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Question

Horizontal component of earth's magnetic field at a place is $$\sqrt{3}$$ times its vertical component. What is the value of angle of dip at that place?


Solution

Given $$H=\sqrt{3}V\Rightarrow \tan\theta=\dfrac{V}{H}=\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow$$ Angle of dip, $$\theta=\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=30^{o}$$

Physics

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