Calculating Self-Inductance of Long Current-Carrying Wire

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SUMMARY

The discussion focuses on calculating the self-inductance per unit length of a long current-carrying wire using two methods. Method 1, which applies Ampere's Circuital Law, yields a result of \(\frac{\mu_{o}}{4\pi}\), while Method 2, based on energy density calculations, results in \(\frac{\mu_{o}}{8\pi}\). The consensus among participants indicates that Method 1's result is incorrect, and the correct self-inductance is given by Method 2. Participants emphasize the complexity of the problem and the need for careful consideration of flux linkage.

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iitjee10
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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}
 
Last edited:
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.
 
  • #10
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.
 
Last edited:

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