Question

Two wires $A$ and $B$ are carrying currents $I_1$ and $I_2$ as shown in the figure. The separation between them is $d$. A third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are

• A. $x=\left(\frac{I_2}{I_1+I_2}\right)d$ and $x=\left(\frac{I_2}{I_1-I_2}\right)d$
• B. $x=\left(\frac{I_1}{I_1-I_2}\right)d$ and $x=\left(\frac{I_2}{I_1+I_2}\right)d$
• C. $x=\left(\frac{I_1}{I_1+I_2}\right)d$ and $x=\left(\frac{I_2}{I_1-I_2}\right)d$
• D. $x=\pm \left(\frac{I_{1}d}{I_1-I_2}\right)$

Answer 1

The correct option is D $x=\pm \left(\frac{I_{1}d}{I_1-I_2}\right)$

Case-1: When the wire-$C$ is placed between the two wires,

Net force on the third wire, carrying current $I$ is given by,

Using Right-hand thumb rule, direction of $B$ at the location of wire-$C$ due to wire-$A$ and wire-$B$ are in the same direction ie inwards.

$F_{13}+F_{23}=0$

$\therefore \frac{{\mu }_0{I}_{1}I}{2\pi x}+\frac{{\mu }_0{I}_{2}I}{2\pi (d-x)}=0$

$\Rightarrow \frac{I_1}{x}+\frac{I_2}{d-x}=0$

$\Rightarrow \frac{I_1}{x}=\frac{I_2}{x-d}$ or

or, $(x-d)I_1=x{I}_2$

$\Rightarrow x(I_1-I_2)=d{I}_1$

$\Rightarrow x=\frac{I_1}{(I_1-I_2)}d.......\left(1\right)$

Case-2: When the wire-$C$ is placed left of wire-$A$


Using Right-hand thumb rule, directions of $B$ at the location of wire-$C$ due to wires $A$ and $B$ will be opposite, so net force will be given by,

$\frac{{\mu }_0{I}_{1}I}{2\pi x}-\frac{{\mu }_0{I}_{2}I}{2\pi (d+x)}$

$\frac{I_1}{x}-\frac{I_2}{d+x}=0$

$\Rightarrow \frac{I_1}{x}=\frac{I_2}{d+x}$

$\Rightarrow (d+x)I_1=x{I}_2$

$\Rightarrow (I_2-I_1)x=d{I}_1$

$\Rightarrow x=-\frac{I_1}{(I_1-I_2)}d.....\left(2\right)$

From Eqs. $\left(1\right)$ and $\left(2\right)$, it is clear that

$x=\pm \frac{I_1}{(I_1-I_2)}d$

Therefore, option $\left(D\right)$ is correct.