Answer 1

The plane of the loop is not along the magnetic field but makes an angle with it. 

Let the dimension of the rectangular coil ABCD, be AB × BC = a × b
The angle between the field and the normal is θ.  
Forces on BC and DA are equal and opposite and they cancel each other as they are collinear.  
Force on AB is F₁ and force on CD is F₂.

F₁ = F₂ = IbB

The magnitude of the torque on the loop as in the figure:

`tau = F_1 a/2 sinθ + F_2 a/2 sinθ`

= IabBsinθ

If there are ‘n’ such turns the torque will be sin nIab sinθ

The magnetic moment of the current m = IA

`vectau = vecm xx vecB`