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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)

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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.

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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

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