At an angle of 30$^\circ$ to the magnetic meridian, the apparent dip is 45$^\circ$. Find the true dip :
Solution
$A\tan \delta = \tan \delta '\cos \theta$<br><br>$= \tan 45^\circ \cos 30^\circ$<br><br>$\tan \delta = 1 \times {{\sqrt 3 } \over 2}$<br><br>$\delta = {\tan ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right)$
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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