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{2{{M}_{d}}}{{{d}^{4}}}=H\tan {{\theta }_{A}} $
$ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2M}{{{d}_{A}}^{3}}=H\tan {{\theta }_{A}} $ $ \left( \begin{align} & \text{Given:}\,{{\theta }_{A}}={{30}^{o}} \\ & {{d}_{A}}=18\,cm \\ \end{align} \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{align} & \text{Given:}{{\theta }_{B}}=? \\ & {{m}_{B}}=16\,{{m}_{A}} \\ & {{d}_{B}}=36\,cm \\ \end{align} \right) $ $ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2l\times 16{{m}_{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}} $