A long wire carries a current of uniform density. Let `i` be the total current carried by the wire. Show that the magnetic energy per unit length stored within the wire equals `(mu_(0)t^(2))/(16pi)` . Note that it does not depend on the wire diameter.

Explanation

Consider a ring of radius `r` and thickness `dr` . `dA=2pirdr`
`B=(mu_(0)ir)/(2piR^(2))`
`u(B^(2))/(2mu_(0))=(mu_(0)^(2)t^(2)r^(2))/(2mu_(0).4pi^(2)R^(4))=(mu_(0)i^(2)r^(2))/(8pi^(2)R^(4))`
Consider unit length of wire
`dV=dAxx1=2pirdr`
`dU=udV=(mu_(0)i^(2)r^(2))/(8pipi^(2)R^(4)).2pirdr`
`=(mu_(0)i^(2)r^(3)dr)/(4piR^(4))`
`U=(mu_(0)i^(2))/(4piR^(4))int_(0)^(R)r^(3)dr=(mu_(0)i^(2))/(16pi)`