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At two different places, the angles of dip are respectively $30^{\circ}$ and $45^{\circ} .$ At these two places the ratio of horizontal component of earth's magnetic field is
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$\sqrt{3}: \sqrt{2}$
The horizontal component of earth's magnetic field is given by, $H=R \cos \delta$
where, $\delta$ is the angle of dip so $H_{1}=R \cos 30^{\circ}$ and $\mathrm{H}_{2}=\mathrm{Rcos} 45^{\circ}$
$\therefore \quad \frac{H_{1}}{H_{2}}=\frac{R \cos 30^{\circ}}{R \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
where, $\delta$ is the angle of dip so $H_{1}=R \cos 30^{\circ}$ and $\mathrm{H}_{2}=\mathrm{Rcos} 45^{\circ}$
$\therefore \quad \frac{H_{1}}{H_{2}}=\frac{R \cos 30^{\circ}}{R \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
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