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

Answer: (A)

Here the sides AB and CD will contribute the force but the sides AD and BC will not because they are perpendicular to wire XY.

We know that the magnetic field at distance r from long wire is $B=\frac{{\mu }_{0}I}{2\pi r}$

and also force on a current (i) flow through line section of legth L due to B is $F=BiL$

Using these, the force on AB is $F_1=\left(\frac{{\mu }_{0}I}{2\pi L/2}\right)iL=\frac{{\mu }_{0}Ii}{\pi }$ and

the force on CD is $F_2=\left(\frac{{\mu }_{0}I}{2\pi (L+L/2)}\right)iL=\frac{{\mu }_{0}Ii}{3\pi }$

As the current directions on AB and CD are opposite so the forces will be opposite.

Thus, net forec $F=F_1-F_2=\frac{{\mu }_{0}Ii}{\pi }-\frac{{\mu }_{0}Ii}{3\pi }=\frac{2{\mu }_{0}Ii}{3\pi }$