Question

Two identical long wires of radius a = 2.80 mm are parallel and carry identical currents of i = 5.00 A in opposite directions. Their center-to-center separation is W = 19.0 cm. Neglect the flux within the wires but consider the flux in the region between the wires. What is the inductance per unit length of the wires?

Answer 1

Inductance per unit length,
`(L)/(l) = 7.02 * 10^(-7) H/m`

Step-by-step explanation:

Radius of the wire, a = 2.80 mm

Currents carried by each of the wires, i = 5.00 A

Center-to-Center Separation, W = 19.0 cm

The flux in the wires is given by the equation, ∅ = Li

The Net flux of the region between the wires is given by the equation:

`\phi = (l \mu_0 i)/(\pi) ln((W-a)/(a))`

Divide both sides by l to get the net flux per unit length

`\phi/l = ( \mu_0 i)/(\pi) ln((W-a)/(a))\\\phi/l = ( 4 \pi * 10^(-7) i)/(\pi) ln((0.019-0.0028)/(0.0028))\\\phi/l = 7.02 * 10^(-7) i\\(Li)/(l) = 7.02 * 10^(-7) i\\(L)/(l) = 7.02 * 10^(-7) H/m`