Alternate Answer
The magnetic field, due to wire – 1, at any point on the wire – 2, is directed normal to the direction of current flowin wire – 2.
Expression for force
B21 =Magnetic field atwire – 2, due to a current I1 in
wire $ - 1 = \frac{\mu_{o}\text{I}_{1}}{2\pi\text{r}}$
Force upon conductor carrying current due to magnetic field
$\overrightarrow{\text{F}} = \text{I}(\overrightarrow{\ell}\times\overrightarrow{\text{B}})$
$\therefore\text{ Force, F}_{21}\text{ on a length l of wire} – 2 = \text{I}_{2} . l.\frac{\mu_{o}\text{I}_{1}}{2\pi\text{r}}$
$ =\frac{\mu_{o}\text{I}_{1}\text{I}_{2}}{2\pi\text{r}}l$
$\text{Similarly,} = \text{F}_{12} = \frac{\mu_{0}\text{I}_{1}\text{I}_{2}}{2\pi\text{r}}l$
Nature
The force is repulsive for currents in opposite direction and attractive when currents flow in the same direction.
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