Inductance of solenoid (length = diameter)

In summary, the homework statement is that a solenoid inductor consists of 200 closely spaced turns, length = diameter = 5cm. Calculate inductance. The magnetic flux needs to be included as a function of position to get the total flux and thereby the total inductance. The most popular approximation is that L = μ0 * N2 * A / L(length), A = cross section area. 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.
  • #1
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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  • #2
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.
 
  • #3
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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Related to Inductance of solenoid (length = diameter)

What is the concept of inductance?

Inductance is the property of an electrical circuit that causes an electromotive force to be generated in that circuit when the current changes. It is measured in units of Henry (H).

How is inductance calculated for a solenoid?

Inductance of a solenoid can be calculated using the formula L = μ0 * N^2 * A/l, where μ0 is the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

What is the relationship between length and inductance for a solenoid?

The length of a solenoid has a direct impact on its inductance. As the length of the solenoid increases, the inductance also increases.

How does diameter affect the inductance of a solenoid?

The diameter of a solenoid also affects its inductance. A larger diameter solenoid will have a higher inductance compared to a smaller diameter solenoid, as it allows for more turns of wire and increases the cross-sectional area.

What are the practical applications of inductance of a solenoid?

Inductance of a solenoid has practical applications in electronic devices such as transformers, motors, and generators. It is also used in wireless charging technology, electromagnets, and inductors used in electronic circuits.

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