Question

A magnet hung at 45° with magnetic meridian makes an angle of 60° with the horizontal. The actual value of the angle of dip is

• A. $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$
• B. $\tan^{-1}\left(\sqrt{6}\right)$
• C. $\tan^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• D. $\tan^{-1}\left(\sqrt{\frac{1}{2}}\right)$

Answer 1

Answer: (A)

$\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$

Answer 2

The correct answer is (A) : $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$
$\tan60°=\frac{B_0\sin\delta}{B_0\cos \delta \cos45°}$
$⇒\tan\delta=\sqrt{\frac{3}{2}}$
$\delta=\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$