The value of horizontal component of the earth-s magnetic field at a point is 0-5- 10-4 tesla and the angle of dip is 45- Find the value of vertical components-

Value of horizontal component of earth's magnetic field #160; B_H = 0.5× 10^ 4 T Angle of dip #160; δ = 45^o Thus value of vertical component of e ...

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Standard XII

Physics

Earth's Magnetism

The value of ...

Question

The value of horizontal component of the earth's magnetic field at a point is $0.5 \times 10^{- 4}$ tesla and the angle of dip is $45^{\circ}$ . Find the value of vertical components.

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Solution

Value of horizontal component of earth's magnetic field $B_{H} = 0.5 \times 10^{- 4} T$
Angle of dip $δ = 45^{o}$
Thus value of vertical component of earth's magnetic field $B_{V} = B_{H} tan δ$
$⟹ B_{V} = 0.5 \times 10^{- 4} \times 1 = 0.5 \times 10^{- 4} T$

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The value of horizontal component of the earth's magnetic field at a point is 0.5×10−4 tesla and the angle of dip is 45∘. Find the value of vertical components.

Value of horizontal component of earth's magnetic field BH=0.5×10−4 TAngle of dip δ=45oThus value of vertical component of earth's magnetic field BV=BH tanδ⟹ BV=0.5×10−4×1=0.5×10−4 T

The value of horizontal component of the earth's magnetic field at a point is $0.5 \times 10^{- 4}$ tesla and the angle of dip is $45^{\circ}$ . Find the value of vertical components.

Value of horizontal component of earth's magnetic field $B_{H} = 0.5 \times 10^{- 4} T$
Angle of dip $δ = 45^{o}$
Thus value of vertical component of earth's magnetic field $B_{V} = B_{H} tan δ$
$⟹ B_{V} = 0.5 \times 10^{- 4} \times 1 = 0.5 \times 10^{- 4} T$