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Consider a hollow cylinder carrying a current $I$ and a wire outside the cylinder carrying a current $I'$. Let's say the cylinder is symmetrical with even current distribution etc.. so the $\mathbf{B}$ field at any point (due to current in cylinder) within the cylinder is zero by Amperes Law. However, this doesn't mean the $\mathbf{B}$ field is zero within the cylinder entirely - there is a $\mathbf{B}$ field contribution from the wire. So my question is: What is the usefulness of Amperes Law?

Does Ampere's Law only tell me something about the $\mathbf{B}$ field from a particular source?

Also say we have a solid cylinder inside a hollow cylinder with radii $a$ and $b$ respectively. They have opposite current directions. Then by Ampere, the $\mathbf{B}$ field at some point $P$ where $a < P < b$ is given as $B = \frac{\mu I}{2\pi r}, I $ the current in the solid cylinder. Is it really? The $\mathbf{B}$ field from the hollow cylinder will be in the opposite direction at $P$ and so acts to cancel the $\mathbf{B}$ field at $P$ from the solid cylinder thus resulting in zero net $\mathbf{B}$ field, no? Yet the $\mathbf{B}$ field at $P$ is in fact nonzero?

I understand how the non zero $\mathbf{B}$ field was obtained using Ampere's Law, but the Amperian loop which coincides with $P$ does not simply shield the $\mathbf{B}$ field from the hollow cylinder. So I am struggling to see why the $\mathbf{B}$ field would be nonzero.

Many thanks.

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  • $\begingroup$ You recognized in the first paragraph that inside a hollow cylinder, there is no B field due to the cylinder. So at point P<b, the hollow cylinder contributes nothing; the entire B field is from the solid cylinder. So what is the confusion? $\endgroup$ – Jim May 1 '13 at 15:08
  • $\begingroup$ What would happen if I had a wire too - so that I have a solid cylinder within the hollow cylinder and the wire outside the hollow cylinder. Suppose the wire is such that the B field (from the wire) is in the opposite direction to the B field due to the solid cylinder at point P. Then by Ampere, the B field is nonzero. So how does this conform with my set up? $\endgroup$ – CAF May 1 '13 at 15:18
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  • What is the usefullness of Amperes Law?Does Ampere's Law only tell me something about the B field from a particular source?

Ampere's law holds for every distribution of currents (this form holds for static currents)

$$ \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} =\mu_0 \int_{\Sigma} \mathbf{J} \cdot \mathrm{d}\mathbf{S} $$

In general, it is not a tool for direct calculation of the magnetic field , but we can use it in some (following) cases to find the field directly.

In all cases , where we can use it to determine the magnetic field, we find a (family of) path(s) , on which $\mathbf{B}$ is constant , and so comes out of the integral . This is the case when we find the field of a wire.

But suppose you want to find the field of two parallel wires. In this case you can not use this Integral relation as easy (and naively) as before , because the field is not constant on a simple path. In such cases we actually use the linearity of Maxwell equations: $$ \mathbf{J}_1 \to\mathbf{B}_1$$ $$ \mathbf{J}_2 \to\mathbf{B}_2$$ $$ \mathbf{J}_1 + \mathbf{J}_2 \to\mathbf{B}_1 +\mathbf{B}_2$$

so we consider one wire at a time and find the field of that wire, which simply can be found by choosing a circle surrounding the wire as our integration path. Then the total field will be sum of two fields.

Using these arguments , the field in a coaxial cable (your second problem) is $ \frac{\mu_0 I}{2\pi r} $.

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  • $\begingroup$ So as my example illustrates, it really only gives you the B field if the problem is highly symmetric and it only really helps in giving the B field locally? (What I mean by this is that even though the Amperian loop within the hollow cylinder gives a zero B field from the cylinder it tells us NOTHING about the B field from the wire - so even though we enclose a certain region within an Amperian loop and the B field happens to be zero for one source, we cannot conclude that the B field is entirely zero). Do you have any thoughts about my second comment posted above? Thank you. $\endgroup$ – CAF May 1 '13 at 16:28
  • $\begingroup$ No. As I said above, if you consider the wire and cylinder simultaneously, the ampere's law will not be incorrect, and considers the field from both the wire and the cylinder .But this field is not constant on your circular amperian loop and you can not get the $B$ out of integral and deduce that it equals 0. $\endgroup$ – Mo_ May 1 '13 at 16:41
  • $\begingroup$ Actually, I think I didn't make myself very clear in my second comment above. Enclose the solid cylinder within the hollow cylinder. Then at some P < b, the B field is only due to the solid cylinder and is nonzero. Let's say I then have a wire outside the hollow cylinder such that the B field it produces is directly opposite the direction of the B field due to the solid cylinder at P. So the net B field is zero (at P) yet by Ampere, we attain a non zero B field. What is wrong with this analysis? $\endgroup$ – CAF May 1 '13 at 17:14
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    $\begingroup$ How do you attain a nonzero field there at P?! If you want to find the field when considering the wire and the inner(field producing) cylinder you can't use Ampere's law to derive the field! As I said above , if you want to find the field using Ampere's law you must consider first the cylinder, without any wire, then the wire without any cylinder, so you can use ampere's law. Then summing these two fields , you will have the total field everywhere. $\endgroup$ – Mo_ May 1 '13 at 17:35
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    $\begingroup$ Yes, that is my confusion. Considering the B field from the inner cylinder first yields a non zero B field. However, the Amperian loop I used to deduce this does not 'shield' the B field from the wire. So even though I used the Amperian loop to deduce a non zero field, in reality there is also the field from this wire. $\endgroup$ – CAF May 1 '13 at 18:26

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