(a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. (b) A proton and a deuteron having equal momenta enter in a region of a uniform magnetic field at right angle to the direction of a field. Depict their trajectories in the field.

a)Let I = current flowing through the coil PQRS
a, b = sides of the coil PQRS
A = ab = area of the coil
θ = angle between the direction of →B and normal to the plane of the coil.
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
τ = Force × perpendicular distance
= IbB × a sin θ
= IBA sin θ
If the rectangular loop has N turns, the torque increases N times i.e.,
τ = NIBA sin θ
But NIA = m, the magnetic moment of the loop, so
τ = mB sin θ
In vector notation, the torque →τ = →m × →B
The direction of the torque τ is such that it rotates the loop clockwise about the axis of suspension.
b)Deuteron has greater mass than a proton so it will experience more deflection than a proton. but since the magnetic field is perpendicular to the velocity of both the particles so both will acquire circular trajectory.. Deuteron will have more radius. Because R is directly proportional to Mass