(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

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 is given as
`vecm=IvecA`
Therefore, `m=I(l)^(2)hatn`
Where, `hatn` is the unit vector along the normal to the surface of the loop. The attractive force on the loop is
`F_(a)=(mu_(0))/(2pi)I_(1)I`
The repulsive force on the loop is
`F_(1)=(mu_(0))/(2pi)(I_(1)I)/(2l)`
`F_("net")=F_(a)-F_(1)` = `(mu_(0))/(2pi)I_(1)I(1-1/2)`
`|F_("net")=(mu_(0)/(4pi)I_(1)I)|`
Since the attractive force is greater than the repulsive forrce, a net force acts on the loop. The torque on the loop is given as
`tau=vecmxxvecB=mB sin theta=IAB sin theta`
`theta=0^(@)` (`beacause` Area vector is parallel to the magnetic field)
`tau=IAB sin 0^(@) , tau=0`
`therefore` The torque acting on the loop is zero.