At a certain place the angle of dip is $30^{\circ}$ and the horizontal component of earth's magnetic field is $0.5 G$ . The earth's total magnetic field (in $G$ ), at that certain place, is :
$\sqrt{3}$
$\frac{1}{\sqrt{3}}$
$1$
$\frac{1}{2}$

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Solution

The correct option is B $\frac{1}{\sqrt{3}}$
Horizontal component of earth's magnetic field $(B_{H}) = B cos 30^{\circ}$

$\Rightarrow B \times \frac{\sqrt{3}}{2} = 0.5 \times G$

$B = \frac{1}{\sqrt{3}} \times G$

$∴$ option $(a)$ is correct choice.

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