Inductance and Magnetic Energy of a Straight Wire

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The discussion focuses on deriving the magnetic energy and inductance of a long wire carrying a uniformly distributed current. It is established that the magnetic energy for a length l of the wire is μ_0 i² l / 16π, with the diameter's influence being negligible due to uniform current distribution. The inductance associated with the flux inside the wire is shown to be μ_0 l / 8π. A participant notes an error in their calculation of magnetic energy, indicating a need to account for the non-uniformity of the magnetic field inside the wire. The conversation emphasizes the importance of integrating to accurately determine total energy in such scenarios.
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Homework Statement



A long wire carries a current i uniformly distributed over a cross section of the wire.

Show that the magnetic energy of a length l equals μ_0 i2 l / 16π. Describe why this doesn't depend on diameter.

Show that the inductance for a length of the wire l associated with the flux inside the wire is μ_0 l / 8π

Homework Equations



u_B = B2/2μ_0

U = L i2 /2

Biot-Savart law for Field in a long wire:

B = μ_0 i / 2 π r

The Attempt at a Solution



Solve for u_B as

u_B = B2/2μ_0

= (μ_0 i / 2 π r ) 2 / 2μ_0

Multiply by volume of wire to find U:

U = ( (μ_0 i / 2 π r ) 2 / 2μ_0 ) * π r^2 l

= μ_0 i2 l / 8π

I'm off by a factor 1/2. Any suggestions?
 
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You need to find B inside the wire.

If uB is not uniform inside, integration will be necessary to find the total energy.
 
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