If B_V and B_H are respectively the vertical | vedclass

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(C) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \delta$ and the vertical component is given by $B_V = B \sin \del

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}} \,B_{V}$

Vedclass · Answer

(C) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \delta$ and the vertical component is given by $B_V = B \sin \delta$, where $B$ is the total magnetic field and $\delta$ is the angle of dip.Given that the angle of dip $\delta = 60^{\circ}$.We know that $B_V = B \sin 60^{\circ} = B \frac{\sqrt{3}}{2}$.Therefore, the total magnetic field $B = \frac{2 B_V}{\sqrt{3}}$.Alternatively, using the relation $B = \sqrt{B_H^2 + B_V^2}$, since $	an \delta = \frac{B_V}{B_H} = 	an 60^{\circ} = \sqrt{3}$, we have $B_V = \sqrt{3} B_H$.Substituting this into the expression for $B$, we get $B = \sqrt{B_H^2 + (\sqrt{3} B_H)^2} = \sqrt{B_H^2 + 3 B_H^2} = \sqrt{4 B_H^2} = 2 B_H$.However, looking at the options provided, the expression $B = \frac{2 B_V}{\sqrt{3}}$ matches option $C$.

If $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^{\circ}$, then the total magnetic field at that place is

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