A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is -
Solution
<p>$\tan 60^\circ = {{{B_0}\sin \delta } \over {{B_0}\cos \delta \cos 45^\circ }}$</p>
<p>$\Rightarrow \tan \delta = \sqrt {{3 \over 2}}$</p>
<p>$\Rightarrow \delta = {\tan ^{ - 1}}\left( {\sqrt {{3 \over 2}} } \right)$</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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