Question

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

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

Answer 1

Answer: (B)

In tan A positions $\frac{{\mu }_0}{4\pi }\times \frac{2Md}{{(d^2-l^2)}^2}=H\tan \theta$ When $d>>>l,$ then $\frac{{\mu }_0}{4\pi }\times \frac{2M_d}{d^4}=H\tan {\theta }_A$
$\frac{{\mu }_0}{4\pi }\times \frac{2M}{{d_A}^3}=H\tan {\theta }_A$ $\left(\begin{array}{rl}& \text{Given:}{\theta }_A={30}^o\\ & d_A=18cm\end{array}\right)$ $\frac{{\mu }_0}{4\pi }\times \frac{2\times 2l\times m_A}{{(18)}^3}=H\tan {30}^o$ ....(i)
In tan B position: $\frac{{\mu }_0}{4\pi }\times \frac{M}{{(d^2+l^2)}^{3/2}}=H\tan {\theta }_B$ $\frac{{\mu }_0}{4\pi }\times \frac{M}{d^3}=H\tan {\theta }_B$ $\frac{{\mu }_0}{4\pi }\times \frac{2l\times m_B}{{d_B}^3}=\tan {\theta }_{B}H$ $\left(\begin{array}{rl}& \text{Given:}{\theta }_B=?\\ & m_B=16m_A\\ & d_B=36cm\end{array}\right)$ $\frac{{\mu }_0}{4\pi }\times \frac{2l\times 16m_A}{{(36)}^3}=H\tan {\theta }_B$ Now, form Eqs. (i) and (ii),we have $\frac{\tan {30}^o}{\tan {\theta }_B}=\frac{2}{{(18)}^3}\times \frac{{(36)}^3}{(16)}=1$ $\tan {\theta }_B=\tan {30}^o$ ${\theta }_B={30}^o$