If the angles of dip at two places are 30^{ci | vedclass

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(A) Let $B_e$ be the net magnetic field of the Earth at both locations. The horizontal component of the Earth's magnetic field $H$ is given by $H = B_

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$\sqrt{3}: \sqrt{2}$

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(A) Let $B_e$ be the net magnetic field of the Earth at both locations. The horizontal component of the Earth's magnetic field $H$ is given by $H = B_e \cos \delta$,where $\delta$ is the angle of dip.For the two places,the ratio of the horizontal components $H_1$ and $H_2$ is:$\frac{H_1}{H_2} = \frac{B_e \cos \delta_1}{B_e \cos \delta_2} = \frac{\cos \delta_1}{\cos \delta_2}$Given $\delta_1 = 30^{\circ}$ and $\delta_2 = 45^{\circ}$,we have:$\frac{H_1}{H_2} = \frac{\cos 30^{\circ}}{\cos 45^{\circ}} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} 	imes \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$Therefore,the ratio $H_1: H_2 = \sqrt{3}: \sqrt{2}$.

If the angles of dip at two places are $30^{\circ}$ and $45^{\circ}$ respectively,then the ratio of horizontal components of earth's magnetic field at the two places will be

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