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In the magnetic meridian at a certainlocation, the horizontal component of earth'smagnetic field is 0.16 G and angle of dip is60°. The earth's magnetic field at thislocation is

Question

In the magnetic meridian at a certainlocation, the horizontal component of earth'smagnetic field is 0.16 G and angle of dip is60°. The earth's magnetic field at thislocation is

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Solution

The Earth's magnetic field (B) can be calculated using the formula:

B = Bh / cos(d)

where Bh is the horizontal component of the Earth's magnetic field and d is the angle of dip.

Given that Bh = 0.16 G and d = 60°, we can substitute these values into the formula:

B = 0.16 G / cos(60°)

Cosine of 60° is 0.5, so:

B = 0.16 G / 0.5 = 0.32 G

Therefore, the Earth's magnetic field at this location is 0.32 G.

This problem has been solved

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