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`
`T sin theta = F =(mu_0)/(4pi) l (2i_1i_2)/(d) " or " T sin theta = (mu_0)/(4pi) l. (2i^2)/(d)`
from the above equation ` tan theta = (mu_0)/(4pi) l. (2i^2)/(d) . (1)/(mg)`
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) kg`
where l = length of conductor in meter Substituting
` (0.5 xx 10^(-2) )/(75 xx 10^(-2) ) = 10^(-7) .I (2i^2)/(1.5 xx 10^(-2) ) xx (1)/(40 xx 10^(-3) 1 xx 9.8)`
Solving, we get i = 14amp.
As conductgors are repelled, the currents in them are in opposite directions.