Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found an equation in theory about magnetic induction in a solenoid: $B_s=\mu_0 I n$. It should be magnetic induction for infinite length solenoid. I wonder if this is anyhow useful. Where can this be used?

($n = \frac {N}{L} $, where $L$ is length of solenoid and $N$ is number of turns... which doesn't make sense to me, if the length is supposed to be infinite)

share|cite|improve this question
up vote 2 down vote accepted

The infinite length is not really infinite but it is infinite relative to the very small radius of loops.

If we consider radius to be relatively comparable, then the field will depend upon the radius and the point where the field is to be calculated.(can be done by adding field due to all loops by integration)

Then field inside at a point comes out to be $$\dfrac12 \mu_0nI\bigg[\dfrac{b}{\sqrt{b^2+r^2}}+\dfrac{a}{\sqrt{a^2+r^2}}\bigg]$$ where $a$ and $b$ are distances from the point to the ends of solenoid.

We can see if $r<<a,b$ then field comes out to be$$\mu_0nI$$ and this condition is called infinite length.

share|cite|improve this answer
In our materials it is $$\dfrac12 \mu_0nI\bigg[\dfrac{b}{\sqrt{b^2+r^2}}-\dfrac{a}{\sqrt{a^2+r^2}}\bigg]$$ and that would be $0$. If I put here $+$, graph is not right. – user50222 Apr 20 '13 at 14:27
@user23125 It must be in vector form or in other words the a,b are taken with directions.I have given the magnitude and direction has to be seen physically. So, in your case , if you see $\vec a$ would be negetive and so magnitude will come out to be $\mu_0nI$ – ABC Apr 20 '13 at 14:33
@user23125 otherwise the point where field induction is calculated is taken as origin and a,b are displacements of the end points of solenoid. , and so it comes out same/ – ABC Apr 20 '13 at 14:39

Your Answer


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.