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.

In the given 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 reason to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.

#### Solution

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,

`vecm = I(l)^2 hatn`

Where, `vecn` is the unit vector along the normal to the surface of the loop.

The attractive force per unit length on the loop is

`F_a = (mu_0)/(2pi) (I_1I)/I`

`F_a = (mu_0)/(2pi)I_1I`

The repulsive force per unit length on the loop is

`F_r/l = (mu _0)/(2pi) (I_1I)/(2l)`

`F_r = (mu _0)/(2pi) (I_1I)/(2l)`

`F_(\text { net}) = F_a - F_t`

= `(mu_0 )/(2pi)I_1I(1-1/2)`

`|F_(text {net}) = (mu_0)/(4pi)I_1I|`

Since the attractive force is greater than the repulsive force, a net force acts on the loop.

The torque on the loop is given as

`tau = vecm xx vecB`

`=mBsintheta`

`=IAB sintheta`

θ = 0° (`because`Area vector is parallel to the magnetic field)

τ = *IAB* sin0°

τ = 0

∴ The torque acting on the loop is zero.