Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.