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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.

Posted by

Safeer PP

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