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

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

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

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(A) The horizontal component of the Earth's magnetic field is given by the formula: $B_H = B \cos \delta$,where $B$ is the total magnetic field and $\delta$ is the angle of dip.Given: $B_H = 0.5 \ G$ and $\delta = 30^{\circ}$.Substituting the values into the formula: $0.5 = B \cos 30^{\circ}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.5 = B 	imes \frac{\sqrt{3}}{2}$.Solving for $B$: $B = \frac{0.5 	imes 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \ G$.

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.5 \ G$. The Earth's total magnetic field (in $G$) at that place is:

Similar Questions

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Choose the correct option regarding the relationship between true dip $(\phi)$ and apparent dip $(\phi^{\prime})$:

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