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The needle of a deflection galvanometer shows a deflection of $60^o $ due to a short bar magnet at a certain distance in $\tan A$ position. If the distance is doubled, the deflection is
${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{8}} \right)$
${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{8}} \right)$
${\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{8}} \right)$
${\cot ^{ - 1}}\left( {\frac{{\sqrt 3 }}{8}} \right)$
Solution
(c) For short bar magnet in tan $A-$ position
$\frac{{{\mu _0}}}{{4\pi }}\frac{{2M}}{{{d^3}}} = H\tan \theta $…..$(i)$
When distance is doubled, then new deflection $\theta '$ is given by
$\frac{{{\mu _0}}}{{4\pi }}\frac{{2M}}{{{{(2d)}^3}}} = H\tan \theta '$…..$(ii)$
$\therefore$ $\frac{{\tan \theta '}}{{\tan \theta }} = \frac{1}{8}$ $==>$ $\tan \theta ' = \frac{{\tan \theta }}{8} = \frac{{\tan 60^\circ }}{8} = \frac{{\sqrt 3 }}{8}$
$==>$ $\theta ' = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 }}{8}} \right)$
