think how to prove this. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 5:08
  • $\begingroup$ @user7348 The definition of a curl should be in terms of a line integral. There are other definitions are curls that only work if the field has partial derivatives in all directions, which then requires continuity. It might help if you say what you do know. $\endgroup$
    – Timaeus
    Commented Aug 23, 2015 at 5:08
  • $\begingroup$ This is an introductory physics book and we have defined curl as a vector derivative, so it has 3 components that go like a cross product. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 5:10
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The magnitude of the B-field is $a/r$ and circulates around the axis. By symmetry, you understand that the magnitude is zero on-axis. But if $a$ is anything but zero, your expression gives an infinite B-field magnitude. Therefore $a$ must be zero and therefore the B-field is also zero everywhere else inside the pipe.

The result also follows from Ampere's law. The line integral of the B-field around a closed circular loop inside the pipe, which encloses no current, should be zero. As the B-field is parallel to the line element (if $a$ is non-zero), you would get a non-zero line integral. Therefore both $a$ and the B-field must be zero.

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  • $\begingroup$ I know how to use Ampere's law to show this. I really wanted to follow the reasoning in the question. The point of the question is to figure it out without Ampere's Law. Also, your answer is not correct. My equation for B applies at all points except r = 0. According to your line of reasoning K in Coulomb's law would have to be zero or else the electric field of a single point particle would be infinite at r = 0. But that's wrong, because Coulomb's law is only valid for non-zero r. $\endgroup$
    – user7348
    Commented Aug 23, 2015 at 19:41
  • $\begingroup$ @user7348 Firstly, you don't specify not using Ampere's law (which is the obvious and simplest thing to do). Second, you should specify (and say why) you think your equation does not apply at $r=0$ (it seems to me that it does as long as $a=0$). The comparison with Coulomb's law is irrelevant, since we can't say the E-field is zero (or any value) at the origin, whereas in this case we can say B=0 at the origin because it is axially symmetric and can't be radial. $\endgroup$
    – ProfRob
    Commented Aug 23, 2015 at 20:05
  • $\begingroup$ @user7348 Possibly more to the point is that the E-field (and its spatial derivatives) in the Coulomb's law example is discontinuous because E-field lines begin and end on charges. The B-field cannot be discontinuous in the absence of currents. $\endgroup$
    – ProfRob
    Commented Aug 23, 2015 at 20:22

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