Question

Two magnets are suspended by a given wire one by one. In order to deflect the first magnet through ${45}^0$, the wire has to be twisted through ${540}^0$ whereas with the second magnet, the wire requires a twist of ${360}^0$ for the same deflection. Then the ratio of magnetic moments of the two is

• A. $11/7$
• B. $3/2$
• C. $4/3$
• D. $7/6$

Answer 1

The correct option is A $11/7$

$Solution:$ If $C$ is torque per unit angular twist of the wire,

then for a twist $\varphi$,

$\tau$ =C$\varphi$=$MB$ $\sin$ $\theta$,

In the $1^{st}$ case, ${\varphi }_1$=${540}^{\circ }$-${45}^{\circ }$=${495}^{\circ },{\theta }_1$=${45}^{\circ }$

In the $2^{\text{nd}}$ case, ${\varphi }_2={360}^{\circ }$-${45}^{\circ }$=${315}^{\circ }$, ${\theta }_2$=${45}^{\circ }$

$\therefore C\left({495}^{\circ }\right)=M_{1}B\sin {45}^{\circ }\dots \left(i\right)$

and $C\left({315}^{\circ }\right)=M_{2}B\sin {45}^{\circ }\dots \left(ii\right)$

Dividing (i) by (ii), we get $\frac{M_1}{M_2}=\frac{495}{315}=\frac{11}{7}$

$So,the$ $correct$ $option:A$