Question

In a deflection magnetometer experiment, in tan A position, a short bar Magnet placed at $18\text{cm}$ from the centre of compass needle produces a deflection of ${30}^o$. If another magnet of same length but 16 times pole strength as that of first magnet is placed in tan B position at $36\text{cm}$, the deflection will be

• A. 0${}^o$
• B. 30${}^o$
• C. 45${}^o$
• D. 60${}^o$

Answer 1

The correct option is B 30${}^o$

$\text{By Tangent law:}B=B_H\tan \theta$

$\text{For Tan A Position:}B=\frac{{\mu }_0}{4\pi }\frac{2M_1}{d_1^3}$

$\Rightarrow \frac{{\mu }_0}{4\pi }\frac{2M_1}{d_1^3}=B_H\tan {\theta }_1$

$\text{For Tan B Postion:}B=\frac{{\mu }_0}{4\pi }\frac{M_2}{d_2^3}$

$\Rightarrow \frac{{\mu }_0}{4\pi }\frac{M_2}{d_2^3}=B_H\tan {\theta }_2$

$\frac{2M_1{d}_2^3}{M_2{d}_1^3}=\frac{\tan {\theta }_1}{\tan {\theta }_2}$

$\text{But}{d}_1=18\text{cm},d_2=36\text{cm},M_2=16M_1,{\theta }_1={30}^o$

$\Rightarrow \frac{2M_1\times (36{)}^3}{16M_1\times (18{)}^3}=\frac{\tan {30}^o}{\tan {\theta }_2}$

$\Rightarrow \tan {\theta }_2=\tan {30}^o\Rightarrow {\theta }_2={30}^o$