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(A) Given,the horizontal component of the earth's magnetic field $H_{E} = 0.26 \ G$ and the angle of dip $\delta = 60^{\circ}$.The relationship betwee

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$0.52$

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(A) Given,the horizontal component of the earth's magnetic field $H_{E} = 0.26 \ G$ and the angle of dip $\delta = 60^{\circ}$.The relationship between the total magnetic field $B_{E}$,the horizontal component $H_{E}$,and the angle of dip $\delta$ is given by the formula:$H_{E} = B_{E} \cos \delta$Rearranging the formula to solve for $B_{E}$:$B_{E} = \frac{H_{E}}{\cos \delta}$Substituting the given values:$B_{E} = \frac{0.26}{\cos 60^{\circ}}$Since $\cos 60^{\circ} = 0.5$:$B_{E} = \frac{0.26}{0.5} = 0.52 \ G$Therefore,the magnetic field of the earth at this location is $0.52 \ G$.

In the magnetic meridian of a certain place,the horizontal component of the earth's magnetic field is $0.26 \ G$ and the dip angle is $60^{\circ}$. What is the magnetic field of the earth at this location (in $G$)?

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