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Q.The horizontal component of the earth's magnetic field at a certain place is 2.6 x 10^-5 T and the angle of dip is 60°. What is the magnetic field of the earth at this location?

Andhra Pradesh BieapBIEAP Intermediate Board 2024Subjective· 2mImportance★★★★★
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Since BH=Bcos⁡δB_H = B\cos\delta, the total field is B=BH/cos⁡δ=2.6×10−5/0.5=5.2×10−5 TB = B_H/\cos\delta = 2.6\times10^{-5}/0.5 = 5.2\times10^{-5}\,\text{T}.

Concept:

The total magnetic field BB of the Earth is related to its horizontal component BHB_H through the angle of dip δ\delta:

BH=Bcos⁡δB_H = B\cos\delta

So the total field is B=BHcos⁡δB = \dfrac{B_H}{\cos\delta} (a standard NCERT/CBSE-aligned magnetism-and-matter relation).

Given: BH=2.6×10−5 TB_H = 2.6\times10^{-5}\,\text{T}, δ=60°\delta = 60°.

Step 1 — put in values: …

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