The angles of dip are 30^{circ} and 45^{circ} | vedclass

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Question

(A) The horizontal component of the Earth's magnetic field $(B_H)$ is given by the formula $B_H = B \cos \phi$,where $B$ is the total magnetic field a

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

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(A) The horizontal component of the Earth's magnetic field $(B_H)$ is given by the formula $B_H = B \cos \phi$,where $B$ is the total magnetic field and $\phi$ is the angle of dip.Assuming the total magnetic field $B$ is the same at both locations,the ratio of the horizontal components is given by:$\frac{(B_H)_1}{(B_H)_2} = \frac{B \cos 30^{\circ}}{B \cos 45^{\circ}}$Substituting the values $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$:$\frac{(B_H)_1}{(B_H)_2} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} 	imes \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be:

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