# Applying the Biot-Savart Law to solenoids

1. Feb 13, 2015

### RawrSpoon

Not really homework but I figured this was the best place to post anyway.

1. The problem statement, all variables and given/known data

I want to find the magnetic field B for an arbitrary solenoid using the Biot-Savart Law. I can find it easily through Ampere's Law, but I'd like mastery over the Biot-Savart Law.

2. Relevant equations
$$B=\frac{μ_{0}}{4\pi}\int \frac{K \times (r-r')}{|r-r'|^3}da'$$
As there's no curly r as used in Griffiths Electrodynamnics, I'll replace curly r with an arbitrary symbol γ such that
$$\gamma=r-r'$$
Thus
$$\hat{\gamma}=\frac {r-r'}{|r-r'|}$$
And so
$$B=\frac{μ_{0}}{4\pi}\int \frac{K \times \hat{\gamma}}{\gamma^2}da'$$
Maybe unnecessary but perhaps not. I just enjoy that notation as it's what I'm used to.

3. The attempt at a solution

As K is the surface charge density, I'll make the supposition that
$$K=\frac{NI}{L}=nI$$
where N is the number of turns on the solenoid, and L is the length of said solenoid.

And this is where I get stuck. Whereas I easily found the magnetic field due to a single loop of wire, the solenoid having length makes me be very unsure as to where to even begin. Whereas a loop of wire has γ easily defined as
$$\gamma=\sqrt {R^2+x^2}$$
where R is the radius of the loop of wire and x is the distance of an arbitrary point M along the same axis as the loop

I've found that γ at the part of the solenoid closest to M is equal to
$$\gamma=\sqrt {R^2+x^2}$$
the part of the solenoid furthest from M gives a γ of
$$\gamma=\sqrt {R^2+(x+L)^2}$$

I'm really lost. This doesn't mean I haven't tried my hardest or I'm being lazy. Any attempt to solve in a similar manner to that of a loop of wire ends up giving me multiple integrals that become a massive headache really quickly. Any nudge in the right direction would be greatly appreciated. If anything is unclear, let me know so I can try to either clarify my language or even attempt to draw a (likely very poor quality) image.

Last edited: Feb 13, 2015
2. Feb 13, 2015

### BvU

Perhaps example 30.33 here ? It's already pretty hefty.
But I suppose you already went through that and now want a real challenge.
However, the general case is probably done numerically most of the time ('using software' this guy says :) ).

3. Feb 15, 2015

### RawrSpoon

I hadn't done one like that problem actually so thank you so much for that! And I'm a little disappointed the integrals are very difficult by hand, but at least now I know a little better when to use Ampere's Law. Thank you.