Calculate inductance of finite Solenoid

AI Thread Summary
The discussion focuses on calculating the inductance of a finite solenoid, which differs from the standard formula used for infinite solenoids. The magnetic field produced by the solenoid is derived using a detailed calculation involving the geometry of the solenoid and the current flowing through it. The participant seeks confirmation on their calculations and guidance on how to determine the total magnetic flux through the solenoid to ultimately calculate the inductance. They also mention existing formulas for various solenoid configurations, including the Nagaoka coefficient for cylindrical air core coils. The conversation highlights the complexity of finding inductance for finite solenoids compared to infinite ones.
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Homework Statement


A finite solenoid with "N" turns of wire, "L" length , "R" is the radius of the solenoid and passes through it a current "I".
The objective is to calculate "L" of a finite solenoid. Not the basic formula ##L=\frac{\mu_0·N^2·S}{Length}## which is for a infinite solenoid.
See picture.
mvhpwm.png


Homework Equations


Magnetic field produced by 1 coil at a point far from the coil a distance "x"
##B=\frac{\mu_0·I·R^2}{2(R^2+x^2)^\frac{3}{2}}##
x=distance from the center of the coil to a point in it's axes

The total magnetic flux into a solenoid is proportional to the current : ##\phi_m=L·I## where L=inductance of the solenoid

The Attempt at a Solution


First of all I calculate the magnetic field produced by the solenoid in a point out of the solenoid as follows:
The elementary magnetic field by a proportion of conductors in the region dx is:
##dB=\frac{\mu_0·I·R^2}{2(R^2+x^2)^\frac{3}{2}}·\frac{N}{L}dx##

And from the figure I find out that: ##x=R·ctg\beta \Rightarrow dx=-R·(cosec\beta)^2·d\beta ## and ##R^2+x^2=R^2(cosec\beta)^2##
So substituing the elementary magnetic field is: ##dB=\frac{\mu_0·N·I}{2L} (-sin\beta d\beta)##

The total magnetic field in that point is:
##B=\frac{\mu_0·N·I}{2L}\int_(\beta_1)^(\beta_2) -sin\beta d\beta=\frac{\mu_0·N·I}{2L}(cos\beta_2 - cos\beta_1)##

And if the point is placed in the center of the first coil --> ##cos\beta_1=0 ; cos\beta_2=\frac{L}{(L^2+R^2)^\frac{1}{2}}##

So the magnetic field in the first coil is : ##B=\frac{\mu_0·N·I}{2L}\frac{L}{(L^2+R^2)^\frac{1}{2}}##

And now to calculate the magnetic flux through the first coil --> ##\phi=\int_S^· BdS##

Before I keep doing my calculations my questions are:
1) It is correct what I have done until now ?
2) How do I calculate the magnetic flux ##\phi_m## through all the solenoid so then I can calculate the inductance ##L=\frac{\phi_m}{I}##
 
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Trying to find out the inductance like that is a difficult task. I thought it would be easy to find it but I am wrong... I have just found out by searching properly (with the accurate words) that there are expressions for most used forms of solenoids so I am going to share them with you... If somebody knows more about this field would be great to post them :wink:.

Cylindrical air core coil : ##L=\frac{\mu_0 ·N^2 ·A·K}{L}## where K=Nagaoka coefficient , A=area of cross section, N=number of turns, L=length of the solenoid

Here is a curve to detremine the Nagoaka coefficient
11s1soy.png

On the x axis: length = length of the solenoid; diameter=diameter of the solenoid (do not confuse it with the diameter of the wire you make the solenoid)

If the length >>>> diameter => Nagoaka coefficient is 1 so it is the case of the infinite solenoid
 
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