Earth's Magnetism

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Question

At a place, the earth's horizontal component of magnetic field is $0.36 x 10^{- 4} Wb / m^{2}$ . If the angle of dip at that place is $60^{o}$ , then the vertical component of earth's field at that place in $Wb / m^{2}$ will be approximately

Moderate

Solution

From the relation, $B_{H} = Bcosϕ and B_{V} = Bsinϕ$
$\frac{B_{V}}{B_{H}} = \tan ϕ o r B_{V} = B_{H} \tan ϕ$
$= 0.36 x 10^{- 4} x \tan 60^{o} = 0.623 x 10^{- 4} Wb / m^{2}$

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