- #1

Nabeshin

Science Advisor

- 2,205

- 16

[tex]B=\mu_oNI[/tex]

N is turns, which I call n per length, which I will call [tex]L_s[/tex]

[tex]B=\frac{\mu_onI}{L_s}[/tex]

From ohms law, I = V/R

[tex]B=\frac{\mu_onV}{L_sR}[/tex]

And R is equal to [tex]\frac{\rho L_w}{A}[/tex] Where [tex]L_w[/tex] is the length of wire and A is its cross-sectional area.

[tex]B=\frac{\mu_onVA}{\rho L_s L_w}[/tex]

Up till here I'm confident, but after this I'm not so sure. Now, I tried to model the helical nature of the wire wrap by creating a vector-valued function:

[tex]r(t)=sin(t)i-cos(t)j+\frac{r_w}{\pi}t k\left[/tex]

Now in doing this I assume that the coils are wrapped as tight as possible. By this I mean that the horizontal spacing between two loops is equal to 2r, or the diameter of the wire. So this function should move a distance of 2r up for every turn (2*pi radians).

Now, arc length is given by the formula: [tex]s(t)=\int||r'(t)||dt[/tex] So...

[tex]s(t)=\int\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt[/tex]

Now, in one turn of the wire we rotate 2pi radians, so let's evaluate from 0 to 2pi..

[tex]s(t)=\int^{2\pi}_{0}\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt=2\sqrt{\pi^{2}+r^{2}}[/tex]

So this is the length of wire per one turn, so we can say:

[tex]\frac{Length\f\:wire}{Turn\f\:wire}=2\sqrt{\pi^{2}+r^{2}}[/tex]

And turns of wire is n, so we finally arrive at:

[tex]L_w = 2n\sqrt{\pi^{2}+r^{2}}[/tex]

Back to the main mission! Substituting in for [tex]L_w[/tex] produces:

[tex]B=\frac{\mu_o A V}{2\rho L_s\sqrt{\pi^{2}+r^{2}}}[/tex]

So I gather that the magnetic field of a solenoid depends on three things: How long the solenoid is, the specific wire you're using (both size and material), and the voltage applied. Well, does this make sense to anyone? :rofl: