First slide

Magnetic elements of Earth

Question

If a magnet is suspended at an angle of $30^{0}$ to the magnetic meridian the dip needle makes an angle of $60^{0}$ with the horizontal. The real dip is

Moderate

A

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

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

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

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

Solution

$\tan θ = B_{v} / B_{H}$ …………………..(1)
[real dip is $θ$ ]
In the plane situated at $30^{0}$ with the magnetic meridian the horizontal component of the earth’s magnetic field will be $B_{H} \cos 30^{0}$ and the vertical component will be $B_{v}$ . The angle of dip in this plane is
$\tan 60^{0} = \frac{B_{v}}{B_{H} \cos 30^{0}}$
$\tan 60^{0} . \cos 30^{0} = \frac{B_{v}}{B_{H}} F r o m e q u a t i o n (1)$
$\Rightarrow \tan θ = \tan 60^{0} . \cos 30^{0}$
$\Rightarrow \tan θ = \sqrt{3} . \frac{\sqrt{3}}{2} \Rightarrow θ = \tan^{- 1} (3 / 2)$

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