The angle of dip at a place is 60^{circ}. At | vedclass

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Question

(D) 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

$0.32$

Vedclass · Answer

(D) 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: Total intensity $B = 0.64 	ext{ units}$ Angle of dip $\phi = 60^{\circ}$ Substituting the values: $B_H = 0.64 	imes \cos 60^{\circ}$ Since $\cos 60^{\circ} = 0.5$ $B_H = 0.64 	imes 0.5 = 0.32 	ext{ units}$ Therefore,the correct option is $D$.

The angle of dip at a place is $60^{\circ}$. At this place,the total intensity of the Earth's magnetic field is $0.64 \text{ units}$. The horizontal intensity of the Earth's magnetic field at this place is . . . . . . $\text{units}$.

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