A) \[10{}^\circ \]
B) \[20{}^\circ \]
C) \[30{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: C
Solution :
For short bar magnet in tan A position \[H\tan \theta =\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}\] ?(i) The magnet is cut into three equal part \[M=m\times \left( \frac{l}{3} \right)\] \[=\frac{ml}{3}\] \[=\frac{M}{3}\] Then new deflection \[\theta \] is given by \[H\tan \theta =\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{3{{d}^{3}}}\] ?(ii) From the Eqs. (i) and (ii), we have \[\frac{\tan \theta }{\tan \theta }=\frac{1}{3}\] ie, \[\frac{\tan \theta }{\tan {{60}^{o}}}=\frac{1}{3}\] or \[\tan \theta =\frac{\tan {{60}^{o}}}{3}=\frac{\sqrt{3}}{3}\]\[\Rightarrow \]\[\frac{\sqrt{3}\times \sqrt{3}}{3\sqrt{3}}\] \[\theta =ta{{n}^{-1}}\left( \frac{1}{\sqrt{3}} \right)\] \[\theta ={{30}^{o}}\]
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