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Earth's Magnetic Field

Horizontal co...

Question

Horizontal component of earths magnetic field at a place is √ 3 times the vertical component. What is the value of angle of dip at this place ?

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Solution

$L e t ϕ i s t h e a n g l e o f d i p . \tan ϕ = \frac{B_{v}}{B_{H}} w h e r e B_{v} i s t h e v e r t i c a l c o m p o n e n t o f e a r t h s m a g n e t i c f i e l d . B_{H} i s t h e H o r i z o n t a l c o m p o n e n t o f e a r t h s m a g n e t i c f i e l d . B_{H} = \sqrt{3} B_{v} t h e r e f o r e \tan ϕ = \frac{B_{v}}{\sqrt{3} B_{v}} = \frac{1}{\sqrt{3}} ϕ = 30^{o} h e n c e a n g l e o f d i p i s 30^{o} .$

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