Derive an expression for the torque acting on a current carrying loop suspended in a uniform magnetic field.
According to Fleming's left hand rule, the magnetic forces on sides PS and QR are equal, opposite and collinear (along the axis of the loop), so their resultant is zero.
The side PQ experiences a normal inward force equal to IbB while the side RS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by
The direction of the torque ττ is such that it rotates the loop clockwise about the axis of suspension.