Question

If a magnet is suspended at an angle ${30}^{\circ }$ to the magnetic meridian, it makes an angle of 45o with the horizontal. The real dip is

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

Answer 1

The correct option is A $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)\ast$

Let the real dip be $\varphi$, then $\varphi =\frac{B_V}{B_H}$​
For apparent dip,
$tan{\varphi }^{\prime }=\frac{B_V}{B_{H}cos\beta }=\frac{B_V}{B_{H}cos{30}^{\circ }}=\frac{2B_V}{\sqrt{3}{B}_H}$
or $tan{45}^{\circ }=\frac{2}{\sqrt{3}}.tan\varphi$ or $\varphi =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$