Maharashtra State BoardHSC Science (General) 11th
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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

Sum

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.

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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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APPEARS IN

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