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 $\theta =dipangle$
$B_H=$ 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 $tan45^o=\frac{B_V}{B_{H}cos30^o}$
$\therefore \frac{tan\theta }{tan45^o}=cos30^o$
$\theta ={\left(tan\right)}^{-1}\left(\frac{\sqrt{3}}{2}\right)$