Two parallel horizontal conductors are suspended by two light vertical threads each 75 cm long. Each conductor has a mass of `40gm`, and when there is no current they are 0.5 cm apart. Equal current in the two wires result in a separation of 1.5 cm. Find the values and directions of currents. Take `g=9.8ms^-2`.

Explanation

The situation is shown in figure,
Here, we have
` T cos theta = mg (i)`
`T sin theta = F = (mu_0)/(4 pi) l (2i_1 i_2)/(d) or `
`T sin theta = (mu_0)/(4pi) l (2i^2)/(d)....(ii)`
from eqs. (i) and (ii)
`tan theta = (mu_0)/(4pi) l . (2i^2)/(d). (1)/(mg)....(iii)`

where `theta` is small, `tan theta ~~ sin theta`
From figure `sin theta = (0.5 xx 10^(-2))/(75 xx 10^(-2))`
`m = 40.0 xx 10^(-3) l kg`
where l = length of conductor in meter Substituting in eq. (iii). we get
`(0.5 xx 10^(-2))/(75 xx 10^(-2)) = 10^(-7)l (2i^2)/((1.5 xx 10^(-2))) xx(1)/((40xx 10^(-3))lxx 9.8)`
Solving , we get `i = 14` amp.
For repulsion , the currents are in opposite direction.