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_(1)i_(2))/(d) or T sin theta=(mu_(0))/(4pi).l.(2i^(2))/(d).` from the above equations `tan theta=(mu_(0))/(4pi).l.(2i^(2))/(d).(1)/(mg)` where `theta` is small, `tan theta~~sin theta`
From figure `sin theta=(0.5xx10^(-2))/(75xx10^(-2)),m=40.0xx10^(-3)1kg`
Where `l=` length of conductor in meter Substituting
`(0.5xx10^(-2))/(75xx10^(-2))=10^(-7).I.(2i^(2))/((1.5xx10^(-2)))xx(1)/((40xx10^(-3))1xx9.8)`
Solving, we get `i=14amp`.
As conductors are repelled , the current in them are in opposite direactions.