A magnet is suspended in the magnetic meridia | vedclass

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(C) Let $M_1$ and $M_2$ be the magnetic moments of the magnets and $H$ be the horizontal component of the Earth's magnetic field.For a magnet in equil

Vedclass · Accepted Answer

$5:8$

Vedclass · Answer

(C) Let $M_1$ and $M_2$ be the magnetic moments of the magnets and $H$ be the horizontal component of the Earth's magnetic field.For a magnet in equilibrium,the restoring torque of the wire equals the magnetic torque: $	au = C\phi = MH \sin 	heta$,where $C$ is the restoring couple per unit twist,$\phi$ is the twist angle of the wire,and $	heta$ is the deflection angle.For the first magnet: $\phi_1 = 180^o - 30^o = 150^o$. Thus,$C(150^o) = M_1 H \sin 30^o$.For the second magnet: $\phi_2 = 270^o - 30^o = 240^o$. Thus,$C(240^o) = M_2 H \sin 30^o$.Dividing the two equations: $\frac{M_1}{M_2} = \frac{150^o}{240^o} = \frac{15}{24} = \frac{5}{8}$.Therefore,the ratio of the magnetic moments is $5:8$.

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