Q. A bar magnet of magnetic moment M is divided into ‘n’ equal parts cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength 2 T and held making an angle 60° with the direction of the field. When the magnet is released the K.E. of the magnet in the equilibrium position is

(a) M/n

(b) M n

(c) M/n^{2}

(d) Mn^{2}

Ans: (a)

Sol: Magnetic moment of each part $ \displaystyle M’ = \frac{M}{n} $

Initial P.E $ \displaystyle U_i = – M’ B cos60 $

$\displaystyle U_i = – \frac{M}{n}\times 2 \times \frac{1}{2} $

$ \displaystyle U_i = – \frac{M}{n}$

Final P.E $ \displaystyle U_f = – M’ B cos0 $

$ \displaystyle U_f = – \frac{M}{n}\times 2 \times 1 $

$ \displaystyle U_f = – \frac{2 M}{n}$

Applying Energy conservation;

Loss in P.E = Gain in K.E

$ \displaystyle U_i – U_f = K.E_f – K.E_i $

$ \displaystyle – \frac{M}{n} – (- \frac{2 M}{n}) = K.E_f – 0 $

$ \displaystyle K.E_f = \frac{M}{n} $