Question

A wire with a weight per unit length of 0.071 N/m is suspended directly above a second wire. The top wire carries a current of 72.8 A, and the bottom wire carries a current of 72.7 A. The permeability of free space is 4π × 10−7 T · m/A . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Answer in units of mm.

Answer 1

d = 15 mm

Step-by-step explanation:

Force of repulsion between two current carrying wire is given by

`F = (\mu_0 i_1 i_2 L)/(2\pi d)`

now this force of repulsion is counterbalanced by the weight of the wire

so we have

`mg = (\mu_0 i_1 i_2 L)/(2\pi d)`

now we have

`d = (\mu_0 i_1 i_2 L)/(2\pi mg)`

so here we can say that

`d = (\mu_0 i_1 i_2)/(2\pi (m/L)g)`

now plug in all values in it

`d = (4\pi * 10^(-7) (72.8)(72.7))/(2\pi (0.071))`

`d = 0.015 m`

d = 15 mm