Question

Two long parallel wires are separated by 15 cm. One of the wires carries a current of 34 A and the other carries a current of 69 A. The permeabilty of free space is 1.257 × 10−6 N · m/A. Determine the magnitude of the magnetic force on a 5.9 m length of the wire carrying the greater current. Answer in units of mN

Answer 1

F = 0.018 N

Step-by-step explanation:

Magnetic force between two parallel current carrying wires is given by

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

here we know that

`i_1 = 34 A`

`i_2 = 69 A`

d = 15 cm

L = 5.9 m

now from above formula we can say

`F = ((4\pi * 10^(-7))(34 A)(69 A)5.9)/(2\pi (0.15))`

now the force between two wires is given as

`F = 0.018 N`