Inductance of solenoid (length = diameter)

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lcr2139
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Homework Statement


a solenoid inductor consists of 200 closely spaced turns, length = diameter = 5cm. calculate inductance.

Homework Equations

The Attempt at a Solution


H = NI/L
L = N (flux) / I - you can't use this because the length is equal to the diameter. What equation do I use?
 
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The magnetic flux needs to be included as a function of position to get the total flux and thereby the total inductance. The simplest solution for the B field for a finite length solenoid (it turns out to be exact), is to use a result from the "pole method" of magnetostatics. The B (M.K.S. units) inside the solenoid will be equal to ## B_z= n*\mu_o*I ## plus a subtractive correction term of the B from poles of surface magnetic charge density ## \sigma_m=n*\mu_o*I ## with a "+" pole on the right and a "-" pole on the left. The magnetic surface charge density in this mathematical solution is considered to be uniform over the opening of the solenoid. I don't have a "link" for you, but I do think you could possibly find this solution in one of the older and more advanced E&M texts. One result you get from the above is that the B field at the openings of any finite length solenoid is exactly half of the value that it takes on for one of infinite length.
 
lcr2139 said:
What equation do I use?
If you google inductance+solenoid+calculator you will find different equations, and by using different calculators, you will get different results because the equations they are using are approximations.

The most popular approximation is:

L = μ0 * N2 * A / L(ength) , A = cross section area.

I think a more precise result is found, calculating the B-field ( at 1 A ) at different locations inside the solenoid, then integrating the B-fields to a flux passing through every turn.

Use Biot-Savart.

Having your program up and running, the calculation can be done within say 20 minutes.
 
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