Calculating Self-Inductance of Long Current-Carrying Wire

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Discussion Overview

The discussion revolves around calculating the self-inductance per unit length of a long current-carrying wire using two different methods. Participants explore theoretical approaches and calculations related to magnetic fields and energy associated with the wire's magnetic field.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculated the self-inductance using two methods, obtaining different results: \(\frac{\mu_{o}}{4\pi}\) for Method 1 and \(\frac{\mu_{o}}{8\pi}\) for Method 2.
  • Another participant requested the detailed calculations for Method 1 to verify the results.
  • A participant provided a derivation for Method 1 using Ampere's Circuital Law, leading to the conclusion that \(L_{per unit length} = \frac{\mu_{o}}{4 \pi}\).
  • There was a request for the energy density calculation used in Method 2.
  • A participant questioned the definition of flux linkage in the context of the problem.
  • One participant derived the energy associated with the magnetic field, leading to the conclusion that \(L_{per unit length} = \frac{\mu_{o}}{8 \pi}\).
  • Another participant pointed out a potential issue in the flux calculation related to the dependence on the variable \(z\).
  • One participant asserted that the answer from Method 1 is incorrect and suggested that the second method's result is the correct one, promising to clarify the mistake in Method 1 later.

Areas of Agreement / Disagreement

Participants do not reach a consensus on which method provides the correct self-inductance value, as there are competing views regarding the validity of the calculations and assumptions made in each method.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the dependence of flux on the variable \(z\) and the inclusion of current in the energy density integral.

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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