The frequency of oscillation of current in the indcutor is

Explanation

See figure. The magnetic field B in the space between the two conductors is given by
`B=(mu_(0)i)/(2pir)`
wherei = current in either of two conductors
r=distance of the point from the axis.Because`ointB^(to).dl^(to)=-mu_(0)xx` current enclosed by the path (Ampere.slaw)
`B^(to).2pir=mu__(0)i`
The energy density in the space between the conductors
`mu=(B^(2)/(2mu_(0)=(1)/(2mu-(0)[(mu_(0)i)/(2pir]^(2)=(mu__(0)i^(2)/8pi^(2)r^(2)(joule)/( metre^(3))`
Consider a volume element dV in the form of a cylindrical shell of radii r and (r+dr) as shown in the figure.Energy `aW=mudV=(mu_(0)/(8pi^(2)r^(2)xx2pirldr=(mu_(0)i^(2)l)/(4pi)((dr)/(r))`
Total magnetic energy can be obtained by integrating this expression between the limits r=a tor= b. Hence`W=intdW=(mu_(0)i^(2)l)/(4pi)(int_(a)^(b)((dr)/(r))=(mu__(0)i^(2)l)/(4pi)log_(e)((b)/(a))`
If L be the self inductance of length l of the cable, then the energy in magnetic field will be `(1/2)Li^(2)`.Hence
`(1)//(2)Li^(2)=(mu__(0)i^(2)l)/(4pi)log_(e)((b)/(a))`
`L=(mu_(0)l)/(2pi0log__(e)((b)/(a))`