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Q.On suspending a magnet at 30° with the magnetic meridian, it makes an angle of 45° with the horizontal. What will be the actual angle of dip?

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 3mImportance★★★★★
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Using tan⁡δ′=tan⁡δ/cos⁡θ\tan\delta'=\tan\delta/\cos\theta with δ′=45∘, θ=30∘\delta'=45^\circ,\ \theta=30^\circ gives δ≈41∘\delta\approx41^\circ.

When a dip needle is set in a vertical plane making an angle θ\theta with the magnetic meridian, only the component BHcos⁡θB_H\cos\theta of the horizontal field acts in that plane, while the vertical field BVB_V is unchanged. The apparent dip δ′\delta' then satisfies

tan⁡δ′=BVBHcos⁡θ=tan⁡δcos⁡θ,\tan\delta'=\frac{B_V}{B_H\cos\theta}=\frac{\tan\delta}{\cos\theta},

where δ\delta is the true dip.

Given: δ′=45∘\delta'=45^\circ, θ=30∘\theta=30^\circ. …

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