Question

A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of 33 A and the bottom wire carries a current of 49 A . The permeablity of free space is 1.25664 × 10−6 N/A 2 . 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

Step-by-step explanation:

Magnetic field near current carrying wire

=
`(\mu_0)/(4\pi) (2i)/(r)`

i is current , r is distance from wire

B = 10⁻⁷ x
`(2*49)/(r)`

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x
`(2*49)/(r)` x 33 x 1

= 3234 x
`(10^(-7))/(r)`

This should balance weight of second wire per unit length

3234 x
`(10^(-7))/(r)` = .075

r =
`(3234)/(.075)` x 10⁻⁷

= .0043 m

= .43 cm .