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A square loop $ABCD$ carrying a current $i,$ is placed near and coplaner with a long straight conductor $XY$ carrying a current $I,$ the net force on the loop will be -

Option: 1
$\frac{2 \mu_0 I i}{3 \pi}$

Option: 2
$\frac{ \mu_0 I i}{2 \pi}$

Option: 3
$\frac{2 \mu_0 I iL}{3 \pi}$

Option: 4
$\frac{ \mu_0 I iL}{2 \pi}$

Answers (1)

best_answer

The direction of current is conductor $XY$ and $AB$ is same.

$\therefore F_{A B}=I l B \text { (attractive) }$

$F_{A B}=i(L) \cdot \frac{\mu_0 I}{2 \pi(L / 2)} (\leftarrow) =\frac{\mu_0 I i}{\pi} (\leftarrow )$

$F_{BC}$ opposite to $F_{AB}$

$F_{BC}(\uparrow)$ and $F_{AD}(\downarrow)\rightarrow$ cancels each other

$\begin{aligned} & F_{C D}=i \ell B(\text { repusilive) } \\ & F_{C D}=i(L) \cdot \frac{\mu_0 I}{2 \pi\left(\frac{3 L}{2}\right)}(\rightarrow)=\frac{\mu_0 I L}{3 \pi}(\rightarrow) \end{aligned}$ $\begin{aligned} & F_{C D}=i \ell B(\text { repusilive) } \\ & F_{C D}=i(L) \cdot \frac{\mu_0 I}{2 \pi\left(\frac{3 L}{2}\right)}(\rightarrow)=\frac{\mu_0 I L}{3 \pi}(\rightarrow) \end{aligned}$

Therefore, the net force on the loop.

$\begin{aligned} F_{\text {net }} & =F_{A B}+F_{B C}+F_{C D}+F_{A B} \\ & =\frac{\mu_0 i I}{\pi}-\frac{\mu_0 i I}{3 \pi}=\frac{2 \mu_0 i I}{3 \pi} \end{aligned}$

Posted by

Pankaj Sanodiya

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