Question

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

• A. 11/7
• B. 3/2
• C. 4/3
• D. 7/6

Answer 1

Answer: (A)

If C is torque per unit angular twist of the wire, then for a twist $\varphi$,
$\tau =C\varphi =MBsin\theta .$
In the $1^{st}$ case, ${\varphi }_1=540^{\circ }-45^{\circ }=495^{\circ },{\theta }_1=45^{\circ }$
In the $2^{nd}$ case, ${\varphi }_2=360^{\circ }-45^{\circ }=315^{\circ },{\theta }_2=45^{\circ }$
$\therefore C\left(495^{\circ }\right)=M_{1}Bsin45^{\circ }...\left(i\right)$
and $C\left(315^{\circ }\right)=M_{2}Bsin45^{\circ }...\left(ii\right)$
Dividing (i) by (ii), we get
$\frac{M_1}{M_2}=\frac{495}{315}=\frac{11}{7}$