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If the angle of dip at places $\mathrm{A}$ and $\mathrm{B}$ are $30^{\circ}$ and $45^{\circ}$ respectively, the ratio of horizontal component of earth's magnetic field at $A$ to that at $B$ will be $\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]$
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The correct answer is:
$\sqrt{3}: \sqrt{2}$
(B)
The horizontal component of earth's magnetic field is given by,$H=R \cos \delta$ where, $\delta$ is the angle of dip so $\mathrm{H}_{1}=\mathrm{R} \cos 30^{\circ}$
and $\mathrm{H}_{2}=\mathrm{R} \cos 45^{\circ}$
$\therefore \frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R} \cos 30^{\circ}}{\mathrm{R} \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
The horizontal component of earth's magnetic field is given by,$H=R \cos \delta$ where, $\delta$ is the angle of dip so $\mathrm{H}_{1}=\mathrm{R} \cos 30^{\circ}$
and $\mathrm{H}_{2}=\mathrm{R} \cos 45^{\circ}$
$\therefore \frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R} \cos 30^{\circ}}{\mathrm{R} \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
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