First slide

Earth's Magnetism

Question

If a magnet is suspended at an angle of 30^o to the magnetic meridian, the dip needle makes an angle of 45^o with the horizontal. The real dip is

Difficult

A

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

$\tan^{- 1} (\sqrt{3})$
C

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

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

Solution

Let the actual angle of dip (in the magnetic meridian) is $θ$ . lf B_H and, B_V be the horizontal and vertical components of earth's magnetic field respectively, then
$\tan θ = (B_{V} / B_{H})$
In the plane situated at 30. with the magnetic meridian, see fig.(9) , the horizontal component of earth's magnetic
field will be B_H cos 30^owhile the vertical component will be B_V. The angle of dip in this plane is
$\begin{array}{l} \tan 45^{\circ} = \frac{B_{V}}{B_{H} \cos 30^{\circ}} \\ ∴ \frac{\tan θ}{\tan 45^{\circ}} = \cos 30^{\circ} or \frac{\tan θ}{1} = \frac{\sqrt{3}}{2} \\ ∴ θ = \tan^{- 1} (\sqrt{3} / 2) \end{array}$

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