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Byju's Answer

Standard XII

Physics

Magnetic Dip

At a place of...

Question

At a place of latitude 5 $^{o}$ , the angle of dip is nearly

5^{o}

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10^{o}

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{7.5}^{o}

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{2.5}^{o}

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Solution

The correct option is A $10^{o}$
$tan δ = 2 tan λ$
For small angles ,
$tan θ ≃ θ$
Hence , For small latitudes , $δ = 2 λ$
$δ = 2 \times 5 = 10^{o}$

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Similar questions

Q. If the angle

2 x - 10^{o}

and

x - 5^{o}

are complementary angles,find

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Q. The correct value of dip angle at a place is

45^{o}

. If the dip circle is rotated by

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out of the meridian, then the tangent of the angle of apparent dip at the place is :

Q. If a dip circle is placed in a vertical plane at an angle of

30^{o -}

to the magnetic meridian, the dip needle makes an angle of

45^{o -}

with the horizontal. The real dip at that place is?

Consider the earth as a short magnet with the centre coinciding with the centre of the earth and dipole moment $M$ . The angle of dip $θ$ at the place where the magnetic latitude is $λ$ is :

Q. A magnet oscillating in a horizontal plane has a time period of 2 seconds at a place where the angle of dip is

30^{o}

and 3 seconds at another place where the angle of dip is

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. The ratio of resultant magnetic fields at the two places is:

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At a place of latitude 5o, the angle of dip is nearly

The correct option is A 10otanδ=2tanλFor small angles , tanθ≃θHence , For small latitudes , δ=2λδ=2×5=10o

At a place of latitude 5 $^{o}$ , the angle of dip is nearly

The correct option is A $10^{o}$
$tan δ = 2 tan λ$
For small angles ,
$tan θ ≃ θ$
Hence , For small latitudes , $δ = 2 λ$
$δ = 2 \times 5 = 10^{o}$