(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 a steady current I is `m=IA`, where `A=l^2`.
(ii)
The currents in AB and EF are flowing in the same direction, So, Ab will be attracted towards EF with a force , say `F_1`.
`therefore F_1=(mu_0) /(2pi)(II_1)/(l)xx "lenghth of AB"`.
The currents in CD and EF are flowing in unlike directions. So, CD would experience a repulsive force, say `F_2`.
`F_1=(mu_0)/(2pi)(II_1)/(2l)xx "length of CD"`.
The forces on the portions BC and DA will cancel out each other's effect.
`therefore` Net force `=F_1-F_2`
`=(mu_0) /(2pi).(II_1)/(l)xx " length of AB" -(mu_0)/(2pi).(II_1)/(2l)xx "length of CD" `.
`=(mu_0)/(2pi).(II_1)/(l)["length of AB" -1//2 "length of CD "]`.