Question

derive expression for the torque on the loop

Answer 1

According to Fleming's left hand rule, the magnetic forces on sides and are equal, opposite and collinear (along the axis of the loop), so their resultant is zero. The direction of the torque is such that it rotates the loop clockwise about the axis of suspension.

Answer 2

Consider a rectangular coil PQRSPQRS suspended in a uniform magnetic field B→,B→, with its axis perpendicular to the field.  

Figure (a) A rectangular loop PQRSPQRS in a uniform magnetic field B→B→.  
(b) Top view of the loop, magnetic dipole moment m→m→ is shown.  

Let II = current flowing through the coil PQRSPQRS 
a,ba,b = sides of the coil PQRSPQRS 
A=abA=ab = area of the coil 
θθ = angle between the direction of B→B→ and normal to the plane of the coil. 

According to Fleming's left hand rule, the magnetic forces on sides PSPS and QRQR are equal, opposite and collinear (along the axis of the loop), so their resultant is zero. 

The side PQPQ experiences a normal inward force equal to IbBIbB while the side RSRS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by 

τ=τ= Force ×× perpendicular distance 
  =IbB×asinθ=IbB×asin⁡θ 
  =IBAsinθ=IBAsin⁡θ 

If the rectangular loop has NN turns, the torque increases NN times i.e., 
                       τ=τ= NIBAsinθNIBAsin⁡θ 
But NIA=m,NIA=m, the magnetic moment of the loop, so  
                     τ=mBsinθτ=mBsin⁡θ 
In vector notation, the torque τ→τ→ is given by  
               τ→=m→×B→τ→=m→×B→ 

The direction of the torque ττ is such that it rotates the loop clockwise about the axis of suspension.