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Solve the following problem. A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place. - Physics

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Question

Solve the following problem.

A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.

Sum

Solution

Let the true value of dip be Φ. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’H = BH cos 30° Whereas, the vertical component remains unchanged.
∴ For an apparent dip of 45°,
tan 45° = `"B"_"V"^′/"B"_"H"^′="B"_"V"/("B"_"H"cos30°)="B"_"V"/"B"_"H"xx1/(cos30°)`
But, real value of dip is,
tan Φ = `"B"_"V"/"B"_"H"`

∴ tan 45° = `tanΦ/(cos30°)`

∴ tan Φ = tan 45° × cos 30°

= `1xxsqrt3/2`

∴ Φ = tan−1 (0.866)

The true value of angle of dip is tan−1 (0.866).

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Chapter 12: Magnetism - Exercises [Page 288]

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Balbharati Physics [English] 11 Standard Maharashtra State Board
Chapter 12 Magnetism
Exercises | Q 4. (ii) | Page 288

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