Question

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

• A. $\frac{{\mu }_{0}Ii}{2\pi }$
• B. $\frac{2{\mu }_{0}IL}{3\pi }$
• C. $\frac{{\mu }_{0}IL}{2\pi }$
• D. $\frac{2{\mu }_{0}Ii}{3\pi }$

Answer 1

Answer: (D)

Consider the given figure,

$F_{BA}=\frac{{\mu }_{0}IiL}{2\pi \left(\frac{L}{2}\right)}$
$\Rightarrow F_{CD}=\frac{{\mu }_{0}IiL}{2\pi \left(\frac{3L}{2}\right)}$
Therefore, net force on the loop $ABCD$ will be
$F_{\text{loop }}=F_{BA}-F_{CD}$
$=\frac{{\mu }_{0}IL}{2\pi }\left[\frac{1}{(L/2)}-\frac{1}{(3L/2)}\right]$
$F_{\text{loop }}=\frac{2{\mu }_{0}iI}{3\pi }$