The horizontal component of earth’s magnetic field at a place is $0.2 G$ whereas its total magnetic field is $0.4 G$ . The angle of dip at the place is
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
$30^{\circ}$

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Solution

The correct option is A $60^{\circ}$
$B_{H} = 0.2 G$
$B_{0} = 0.4 G$
$∵ B_{H} = B_{0} cos θ$
$\Rightarrow 0.2 = 0.4 cos θ$
$\Rightarrow cos θ = \frac{1}{2}$
$\Rightarrow θ = 60^{\circ}$
$∵ dip angle = 60^{\circ}$

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The horizontal component of earth’s magnetic field at a place is 0.2 G whereas its total magnetic field is 0.4 G. The angle of dip at the place is60∘45∘90∘30∘

The correct option is A 60∘BH=0.2 G B0=0.4 G ∵ BH=B0 cosθ ⇒ 0.2=0.4 cosθ ⇒ cosθ=12 ⇒ θ=60∘ ∵ dip angle=60∘

The horizontal component of earth’s magnetic field at a place is $0.2 G$ whereas its total magnetic field is $0.4 G$ . The angle of dip at the place is
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
$30^{\circ}$

The correct option is A $60^{\circ}$
$B_{H} = 0.2 G$
$B_{0} = 0.4 G$
$∵ B_{H} = B_{0} cos θ$
$\Rightarrow 0.2 = 0.4 cos θ$
$\Rightarrow cos θ = \frac{1}{2}$
$\Rightarrow θ = 60^{\circ}$
$∵ dip angle = 60^{\circ}$