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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.
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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