Magnetic Field Components Explained

The Earth's magnetic field ($B$) at any location can be represented by its horizontal ($B_H$) and vertical ($B_V$) components. These are related to the total field by the Pythagorean theorem: $B^2 = B_H^2 + B_V^2$

Applying Equal Components Condition

The problem states that at a particular place, the vertical component equals the horizontal component: $B_V = B_H$

Substitute this condition into the field equation: $B^2 = B_H^2 + B_H^2$ $B^2 = 2 B_H^2$ Solving for $B$, we get: $B = \sqrt{2 B_H^2} = B_H \sqrt{2}$

Equatorial Field Relation

At the Earth's equator, the magnetic field has a dip angle ($\delta$) of $0^\circ$. Therefore, the horizontal component ($B_{H_{eq}}$) is equal to the total field ($B_{eq}$): $B_{H_{eq}} = B_{eq} \cos(0^\circ) = B_{eq}$ Given $B_{eq} = 0.30 \text{ Oe}$, we find $B_{H_{eq}} = 0.30 \text{ Oe}$.

We assume the horizontal component $B_H$ at the location of interest is approximately the same as at the equator. Therefore, $B_H = 0.30 \text{ Oe}$.

Calculating Total Field Value

Using the relationship $B = B_H \sqrt{2}$ and the value $B_H = 0.30 \text{ Oe}$: $B = (0.30 \text{ Oe}) \times \sqrt{2}$

Calculate the numerical value using $\sqrt{2} \approx 1.4142$: $B \approx 0.30 \times 1.4142 \text{ Oe}$ $B \approx 0.42426 \text{ Oe}$

Rounding to two decimal places, the value of the Earth's magnetic field is $0.42 \text{ Oe}$.