Electromagnetism – Deriving the Self-Inductance of a Long Wire

electromagnetismhomework-and-exercisesinductance

Currently I am stuck, trying to derive the self-inductance of a long wire. According to literature it should be

$$L=\frac{\mu_r\mu_0l}{8\pi}$$

and in literature its derived by looking at the energy of the magnetic field. I tried to derive this formula via the magnetic flux and I am getting $4\pi$ instead of $8\pi$. This are my considerations:

   _ _
 /  |R \
|   '   | a wire with radius R and length l
 \ _ _ /

The magnetic flux density $B(r)$ is given according to Ampere's law:

$$B(r) 2\pi r=\mu_0\mu_r\frac{r^2}{R^2}I$$

$$B(r)=\mu_0\mu_r\frac{r}{2\pi R^2}I$$

where $r$ is the distance from the center of the wire, $R$ is its radius and $I$ is the total current through the wire. Now I know that the magnetic flux $\phi$ through the upper part of a longitudinal section is

$$\phi = \int_A B dA = \int_0^R B(r)l dr = \frac{\mu_0\mu_rIl}{4\pi}$$

where $l$ is the wire's length. No I use $\phi = LI$ and arrive at $4\pi$.

What am I doing wrong? Where is the mistake in my considerations?

Moreover I have the following problem. If I look at an entire longitudinal section of the wire and not only at its upper half the magnetic flux is zero:

   _ _
 /  |  \
|   |2r | => Magnetic flux is zero (the magnetic field 
 \ _|_ /     penetrating the upper half of the longitudinal cross section is 
             exactly opposite to the magnetic field penetrating the lower
             half)

Hopefully I formulated my problem clear enough. If not please ask me for further details.

Best Answer

The original question talked about a discrepancy between the result obtained by calculating the flux directly and using the definition $L= \Phi/I$. The confusion arises because of a concept known as "flux linkage". When you calculate the flux enclosed by the region of unit length between r and r+dr, you calculated an expression for flux which you integrated to get the total flux. However, the entire flux calculated by you is not "linked" to this area since the current enclosed by the contour of this radius is a fraction $\pi r^2/R^2$ of the total current. Thus the linked flux is $d\Phi= \dfrac{\mu_0 I r dr}{2\pi R^2} \dfrac{\pi r^2}{\pi R^2}$. If you integrate this expression, you would get the correct result. $$\Phi = \dfrac{\mu_0 I}{2\pi R^4}\int_0^R r^3 dr= \mu_0 I/8\pi$$ "Flux linkage' is not a very easy concept but consider what happens when you have N turns of the wire through which the same flux passes. In order to calculate emf using Faraday's law, you will need N times the flux to get the correct emf.

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