Question

The given figure shows three long straight wires carrying identical current $i$ and kept parallel at equally spaced distance. The direction of current in each wire has been shown in figure.The magnetic force experienced by the wire $a$, $b$ and $c$ are $F_a$, $F_b$ and $F_c$ respectively. The forces experienced by the wire will have the order of strength as:

• A. $F_a>F_b>F_c$
• B. $F_b>F_c>F_a$
• C. $F_c>F_a>F_b$
• D. $F_b>F_a>F_c$

Answer 1

The correct option is B $F_b>F_c>F_a$

The direction of current in wires $a$ and $b$ is in same direction hence they will attract each other. Let the magnitude of force of attraction b/w $a$ & $b$ be $F_1$

Again, wire $b$ and $c$ will repel each other due to current being in opposite direction. Let the magnitude of force of repulsion be $F_2$

Similarly, let the wire $a$ and $c$ repel each other with a force with magnitude $F_3$

The separation b/w each pair of wire is given by,

$d_{ac}=2d_{ab}=2d$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi x}$

$(\because d_{ac}>d_{ab})$

Thus $F_3<F_2$
The magnitude of net force on respective wire is,

$F_a=F_1-F_3$

$F_b=F_1+F_2$

$F_c=F_2+F_3$

Here the value of $(F_1+F_2)>(F_2+F_3)$, because $|F_2|=|F_1|$ and $|F_2|>|F_3|$

Thus order of forces:

$F_b>F_c>F_a$

Hence, option $\left(B\right)$ is the correct answer.