Question

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

Let $M_1$ and $M_2$ be the magnetic moments of magnets and H the horizontal component of earth's field.
We have $\tau =MHsin\theta$. If $\varphi$ is the twist of wire, then $\tau =C\varphi$ C being restoring couple per unit twist of wire
$\Rightarrow C\varphi =MHsin\theta$
Here ${\varphi }_1=({180}^{\circ }-{30}^{\circ })={150}^{\circ }=150\times \frac{\pi }{180}rad$
${\varphi }_2=({270}^{\circ }-{30}^{\circ })={240}^{\circ }=240\times \frac{\pi }{180}rad$
So, $C{\varphi }_1=M_{1}Hsin\theta$ (For deflection $\theta ={30}^{\circ }$ of I magnet)
$C{\varphi }_2=M_{2}Hsin\theta$ (For deflection $\theta ={30}^{\circ }$ of II magnet)
Dividing
$\frac{{\varphi }_1}{{\varphi }_2}=\frac{M_1}{M_2}$
$\Rightarrow \frac{M_1}{M_2}=\frac{{\varphi }_1}{{\varphi }_2}=\frac{150\times \left(\frac{\pi }{180}\right)}{240\times \left(\frac{\pi }{180}\right)}=\frac{15}{24}=\frac{5}{8}$
$\Rightarrow M_1:M_2=5:8$