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In the magnetic meridian of a certain place, the vertical component of the earth's magnetic field is $0.3464 \mathrm{G}$ and the dip angle is $30^{\circ}$. The horizontal component of the earth's magnetic field at this location is
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0.6 G
$\begin{aligned} & \tan \delta=\frac{B_V}{B_H} \Rightarrow B_H=\frac{B_V}{\tan \delta}=\frac{0.3464}{(1 / \sqrt{3})} \\ & =0.3464 \times 1.73=0.6 \mathrm{G}\end{aligned}$
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