$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three parallel conductors of equal lengths and carry currents…
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three parallel conductors of equal lengths and carry currents I, I and 2I respectively as shown in figure. Distance $A B$ and $B C$ is same as ' $d$ '. If ' $F_1$ ' is the force exerted by $\mathrm{B}$ on $\mathrm{A}$ and $\mathrm{F}_2$ is the force exerted by $\mathrm{C}$ on $\mathrm{A}$, then
$\mathrm{F}_1=\mathrm{F}_2$
$\mathrm{F}_1=-\mathrm{F}_2$
$\mathrm{F}_1=2 \mathrm{~F}_2$
$\mathrm{F}_1=\frac{1}{2} \mathrm{~F}_2$
Solution
Force per unit length exerted by $\mathrm{B}$ on $\mathrm{A}$, $\mathrm{F}_1=\frac{\mu_0(\mathrm{I})(\mathrm{I})}{2 \pi \mathrm{d}}=\frac{\mu_0 \mathrm{I}^2}{2 \pi \mathrm{d}}$ (outside the plane of paper) Force per unit length exerted by $\mathrm{C}$ on $\mathrm{A}$, $\mathrm{F}_2=\frac{\mu_0(\mathrm{I})(2 \mathrm{I})}{2 \pi(2 \mathrm{~d})}=\frac{\mu_0 \mathrm{I}^2}{2 \pi \mathrm{d}}$ (Inside the plane of paper) $\therefore \quad F_1=-F_2$
