Question

A bar magnet is suspended horizontally by a torsionless wire in the magnetic meridian. In order to deflect the magnet through $30^{\circ}$ from the magnetic meridian, the upper end of the wire has to be rotated by $180^{\circ}$. Now this magnet is replaced by another magnet. If it has to be deflected by $30^{\circ}$ from the magnetic meridian, upper end of the wire has to be rotated by $270^{\circ}$. Ratio of the magnetic moments of two magnets is

• A. $\frac{8}{5}$
• B. $\frac{7}{6}$
• C. $\frac{6}{7}$
• D. $\frac{5}{8}$

Answer 1

Answer: (D)

$M_{1} B \sin 30^{\circ} \propto\left(180^{\circ}-30^{\circ}\right)$
and $ M_{2} B \sin 30^{\circ} \propto\left(270^{\circ}-30^{\circ}\right)$
$\therefore \frac{M_{1}}{M_{2}}=\frac{150^{\circ}}{240^{\circ}}=\frac{5}{8}$