MCQ
Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will:
- ARemain stationary.
- ✓Move towards the wire.
- CMove away from the wire.
- DRotate about the wire.
✓
Answer
Correct option: B.
Move towards the wire.
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.
$\overrightarrow{\text{F}}_\text{AD}+\overrightarrow{\text{F}}_\text{BC}=0$
$\overrightarrow{\text{F}}_\text{AB}>\overrightarrow{\text{F}}_\text{CD}$
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.
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