electromagnetism
(source: gsu.edu)
To calculate the magnetic field, a rectangle amperian loop was drawn, and since the sides of the rectangle are perpendicular to the magnetic field, and the top is too far away to have any field lines cross it, only BL makes a contribution.
(source: gsu.edu)
I don't understand why this magnetic field is only contained within the solenoid. Shouldn't the magnetic field at any point on the amperian loop equal μnI, according to Ampere's law?
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} $.
Mostly, yes on both counts.
It depends whether the solenoid also gets longer as it gets wider. The approximation that the field outside the solenoid vanishes is valid for points whose distance to the solenoid's centre is much smaller than the distance to both ends. This is impossible for a point outside a solenoid that's wider than it is tall.
To answer your more specific question about opposite points, the reason these points still cancel is because there is more current per unit angle as the solenoid gets wider if the distance remains fixed. It's more complicated than this - the field really vanishes because of the cancellation of the complete integral over the infinite surface - but one good picture to keep in mind is the following.
The field of a solenoid that's considerably stretched, or of a series of circles that are widely separated, is indeed different to that of an 'ideal' solenoid, which is usually modelled as a surface current, with no interruptions, over a cylinder. For a real solenoid with spacings between the loop, the approximation will hold for points whose distance to the nearest point on the solenoid is greater than the inter-loop spacings.
Best Answer
Solenoids of finite length do have a non-zero magnetic field outside. As the length increases, this magnetic field is significantly smaller than the magnetic field inside the solenoid. An ideal solenoid has infinite length and this magnetic field is zero. One is usually dealing with this idealisation. This is similar to assuming that there is no electric field outside a parallel plate capacitor. It is a fun (not necessarily easy) problem to estimate the ratio of the interior and exterior magnetic fields for a finite solenoid.