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At certain place, the horizontal component of earth's magnetic field is \( 3.0 \mathrm{G} \) and the angle dip
at that place is \( 30^{\circ} \). The magnetic field of earth at that location
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at that place is \( 30^{\circ} \). The magnetic field of earth at that location
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The correct answer is:
\( 3.5 \mathrm{G} \)
Given, horizontal component of Earth's magnetic field $=3.0 \mathrm{G} ;$ angle of dip $=30^{\circ}$
Now, horizontal component = B $\cos$ (angle of dip)
$\Rightarrow 3.0 G=B \cos 30^{\circ}$
$\Rightarrow B=\frac{3.0}{\cos 30^{\circ}}=\frac{3.0 \times 2}{\sqrt{3}}$
$\Rightarrow B=2 \sqrt{3}=3.464 \sim 3.5 \mathrm{G}$
Therefore, magnetic field of Earth at that location $=3.5 \mathrm{G}$
Now, horizontal component = B $\cos$ (angle of dip)
$\Rightarrow 3.0 G=B \cos 30^{\circ}$
$\Rightarrow B=\frac{3.0}{\cos 30^{\circ}}=\frac{3.0 \times 2}{\sqrt{3}}$
$\Rightarrow B=2 \sqrt{3}=3.464 \sim 3.5 \mathrm{G}$
Therefore, magnetic field of Earth at that location $=3.5 \mathrm{G}$
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