Two infinitely long straight wires A1 and A2 carrying currents I and 2I flowing in the same directions are kept ‘d’ distance apart. Where should a third straight wire A_{3} carrying current 1.5 I be placed between A_{1} and A_{2} so that it experiences no net force due to A_{1} and A_{2}  ? Does the net force acting on A_{3}  depend on the current flowing through it ?

Answers (1)

As we know the force per unit length on two infinitely current-carrying wires is given by \frac{\mu _{o}I_{a}I_{b}}{2\pi d} Where I_{a} is current carrying by wire A  and I_{b} by wire A_{2} and d is the distance between parallel wires.

Now, force per unit length on wire A_{3} due to A_{1}=\frac{\mu _{o}I\times 1.5I}{2\pi x}

In -ve x-direction.

and force per unit length on wire A_{3} due to A_{2}.

=\frac{\mu _{o}2I\times 1.5I}{2\pi (d-x)}

In +ve x- direction.

So, for net force on A_{3}, both the equations should be equal.

therefore, \frac{\mu _{o}I\times 1.5I}{2\pi x} =\frac{\mu _{o}2I\times 1.5I}{2\pi (d-x)}

\Rightarrow 2x=d-x

3x=d

\therefore if x=d/3 the net force on A_{3} will be zero.

Net force depends on the current flowing.

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