New Technique!!!-Part 2

I was thinking about some stuffs while taking a piss… I imagined something in my head without really taking it seriously. Then suddenly BAM!! I realized that actually this thing might work, it turns out that I found another application of the Awesome Technique! Holy sh*t, I felt like it was a piss which would determine the fate of the world.

Here is the problem (actually I made it up after I found the solution):
Most electrodynamics textbooks simply state that the magnetic field outside a solenoid is negligibly small without giving clear reasonings\. How small is negligibly small? We need to know exactly what are the cases where we can still safely neglect the external field of a solenoid! Otherwise when the time comes we may accidentally neglect some important terms!

Okay, let’s welcome our guest star today:

A long solenoid with length L, and cross sectional area A. Such that L>>\sqrt{A}.

Our mission is to calculate the magnetic field outside of a solenoid, at a point with a distance r from the axis of the solenoid with r<<L. And not too close to any of the ends of the solenoid. Because, you know, things get insane there.

Now here is the time when the hero comes in, let’s use special technique! We need to make a bada$$  loop so that the flux through this loop due to the solenoid’s outer field depends only on B_{out}(r) and r. Does such kind of loop really exist?? Indeed there is a way to do that, by combining two current loops. It is kind of hard to describe, so I’ll just show you how it looks like.

The flux through this loop due to the solenoid’s outer field is

\phi_{RS}=B_{OUT}(r).2\pi r\delta r

Note that there is a component of B_{OUT} which is parallel to the loop, but don’t worry about that. Since we already chosen a point where r<<L, we can only have field lines with small slopes. Which means that B_{parallel with loop} compared to B_{OUT} is like an ant compared to an elephant.

On the other hand the awesome loop also produces it’s own magnetic field , we can neglect the field due to the “connecting wires” because they are so tiny. So basically this field is just the difference of magnetic field due to circular loops of radius r and r+\delta r. The magnetic field due to one ring only is

B_{ring}=\frac {\mu_0 r^2I}{2}\frac{1}{(x^2+r^2)^{3/2}}

The difference of magnetic flux through the whole solenoid due to two such rings with slightly different radius is

\phi_{SR}=\delta ({\int_{-x}^{L-x}\frac{\mu_0 I_Rr^2}{2(x^2+r^2)^{3/2}}Andx})

Kick out all the constant factors!

\phi_{SR}=\frac{\mu_0I_RAnr^2}{2}\delta(\int_{-x}^{L-x}\frac{1}{(x^2+r^2)^{3/2}}dx)

Integrate!

\phi_{SR}=\frac{\mu_0I_RAn}{2}\delta(\frac{L-x}{\sqrt{(L-x)^2+r^2}}+\frac{x}{\sqrt{x^2+r^2}})

The square root of blah blah blah makes me sick, anyone will get tired of seeing this. So let’s take the first order approximation using L>>r. But actually we HAVE to do this, because we have used the approximation once when calculating the field due to the ring, we should be consistent. Here we go

\phi_{SR}=\frac{\mu_0I_RAn}{2}\delta(2-\frac{1}{2}[\frac{r^2}{(L-x)^2}+\frac{r^2}{x^2}])

Taking the differential

\phi_{SR}=-\frac{\mu_0I_RAn}{2}(\frac{1}{(L-x)^2}+\frac{1}{x^2})r\delta r

Then using the mutual inductance symmetry relation

\frac{\phi_{RS}}{I_S}=\frac {\phi_{SR}}{I_R}

Finally we get

B_{OUT}(r)=-\frac{\mu_0I_SAn}{4\pi}(\frac{1}{(L-x)^2}+\frac{1}{x^2})

This only holds at the points which aren’t near the ends of the solenoid, so x is about the same order as L. In other words B_{OUT} is in the order of B_{IN}\frac{A}{L^2}. We can check that if L>>\sqrt{A} then B_{OUT}(r) is vanishingly small.

That was seriously cool, I thought we would never be able to calculate solenoid’s outer magnetic field analytically. My computer is about to explode because of the awesomeness.

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