Force between two current carrying wires( wire 1 and wire 2)

\(\displaystyle{F}_{{{12}}}={\frac{{{K}{i}_{{1}}{i}_{{2}}}}{{{2}\pi{r}}}}\)

as both the wires carrying currents in opposite directions thenature of the fore on 2nd due to 1st is repulsive. hence to nullify the net force acting on the second conductorthird conductor should carry current in upward direction.

Then again 2nd and 3rd conductors carrying currents inopposite directions. hence again 2nd capacitor will experiencerepusive force.

If both the forces are equal then net force acting on thesecond is zero.

\(\displaystyle{F}_{{{23}}}={\frac{{{K}{i}_{{2}}{i}_{{3}}}}{{{2}\pi{x}}}}\)

x is distance

given net force should be zero, hence \(\displaystyle{F}_{{{12}}}={F}_{{{23}}}\)

\(\displaystyle{\frac{{{i}_{{1}}{i}_{{2}}}}{{{2}\pi{r}}}}={\frac{{{i}_{{2}}{i}_{{3}}}}{{{2}\pi{x}}}}\)

\(\displaystyle{i}_{{1}}={1.50}{A};{i}_{{2}}={3.40}{A};{r}={0.16}{m}\)

similarly calculate the net force on the third due to 1st and2nd and following the same procedure we get

\(\displaystyle{\frac{{{i}_{{1}}{i}_{{3}}}}{{{\left({16}+{x}\right)}\times{10}^{{-{12}}}}}}={\frac{{{i}_{{2}}{i}_{{3}}}}{{{x}}}}\)

so we have two unknowns and two equations try to calculatex and \(\displaystyle{i}_{{3}}\); the directions of current \(\displaystyle{i}_{{3}}\) is upwards as mentioned above

\(\displaystyle{F}_{{{12}}}={\frac{{{K}{i}_{{1}}{i}_{{2}}}}{{{2}\pi{r}}}}\)

as both the wires carrying currents in opposite directions thenature of the fore on 2nd due to 1st is repulsive. hence to nullify the net force acting on the second conductorthird conductor should carry current in upward direction.

Then again 2nd and 3rd conductors carrying currents inopposite directions. hence again 2nd capacitor will experiencerepusive force.

If both the forces are equal then net force acting on thesecond is zero.

\(\displaystyle{F}_{{{23}}}={\frac{{{K}{i}_{{2}}{i}_{{3}}}}{{{2}\pi{x}}}}\)

x is distance

given net force should be zero, hence \(\displaystyle{F}_{{{12}}}={F}_{{{23}}}\)

\(\displaystyle{\frac{{{i}_{{1}}{i}_{{2}}}}{{{2}\pi{r}}}}={\frac{{{i}_{{2}}{i}_{{3}}}}{{{2}\pi{x}}}}\)

\(\displaystyle{i}_{{1}}={1.50}{A};{i}_{{2}}={3.40}{A};{r}={0.16}{m}\)

similarly calculate the net force on the third due to 1st and2nd and following the same procedure we get

\(\displaystyle{\frac{{{i}_{{1}}{i}_{{3}}}}{{{\left({16}+{x}\right)}\times{10}^{{-{12}}}}}}={\frac{{{i}_{{2}}{i}_{{3}}}}{{{x}}}}\)

so we have two unknowns and two equations try to calculatex and \(\displaystyle{i}_{{3}}\); the directions of current \(\displaystyle{i}_{{3}}\) is upwards as mentioned above