Question

A bar magnet of magnetic moment $M_1$ is suspended by a wire in a magnetic field. The top of the wire is rotated through ${180}^o$ , then the magnet rotated through ${45}^o$. Under similar conditions, another magnet of magnetic moment is rotated through ${30}^o$. The ratio $M_1:M_2$ is

• A. $9:10\sqrt{2}$
• B. $1:\sqrt{2}$
• C. $1:1$
• D. $1:3$

Answer 1

The correct option is A $9:10\sqrt{2}$

The torque applied by the wire is proportional to the angle turned.
Hence,
$k(180-45)=k\left(135\right)=M_{1}B\sin \left({45}^o\right)$
$k(180-30)=k\left(150\right)=M_{2}B\sin \left({30}^o\right)$
$\therefore \frac{M_1}{M_2}=\frac{135\sin \left({30}^o\right)}{150\sin \left({45}^o\right)}$
$\frac{M_1}{M_2}=9:10\sqrt{2}$