Sol.

Fig.

Due to flow of current through the cable, shown in Fig. , magnetic field is established in the space between cylinders. Energy is stored in this space due to magnetic field.

Since, energy stored per unit volume in magnetic field of induction B is equal to B 2 /2µ 0 , therefore, magnetic induction in the space must be known. But magnetic field in the space is not uniform.

Hence, consider a thin cylindrical coaxial shell in the space. Let radius of the shell be x and radial thickness dx as shown in Fig .

Fig.

According to Ampere’s circuital law, magnetic induction B is given by B 2 π x = µ 0 I

or B =

∴ Energy stored per unit length of the shell considered is

dU = (2 π x dx) = dx

∴ Energy stored per unit length of the cable,

U = = dx = log e 2

Since, energy stored in magnetic field is U = LI 2 where L is self inductance, therefore, inductance per unit length of the cable is

L = = log e 2.