Consider the situation shown in figure. The straight wire is fixe…

MCQ

Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will:

A
Remain stationary.
✓
Move towards the wire.
C
Move away from the wire.
D
Rotate about the wire.

✓

Answer

Correct option: B.

Move towards the wire.

$\overrightarrow{\text{F}}_\text{AD}+\overrightarrow{\text{F}}_\text{BC}=0$
$\overrightarrow{\text{F}}_\text{AB}>\overrightarrow{\text{F}}_\text{CD}$
Force acting on the wire per unit length carrying current $i_2$ due to the wire carrying current $i_1$ placed at a distance $d$ is given by,
$\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}$
So, forces per unit length acting on sides $AB$ and $CD$ are as follows:
$\text{F}_\text{AB}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}} ($Towards the wire$)$
$\text{F}_\text{CD}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi(\text{d}+\text{a})} ($Away from the wire$)$
Here, $F_{AB} > F_{CD}$ because force is inversly proportional to the distance from the wire and wire $AB$ is closer to the wire carrying current $i_1$.
The forces per unit length acting on sides $BC$ and $DA$ will be equal and opposite, as they are equally away from the wire carrying current $i_1$, with current $i_2$ flowing in the opposite direction.
$\therefore\text{F}_\text{BC}=-\text{F}_\text{DA}$
Now,
Net force:
$\text{F} = \text{F}_\text{AB}+\text{F}_\text{BC}+\text{F}_\text{CD}+\text{F}_\text{DA}$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}+\text{F}_\text{BC}-\frac{\mu_0\text{i}_1\text{i}_2}{2\pi(\text{d+a})}-\text{F}_\text{BC}$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi}\Big(\frac{1}{\text{d}}-\frac{1}{\text{d+a}}\Big)$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2\text{a}}{2\pi\text{d}(\text{d+a})}$
$($Towards the wire$)$
Therefore, the loop will move towards the wire.