(i) Write the expression for the magnetic moment `(vecm)` due to a planar square loop of side 'l' carrying a steady current I in a vector form.
(ii) In figure., this loop is placed in a horizontal plane near a long straight conductor carrying a steady current `I_1` at a distance l as shown. Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.

Explanation

(i) The magnetic moment `(vecm)` due to a planar square loop of side l carrying current I is
`vecm=IvecA`, where `A=l^2`
(ii) The currents in AB and EF are in the same direction. So AB will be attracted towards EF with a foce `F_1`, given by
`F_1=(mu_0)/(4pi)(2II_1)/(l)xx(l eng th AB)=(mu_0)/(4pi)(2II_1)/(l)l=(mu_0)/(4pi)2II_1`.
The currents in CD and EF are in opposite diretions. So CD will experience repulsion away from EF, with a force `F_2`, given by
`F_2=(mu_0)/(4pi)(2II_1)/(2l)(l eng th CD)=(mu_0)/(4pi)(2II_1)/(2l)l=(mu_0)/(4pi)II_1`
The forces on the portions BC and DA will cancel out each other's effect. Therefore, net force on loop is
`=F_1-F_2=(mu_0)/(4pi)2II_1-(mu_0)/(4pi)II_1=(mu_0)/(4pi)II_1`
towards EF
As the area vector is parallel to the magnetic field, `theta=0^@`, so torque on the square loop, `tau=IAB sin theta=I AB sin 0^@=0`. Also `F_1!=F_2`, so no torque acts on the loop.