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प्रश्न
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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उत्तर
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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