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 $\phi$,
$\quad\tau=C\,\phi= MB\, sin \,\theta.$
In the $1^{st}$ case, $\phi_{1}=540^{\circ}-45^{\circ}=495^{\circ},\,\theta_{1}=45^{\circ}$
In the $2^{nd}$ case, $\phi_{2}=360^{\circ}-45^{\circ}=315^{\circ},\,\theta_{2}=45^{\circ}$
$\therefore \quad C\left(495^{\circ}\right)=M_{1}\,B\,sin\,45^{\circ}\quad\quad\quad\quad\quad...\left(i\right)$
and $\,C\left(315^{\circ}\right)=M_{2}\,B\,sin\,45^{\circ}\quad\quad\quad\quad\quad...\left(ii\right)$
Dividing (i) by (ii), we get
$\quad\quad \frac{M_{1}}{M_{2}}=\frac{495}{315}=\frac{11}{7}$