Calculate inductance of finite Solenoid

In summary, the objective is to calculate the length of a finite solenoid with N turns of wire, radius R, and current I passing through it. The basic formula for an infinite solenoid does not apply in this case. To find the magnetic field produced by the solenoid at a point outside of it, the equation dB = (μ0NI/2L)(-sinβdβ) is used, where β is the angle between the axis of the solenoid and the point. The total magnetic field at this point can be found by integrating this equation. To find the magnetic flux through the solenoid, the equation φ = ∫BdS is used. There are specific expressions for finding the
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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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  • #2
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
 

1. How do you calculate the inductance of a finite solenoid?

The inductance of a finite solenoid can be calculated using the formula L=N²μA/l, where N is the number of turns, μ is the permeability of the material, A is the cross-sectional area, and l is the length of the solenoid.

2. What is the difference between a finite solenoid and an infinite solenoid?

A finite solenoid has a limited length and the magnetic field lines are confined within the solenoid. In contrast, an infinite solenoid has an infinite length and the magnetic field lines extend infinitely in both directions.

3. How does the number of turns affect the inductance of a finite solenoid?

The inductance of a finite solenoid is directly proportional to the square of the number of turns. This means that doubling the number of turns will result in a four-fold increase in inductance.

4. Can the inductance of a finite solenoid be changed?

Yes, the inductance of a finite solenoid can be changed by altering its physical dimensions, such as the number of turns, length, and cross-sectional area. It can also be changed by varying the permeability of the material used.

5. What is the unit of measurement for inductance?

The unit of measurement for inductance is the Henry (H), which is equivalent to volts per ampere (V/A). It is named after the physicist Joseph Henry, who made significant contributions to the study of electromagnetism.

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