Calculating Self-Inductance of Long Current-Carrying Wire

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In summary, Ampere's circuital law states that the flux (energy) through a wire is proportional to the cross sectional area.
  • #1
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 [tex]\Phi = LI[/tex]
The answer came out to be [tex]\frac{\mu_{o}}{4\pi}[/tex]

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

Which one is correct and where is the mistake in the wrong one ?
 
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  • #2
Can you show your work for method 1?
 
  • #3
Using Ampere's Circuital Law,
[tex] B_{inside}*2 \pi s = \mu_{o}\frac{I \pi s^{2}}{\pi a^{2}} [/tex]

[tex] => B_{inside} = \frac{\mu_{o}Is}{2\pi a^{2}} [/tex]

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

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

[tex] => L_{per unit length} = \frac{\mu_{o}}{4 \pi} [/tex]
 
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  • #4
anyone ??
 
  • #5
Your Method #1 calculation looks OK. Show how you calculated the energy density for Method #2.
 
  • #6
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?
 
  • #7
For the energy method

[tex]U = \frac{1}{2\mu _{o}}\int B^{2}dV[/tex]

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

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
[tex] L_{perunitlength} = \frac{\mu _{o}}{8 \pi} [/tex]
 
  • #8
Should there be an I in your integral .
 
  • #9
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.
 
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Related to Calculating Self-Inductance of Long Current-Carrying Wire

1. What is self-inductance?

Self-inductance is the property of a circuit that causes a change in the current flowing through it to induce a voltage in the same circuit.

2. How is self-inductance calculated?

Self-inductance is calculated using the formula L = μ0 * μr * N^2 * A/l, where μ0 is the permeability of free space, μr is the relative permeability of the material surrounding the wire, N is the number of turns in the wire, A is the cross-sectional area of the wire, and l is the length of the wire.

3. What factors affect the self-inductance of a wire?

The self-inductance of a wire is affected by the permeability of the material surrounding the wire, the number of turns in the wire, the cross-sectional area of the wire, and the length of the wire. Additionally, the frequency and current flowing through the wire can also impact its self-inductance.

4. Why is self-inductance important?

Self-inductance is important because it affects the behavior of circuits and can cause voltage spikes and delays in the flow of current. Understanding self-inductance is crucial in designing and analyzing circuits and can help prevent unwanted effects such as electromagnetic interference.

5. How can the self-inductance of a wire be measured?

The self-inductance of a wire can be measured using a variety of methods, such as using an inductance meter, a bridge circuit, or by measuring the voltage and current in the circuit and using the formula L = V/I. It can also be calculated using the dimensions and material properties of the wire.

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