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.
Best Answer
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.