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

Answer 1

The correct option is A $\frac{2{\mu }_{0}Ii}{3\pi }$


The direction of currents in a long straight conductor XY and arm AB of a square loop ABCD are in the same direction.
So, there exists a force of attraction between the two conductors which is given by $F_{AB}$ as
$F_{AB}=BiL=\frac{{\mu }_{0}IiL}{2\pi \left(\frac{L}{2}\right)}$ $(\because B_{AB}=\frac{{\mu }_{0}I}{2\pi \left(\frac{L}{2}\right)})$

The direction of currents in a long straight conductor XY and arm CD, the direction of currents are in the opposite direction. There exists a force of repulsion which will be experience by CD as
$F_{CD}=BiL=\frac{{\mu }_{0}IiL}{2\pi \left(\frac{3L}{2}\right)}$$(\because B_{CD}=\frac{{\mu }_{0}I}{2\pi \left(\frac{3L}{2}\right)})$

Now, As arms BC and CD are perpendicular to straight conductor XY and carry current in opposite directions, net force on both of them will cancel each other.

Net force on the loop ABCD is
$F_{loop}=F_{BA}-F_{CD}=\frac{{\mu }_{0}IiL}{2\pi }(\frac{1}{\left(\frac{L}{2}\right)}-\frac{1}{\left(\frac{3L}{2}\right)})$
$F_{loop}=\frac{2{\mu }_{0}Ii}{3\pi }$