Question

Calculate the real angle of dip, if a magnet is suspended at an angle of $30^\circ $ to the magnetic meridian, the dip needle makes an angle of $45^\circ $ with horizontal.

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

Answer 1

Answer: (A)

Let $\theta =dip \, angle$
$B_{H}=$ Horizontal component of magnetic field of earth
$B_{V}=$ Vertical component of magnetic field of earth
Also, $tan \theta = \left(\frac{B_{v}}{B_{H}}\right)$
But $tan 45^{o} = \frac{B_{V}}{B_{H} cos ⁡ 30^{o}}$
$\therefore \, \, \, \frac{tan \theta }{tan ⁡ 45^{o}}=cos ⁡ 30^{o}$
$\theta =\left(tan\right)^{- 1} \left(\frac{\sqrt{3}}{2}\right)$