At a certain place,the angle of dip is 30^{ci | vedclass

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(C) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the f

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	ext{ Oersted}$

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(C) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the formula: $B_H = B \cos \phi$. Given: $B_H = 0.50 	ext{ Oersted}$ and $\phi = 30^{\circ}$. Substituting the values: $0.50 = B \cos(30^{\circ})$. Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $0.50 = B 	imes \frac{\sqrt{3}}{2}$. Therefore,$B = \frac{0.50 	imes 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} 	ext{ Oersted}$.

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.50 \text{ Oersted}$. The Earth's total magnetic field is:

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