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At a location, the horizontal component of the earth's magnetic field is $0.3 \mathrm{G}$ in the magnetic meridian and the dip angle is $60^{\circ}$. The earth's magnetic field at this location in $G$ is
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The correct answer is:
0.6
Using, $\beta_H=B \cos \delta$, we get
$\begin{aligned}
& 0.3=B \cos 60^{\circ} \\
& 0.3=B \times \frac{1}{2} \text { or } B=0.6 \mathrm{G}
\end{aligned}$
$\begin{aligned}
& 0.3=B \cos 60^{\circ} \\
& 0.3=B \times \frac{1}{2} \text { or } B=0.6 \mathrm{G}
\end{aligned}$
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