First slide

Magnetic elements of Earth

Question

A magnetic needle is free to rotate in a vertical plane, which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal , what would be the dip at that place?

Moderate

A

$60^{0}$
B

$30^{0}$
C

$45^{0}$
D

$90^{0}$

Solution

We know the relationship between true dip and apparent dip
$\tan δ^{1} = \frac{\tan δ}{\cos θ}$
Given : Apparent dip
$δ^{1} = \tan^{- 1} (\frac{2}{\sqrt{3}}) \Rightarrow \tan δ^{1} = \frac{2}{\sqrt{3}}$
True dip $δ = ?$
$\begin{array}{l} A L s o, θ = 60 ° \\ ∴ \frac{2}{\sqrt{3}} = \frac{\tan δ}{\cos 60 °} \\ \Rightarrow \tan δ = \frac{1}{\sqrt{3}} \\ \Rightarrow \tan δ = \tan 30 ° \\ \Rightarrow δ = 30 ° \end{array}$

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