Question

(a) Derive the expression for the torque on a rectangular current carrying lop suspend 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.

Answer 1

a)

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 $\overrightarrow{B}$ 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 $\overrightarrow{\tau }$
$\tau =\overrightarrow{m}\times \overrightarrow{B}$
The direction of the torque $\tau$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