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

Question

A bar magnet is suspended horizontally by a torsionless wire in the magnetic meridian. In order to deflect the magnet through 30° from the magnetic meridian, the upper end of the wire has to be rotated by 180°. Now this magnet is replaced by another magnet. If it has to be deflected by 30° from the magnetic meridian, upper end of the wire has to be rotated by 270°. Ratio of the magnetic moments of two magnets is (A) (8/5) (B) (7/6) (C) (6/7) (D) (5/8)

Tardigrade Admin · Accepted Answer

M1 B sin 30° propto(180°-30°) and M2 B sin 30° propto(270°-30°) ∴ (M1/M2)=(150°/240°)=(5/8)

$\frac{8}{5}$

$\frac{7}{6}$

$\frac{6}{7}$

$\frac{5}{8}$

$M_{1} B sin 3 0^{\circ} \propto (18 0^{\circ} - 3 0^{\circ})$
and $M_{2} B sin 3 0^{\circ} \propto (27 0^{\circ} - 3 0^{\circ})$
$∴ \frac{M _{1}}{M _{2}} = \frac{15 0 ^{\circ}}{24 0 ^{\circ}} = \frac{5}{8}$

58​

67​

76​

85​

M1​Bsin30∘∝(180∘−30∘) and M2​Bsin30∘∝(270∘−30∘) ∴M2​M1​​=240∘150∘​=85​

$\frac{8}{5}$

$\frac{7}{6}$

$\frac{6}{7}$

$\frac{5}{8}$

$M_{1} B sin 3 0^{\circ} \propto (18 0^{\circ} - 3 0^{\circ})$
and $M_{2} B sin 3 0^{\circ} \propto (27 0^{\circ} - 3 0^{\circ})$
$∴ \frac{M _{1}}{M _{2}} = \frac{15 0 ^{\circ}}{24 0 ^{\circ}} = \frac{5}{8}$