Find the inductance of a unit length of two parallel wires, each of radius `a`, whose centers are a distance `d` apart and carry equal currents in opposite direction. Neglect the flux within the wire.

Options

A. `(mu_(0))/(2pi)In((d-a)/(a))`
B. `(mu_(0))/(pi)In((d-a)/(a))`
C. `(3mu_(0))/(pi)In((d-a)/(a))`
D. `(3mu_(0))/(3pi)In((d-a)/(a))`

Answer

b

Explanation

Since the wires are infinite, so the system of these two wires can be considered as a closed rectangle of infinite length and breadth equal to d . Flux through the strip of area `ldr`, due to current follwing in one wire is given by
`phi int_(a)^(d-a)(mu_(0)-I)/(2pir)(ldr)=(mu_(0)Il)/(2pi) ln ((d-a)/(a))`

The other wire produces the same result, so tha total flux through hte dotted rectangle is
`phi_("total")=(mu_(0)lI)/(pi)ln((d-a)/(a))`
The total inductance of length l,
`L=(phi_("total"))/(I)=(mu_(0)I)/(pi)ln((d-a)/(a))`
Inductance per unit length `(L)/(l)=(mu_(0))/(pi)ln((d-a)/(a))`