# Homework Help: 2 Derivations of Solenoid's Magnetic Field - Problem

1. Apr 4, 2007

### 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?

Last edited: Apr 4, 2007