In the magnetic meridian of a certain place, the vertical component of the earth's magnetic field is $0.3464…

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
  1. 0.4 G
  2. 0.6 G
  3. 0.7 G
  4. 0.8 G

Solution

$\begin{aligned} & \tan \delta=\frac{B_V}{B_H} \Rightarrow B_H=\frac{B_V}{\tan \delta}=\frac{0.3464}{(1 / \sqrt{3})} \\ & =0.3464 \times 1.73=0.6 \mathrm{G}\end{aligned}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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