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
`Tcostheta=mg`

`Tsintheta=F=(mu_(0))/(4pi)l.(2i_(1)i_(2))/d` or `Tsintheta=(mu_(0))/(4pi)l.(2i^(2))/d`
from the above equations `tantheta=(mu_(0))/(4pi).1.(2i^(2))/d.(1)/(mg)`
where `theta` is small, `tantheta~~sintheta`
From figure `sinteta=(0.5xx10^(-2))/(75xx10^(-2))`
`m=40.0xx10^(-3)` 1kg
where l = length of conductor in meter
Substituting in eq. (), we get `(0.5xx10^(-2))/(75xx10^(-2))`=
`10^(-7).1.(2i^(2))/((1.5xx10^(-2)))xx1/((40xx10^(-3))1xx9.8)`
Solving, we get i = 14 amp.
As conductors are repelled, the currents in them are in opposite directions .