Magnetic energy density & inductance of a current carrying wire

[physicsidiot1]

Homework Statement

A long straight wire of radius r=a carries a current I uniformly distributed over its cross section. Show that the magnetic energy per unit length (energy density u=B^2/2mu) is given by $$\mi_{o}\ =\ 4\pi\ \timesI^2\ 10^{-7}$$/16pi. What is the contribution that a single long wire makes to the inductance per unit length if we consider only the internal magnetic energy?

Homework Equations

U=.5$$\mi_{o}\ =\ 4\pi\ \timesl*A*I^2\ 10^{-7}$$
u=B^2/$$\mi_{o}\ =\ 4\pi\ \times\ 10^{-7}$$

The Attempt at a Solution

I did the obvious and plugged in B=$$\mi_{o}\ =\ 4\pi\ \timesI\ 10^{-7}$$/2piR but I ended up with a result that was off by 2 and the radius term was still there. I cannot figure out where I am going wrong. As for the second part of the question concerning the contribution to inductance, I am completely lost.

 

[anathema]

Any help would be appreciated.

Regarding the first part of your question, it seems like you have the right approach but there may be a small error in your calculation. The correct expression for magnetic energy per unit length is u=B^2/(2\mi_{o}\ ), so the final answer should be u=\mi_{o}\ =\ 4\pi\ \timesI^2\ 10^{-7}/16pi. As for the second part of the question, the contribution of a single long wire to inductance per unit length can be calculated using the formula L=\mi_{o}\ =\ 4\pi\ \timesN^2A\ 10^{-7}/l, where N is the number of turns per unit length and A is the cross-sectional area of the wire. Since we are only considering the internal magnetic energy, we can assume that N=1 and A=\pi*r^2, where r is the radius of the wire. This gives us an inductance per unit length of L=\mi_{o}\ =\ 4\pi\ \times\pi*r^2\ 10^{-7}/l. I hope this helps! Let me know if you have any further questions.