Question

Prove that the inductance of parallel wires of length l in the same circuit is given by L = `((mu_0l)/pi) "in" (d/a)` where 'a' is the radius of wire and d is seperation between wire axes.

Answer 1

Using Ampere's law,

`B = (mu_0I)/(2pir)`

dΦ = BdA = `(mu_0I)/(2pir)ldr`

 

Total flux is two times flux generated by one wires, (due to symmetry)

Φ_total = `2int_a^(d-a)BdA = 2int_a^(d-a)((mu_0I)/(2pir))ldr`

`=(mu_0Il)/pi int _a^(d-a)(dr)/r`

`=(mu_0Il)/pi[lnr]_a^(d-a)`

`=(mu_0Il)/piln[(d-a)/a]`

If a < < d, then d - a ≈ d

`L= (mu_0l)/piln(d/a)`