Calculating Self-Inductance of Long Current-Carrying Wire

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The discussion centers on calculating the self-inductance per unit length of a long current-carrying wire using two methods. Method 1, which calculates the magnetic flux and derives inductance as μ₀/4π, is deemed incorrect. Method 2, based on energy associated with the magnetic field, yields a correct inductance of μ₀/8π. Participants highlight potential errors in Method 1, particularly regarding the assumption of flux linkage and the integration process. The consensus suggests that the second method provides the accurate result, with further clarification on Method 1 needed.
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I was trying to calculate the self inductance (per unit length) of the following system using two methods:

System : A long current carrying wire of radius R carrying uniform current density and the same current returning along the surface. (Assuming that the surface is insulated by a very thin sheet).

Method 1 : I calculated B inside and found out flux. Then I used \Phi = LI
The answer came out to be \frac{\mu_{o}}{4\pi}

Method 2 : I calculated the energy associated with the magnetic field and equated it to \frac{1}{2}LI^{2}. The answer came out to be \frac{\mu_{o}}{8\pi}

Which one is correct and where is the mistake in the wrong one ?
 
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Can you show your work for method 1?
 
Using Ampere's Circuital Law,
B_{inside}*2 \pi s = \mu_{o}\frac{I \pi s^{2}}{\pi a^{2}}

=> B_{inside} = \frac{\mu_{o}Is}{2\pi a^{2}}

=> d \Phi = B_{inside}.da = \frac{\mu_{o}Is}{2\pi a^{2}}dsdz

=> \Phi = \int B.da = \frac{\mu_{o}Il}{4 \pi} = LI

=> L_{per unit length} = \frac{\mu_{o}}{4 \pi}
 
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anyone ??
 
Your Method #1 calculation looks OK. Show how you calculated the energy density for Method #2.
 
Hey Hi! I am not here to answer your question and sorry if that disappoints you but how do you define flux linkage in this particular situation?
 
For the energy method

U = \frac{1}{2\mu _{o}}\int B^{2}dV

=> U = \frac{1}{2\mu _{o}} \int \frac{\mu _{o}^{2}s^{2}}{4\pi ^{2}a^{4}}sdsd\phi dz

phi varies from 0 to 2pi, s from 0 to a and integral of dz = l

equating value of U with 0.5LI^2 we get
L_{perunitlength} = \frac{\mu _{o}}{8 \pi}
 
Should there be an I in your integral .
 
iitjee10,

Not an easy problem.
I think there is a problem in your third equation which says that flux is linearly proportional to z. Finding the inductance of a straight length of wire is complex. Here is a reference http://www.ee.scu.edu/eefac/healy/indwire.html. Good luck.
 
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The answer you got through method 1 is wrong.
The correct answer is the one you got in your second method. Once I figure out why method 1 is wrong I will post it here.
 
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