2 Derivations of Solenoid's Magnetic Field - Problem

[sprinkler]

Hello,

I have been asked to find the magnetic field inside a solenoid (along its axis)using two methods. One is using Ampere's law (and the approximation that the B field is zero outside the solenoid). Here is what I get:

B = (uIN)/(L)

where I is current, N is the number of loops, and L=length of solenoid.

The other method is integrating the equation for the B field of a single ring, over the length of the solenoid. Here is what i get:

B = (uIN)/(2L) * [ Z/(R^2 + Z^2)^(1/2) - (Z-L)/(R^2 + (Z-L)^2)^(1/2) ]

where R is the radius of the solenoid, Z is the location on the "z" axis where I wish to calculate the magnetic field. (Can also be seen here: http://www.netdenizen.com/emagnettest/solenoids/?thinsolenoid)

Now with this second equation, if I make the approximation that L>>R, it boils down to this:

B = (uIN)/(2L)

(basically only the initial constants remain, the rest of the equation ends up equaling one)

However this is not what I found with amperes law, there is an extra factor of (1/2). Can anyone explain why?

 

[kuruman]

However this is not what I found with amperes law, there is an extra factor of (1/2). Can anyone explain why?

The reason why is subtle and has to do with how you did the integral and what it means to let the length become large relative to the radius. You took the limits of integration from zero to $L$ and then allowed $L$ to become large. That is not an infinite solenoid, it is a semi-infinite solenoid because it starts at z = 0 and points closer to that end have a different field than points farther in because of the asymmetry.

In an "infinite" solenoid every point on the axis must be equivalent to any other point, i.e. all points must be equally far away from the ends. That is achieved if you place the origin in the middle and integrate over symmetric limits from $(-L/2)$ to $(+L/2)$. In that case you get $$B(z)=\frac{\mu_0NI}{2L}\left(\frac{z+\frac{L}{2}}{\sqrt{\left(z+\frac{L}{2}\right)^2+R^2}}-\frac{z-\frac{L}{2}}{\sqrt{\left(z-\frac{L}{2}\right)^2+R^2}}\right).$$ When $L>>z,R$, this expression reduces to the one obtained using Ampere's law.