Question

Two magnets of magnetic moments $M_1$ and $M_2$ ($M_1$ > $M_2$) are placed one by one on a deflection magnetometer at a distance $d$ from the centre of the compass. The deflections produced are ${60}^o$ and ${30}^o$ . $M_1$ is placed at a distance $10cm$ from the compass. To get null deflection, $M_2$ to be placed at a distance of $(3^{1/3}=1.5)$

• A. $6.6cm$
• B. $3.3cm$
• C. $4cm$
• D. $2cm$

Answer 1

The correct option is A $6.6cm$

For null deflection,
$\frac{M_1}{M_2}=\frac{d_1^3}{d_2^3}$ ........(1)
But,
$\frac{M_1}{d_1^3}=\frac{4\pi B_H\left(\tan {30}^{\circ }\right)}{{\mu }_0}$ and $\frac{M_2}{d_2^3}=\frac{4\pi B_H\left(\tan {60}^{\circ }\right)}{{\mu }_0}$

comparing above equations, we get
$\frac{d_1^3}{d_2^3}=\frac{M_1\left(\tan {60}^{\circ }\right)}{M_2\left(\tan {30}^{\circ }\right)}$
$\therefore \frac{d_2^3}{d{1}^3}=\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}$ ........from(1)
$\therefore d_2^3=\left(\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}\right)d_1^3$
$\therefore d_2^3=\left(\frac{0.5773}{1.732}\right)(10{)}^3$
$\therefore d_2^3=\left(0.3\right)(10{)}^3$
$\therefore d_2=0.66\times 10=6.6cm$