Geophysical Anomaly Calculation

This solution determines the ratio between the vertical ($\Delta Z$) and horizontal ($\Delta H$) magnetic anomalies under specific conditions.

Magnetic Anomaly Component Relationship

The total field magnetic anomaly ($\Delta T$) is the projection of the anomaly vector onto the direction of the Earth's main magnetic field. At a given magnetic latitude ($\lambda$), the relationship between the vertical anomaly ($\Delta Z$), the horizontal anomaly ($\Delta H$), and the inclination ($I$) is given by:

$ \Delta T = \Delta Z \sin I + \Delta H \cos I $

We are given that the total field anomaly is zero ($\Delta T = 0$):

$ \Delta Z \sin I + \Delta H \cos I = 0 $

The inclination ($I$) is related to the magnetic latitude ($\lambda$) by:

$ \tan I = 2 \tan \lambda $

Calculating the Ratio

First, we rearrange the $\Delta T = 0$ equation to find the ratio $\Delta Z / \Delta H$:

$ \frac{\Delta Z}{\Delta H} = -\frac{\cos I}{\sin I} = -\cot I $

Next, we calculate the inclination ($I$) using the given magnetic latitude $\lambda = 30^\circ N$:

$ \tan I = 2 \tan(30^\circ) = 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} $

Now, we find the cotangent of the inclination:

$ \cot I = \frac{1}{\tan I} = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2} $

Finally, substitute the value of $\cot I$ back into the ratio equation:

$ \frac{\Delta Z}{\Delta H} = -\cot I = -\frac{\sqrt{3}}{2} $

Therefore, the ratio of the anomalies in the vertical and horizontal components ($\Delta Z : \Delta H$) is:

$ -\sqrt{3} : 2 $