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  4. The horizontal component of the earth’s magnetic field is 3.6 × 10-5 tesla where the dip angle is 60°. The magnitude of the earth’s magnetic field is

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Q. The horizontal component of the earth’s magnetic field is $3.6 \times 10^{-5}$ tesla where the dip angle is $60^{\circ}$. The magnitude of the earth’s magnetic field is

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Solution:

Horizontal component of earth's field,
$H=B \cos \theta$,
since, $\theta=60^{\circ}$
image
$3.6 \times 10^{-5}= B \times \frac{1}{2} $
$\Rightarrow B =7.2 \times 10^{-5}$ Tesla