# Magnetic field produced by a solenoid

Ok, so anyone who has studied magnetism knows that the magnetic field due to a solenoid is given by the equation $$B=\mu_oNI$$ Where N is the number of turns in a given length. Well, in thinking about these, I tend not to make measurements and would rather like to predict the magnetic field of the solenoid, so I tried to "simplify" the formula. Here goes:

$$B=\mu_oNI$$
N is turns, which I call n per length, which I will call $$L_s$$
$$B=\frac{\mu_onI}{L_s}$$
From ohms law, I = V/R
$$B=\frac{\mu_onV}{L_sR}$$
And R is equal to $$\frac{\rho L_w}{A}$$ Where $$L_w$$ is the length of wire and A is its cross-sectional area.
$$B=\frac{\mu_onVA}{\rho L_s L_w}$$

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:
$$r(t)=sin(t)i-cos(t)j+\frac{r_w}{\pi}t k\left$$
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: $$s(t)=\int||r'(t)||dt$$ So...

$$s(t)=\int\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt$$
Now, in one turn of the wire we rotate 2pi radians, so let's evaluate from 0 to 2pi..
$$s(t)=\int^{2\pi}_{0}\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt=2\sqrt{\pi^{2}+r^{2}}$$
So this is the length of wire per one turn, so we can say:

$$\frac{Length\ f\:wire}{Turn\ f\:wire}=2\sqrt{\pi^{2}+r^{2}}$$

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

$$L_w = 2n\sqrt{\pi^{2}+r^{2}}$$

Back to the main mission! Substituting in for $$L_w$$ produces:

$$B=\frac{\mu_o A V}{2\rho L_s\sqrt{\pi^{2}+r^{2}}}$$

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:

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