Question

Three long straight conductors are arranged parallel to each other in the same plane and carry currents of $1$$A$, $2$$A$ and $3$$A$ all in the same direction. The distance between the first two conductors is $x$ and the distance between the second and third conductors is $y$. If the middle conductor is in equilibrium, then the ratio $x:y$ is (assume gravitation force is negligible):

• A. $1:3$
• B. $3:1$
• C. $1:\sqrt{3}$
• D. $\sqrt{3}:1$

Answer 1

Answer: (A)

$1:3$

Force on $C_2$ due to $C_1$

$F_{21}=\frac{{\mu }_0{i}_1\cdot i_2}{2\pi r}dl$

$=\frac{{\mu }_{0}2\times 1\times dl}{2\pi x}$

Force on $C_2$ due to $C_3$

$F_{23}=\frac{{\mu }_0\times 2\times 3}{2\pi y}dl$

$C_2$ is in equilibrium

So, $=F_{21}=F_{23}$

$\frac{{\mu }_0\times 2\times 1\times dl}{2\pi x}=\frac{{\mu }_0\times 2\times 3\times dl}{2\pi y}$

$x:y=1:3$