At the place where the angle of dip is 60^{ci | vedclass

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Question

(B) The angle of dip $(I)$ is defined by the ratio of the vertical component $(Z_{E})$ to the horizontal component $(H_{E})$ of the Earth's magnetic f

Vedclass · Accepted Answer

$Z_{E} = \sqrt{3} H_{E}$

Vedclass · Answer

(B) The angle of dip $(I)$ is defined by the ratio of the vertical component $(Z_{E})$ to the horizontal component $(H_{E})$ of the Earth's magnetic field as follows:$	an I = \frac{Z_{E}}{H_{E}}$Given that the angle of dip $I = 60^{\circ}$,we substitute this value into the equation:$	an 60^{\circ} = \frac{Z_{E}}{H_{E}}$Since $	an 60^{\circ} = \sqrt{3}$,we have:$\sqrt{3} = \frac{Z_{E}}{H_{E}}$Therefore,the relationship is:$Z_{E} = \sqrt{3} H_{E}$

At the place where the angle of dip is $60^{\circ}$,what is the relation between the vertical component $Z_{E}$ and the horizontal component $H_{E}$ of the Earth's magnetic field?

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