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You are given a long length W of copper wire. How would you arrange it to obtain the maximum self-inductance? Why?

I am trying to use the equation

$$L=\mu_o n^2 l A$$

I try to solve it using a fixed length wire of 10 units, width 1mm and winding it into a solenoid. I plug in values of circumference 10, 5, 2.5 and finding the inductance through number crunching. However, I am getting a larger values for multiple loops but the answer is a single loop (ie. a circle) rather then a solenoid.

Here are the sample values I got:

$n=1; C=10; r=1.59; L=0.079 \mu_o$

$n=2; C=5; r=0.79; L=0.156\mu_o$

$n=4; C=2.5; r=0.3978; L=0.318\mu_o$

If anyone could enlighten me on the proper way of solving this, I'd appreciate it.

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up vote 1 down vote accepted

I believe you have a mistake in your formula as the self-inductance of a coil is given by $$L\approx\mu_0 \frac{n^2 A}{\ell};$$ here $n$ is the number of windings, $A$ is area of the cross-section, and $\ell$ is the length of the coil.

Your task is to maximize $L$ with the constraint that the length of the copper wire is $W$. Assuming that the solenoid is a cylinder, the cross-section read $A=\pi R^2$ with $R$ the radius of the cylinder.

A solenoid with $n$ windings needs a wire of length $W= 2\pi Rn$. Thus, $$ L \approx \mu_0 \frac{W^2}{\ell}.$$ We see that the inductance of the solenoid decreases with increasing length (keeping the total length of the wire fixed). Thus, we obtain the largest self-inductance having the smallest length which is a single loop with $n=1$. For a single loop the formula given above is not correct (as it assumes $\ell \gg \sqrt{A}$) and thus we have $$L\approx \mu_0 R \ln (R/r) \approx \mu_0 \frac{W}{2\pi} \ln(W/r) $$ with $r$ the radius of the wire.

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Bah it seems that your equation is correct according to other internet sources. However I found the solution the prof posted, and he uses the equation I gave :/ – mrcharlie Apr 8 '12 at 23:12
how does W^2 = n^2 A? – mrcharlie Apr 8 '12 at 23:35
now how can both equation be correct?? at least one of them is dimensionally incorrect.... :( – Vineet Menon Apr 9 '12 at 7:57

Aright, I actually found the solution ..

$$W=(2\pi r) \cdot N = (2 \pi r)(nl)$$

so $$r=W/2 \pi n l$$

also, $A = \pi r^2$ so this leads to

$$L = \mu_o n^2 l \pi r^2$$ $$ = \mu_o n^2 l \pi (W/2 \pi n l)^2$$ $$=\mu_o (W/4 \pi l)$$

so to maximize $L$, you want $l$ to be as small as possible, ie. 1.

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