Question

A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of the wire is rotated through ${180}^{\circ }$ to deflect the magnet by ${30}^{\circ }$ from the magnetic meridian. When this magnet is replaced by another magnet, the upper end of the wire is rotated through ${270}^{\circ }$ to deflect the magnet ${30}^{\circ }$ from the magnetic meridian. The ratio of magnetic moments of magnets is

• A. $1:5$
• B. $1:8$
• C. $5:8$
• D. $8:5$

Answer 1

The correct option is C $5:8$

Given,
${\alpha }_1={180}^{\circ };{\alpha }_2={270}^{\circ }{\theta }_1={30}^{\circ };{\theta }_2={30}^{\circ }$

Let $M_1$ and $M_2$ be the magnetic moments of magnets and $H$ the horizontal components of earth's magnetic field.

In equilibrium position,

$\text{Torque in the wire}=\text{Restoring torque}$

$M{B}_H\sin \theta =C(\alpha -\theta )$

$\Rightarrow M_1{B}_H\sin {\theta }_1=C({\alpha }_1-{\theta }_1)$

$\Rightarrow M_1{B}_H\sin {30}^{\circ }=C({180}^{\circ }-{30}^{\circ }).....\left(1\right)$

Similarly,

$\Rightarrow M_2{B}_H\sin {30}^{\circ }=C({270}^{\circ }-{30}^{\circ }).....\left(2\right)$

From (1) and (2) we get,

$\Rightarrow \frac{M_1}{M_2}=\frac{150\times \left(\frac{\pi }{180}\right)}{240\times \left(\frac{\pi }{180}\right)}=\frac{15}{24}=\frac{5}{8}$

$\Rightarrow M_1:M_2=5:8$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.