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.
Best Answer
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} $.