First slide

Magnetic elements of Earth

Question

The horizontal component of the earth’s magnetic field is 0.22G and total magnetic field is 0.4G. The angle of dip is

Moderate

A

$t a n^{- 1} (1)$
B

$t a n^{- 1} (\infty)$
C

$t a n^{- 1} (1.518)$
D

$t a n^{- 1} (3.14)$

Solution

$L e t A n g l e o f d i p = ϕ W e k n o w t h a t,$
$T o t a l m a g n e t i c f i e l d = B = \sqrt{{B_{H}}^{2} + {B_{V}}^{2}} \Rightarrow B_{V} = \sqrt{B^{2} - {B_{H}}^{2}} = \sqrt{0 . 4^{2} - 0 . 22^{2}} = \sqrt{0.16 - 0.0484} = \sqrt{0.1116} = 0.334$
$A l s o, t a n ϕ = \frac{B_{V}}{B_{H}}$
$\Rightarrow t a n ϕ = \frac{0.334}{0.22} = 1 .518$
$\Rightarrow ϕ = t a n^{- 1} (1.518)$

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