Earth's magnetism — magnetic elements, declination, inclination, horizontal component

Question

At a certain location on the Earth’s surface, the angle of dip is 60° and the horizontal component of the Earth’s magnetic field is 0.3 G. Find:

The vertical component of the field
The total intensity of the Earth’s magnetic field

Solution — Step by Step

The Earth’s magnetic field $\vec{B}$ at any point can be resolved into two components:

Horizontal component $B_H = B\cos\delta$ (where $\delta$ is the angle of dip/inclination)
Vertical component $B_V = B\sin\delta$

The ratio gives us: $\tan\delta = B_V / B_H$

We know $\delta = 60°$ and $B_H = 0.3$ G.

B_V = B_H \tan\delta = 0.3 \times \tan 60° = 0.3 \times \sqrt{3}

B_V = 0.3\sqrt{3} \approx \mathbf{0.52 \text{ G}}

B = \frac{B_H}{\cos\delta} = \frac{0.3}{\cos 60°} = \frac{0.3}{0.5} = \mathbf{0.6 \text{ G}}

Alternatively: $B = \sqrt{B_H^2 + B_V^2} = \sqrt{0.09 + 0.27} = \sqrt{0.36} = 0.6$ G. Same answer.

Why This Works

The Earth behaves like a giant bar magnet with its magnetic south pole near the geographic north pole. At any location, the field vector points into the Earth at some angle — this angle with the horizontal is called the angle of dip (or inclination).

graph TD
    A["Earth's Magnetic Field B"] --> B["Horizontal Component B_H = B cos δ"]
    A --> C["Vertical Component B_V = B sin δ"]
    B --> D["Points toward magnetic north"]
    C --> E["Points into the Earth in northern hemisphere"]
    A --> F["Three Magnetic Elements"]
    F --> G["Declination: angle between geographic and magnetic north"]
    F --> H["Inclination / Dip: angle of B with horizontal"]
    F --> I["Horizontal Component B_H"]

The three magnetic elements — declination, inclination, and horizontal component — completely describe the Earth’s field at any location. If you know any two of the three quantities ( $B$ , $B_H$ , $B_V$ , $\delta$ ), you can find the rest.

At the equator, $\delta = 0°$ so $B_H = B$ and $B_V = 0$ . At the magnetic poles, $\delta = 90°$ so $B_H = 0$ and $B_V = B$ . This appeared as a conceptual MCQ in NEET 2022.

Alternative Method

If you forget the component relations, just draw a right triangle with $B$ as the hypotenuse, $B_H$ as the base, and $B_V$ as the height. The angle between $B$ and $B_H$ is the dip angle $\delta$ . All the relations fall out from basic trigonometry.

For quick calculations: when $\delta = 45°$ , $B_H = B_V$ and $B = B_H\sqrt{2}$ . When $\delta = 60°$ , $B_V = \sqrt{3} \cdot B_H$ .

Common Mistake

Confusing declination with inclination. Declination is the angle between geographic north and magnetic north (measured in the horizontal plane). Inclination (dip) is the angle of the field with the horizontal (measured in the vertical plane containing the field). These are two completely different angles — mixing them up gives wrong answers and costs marks in boards.

Another trap: some problems give the angle of dip at a place and ask what happens if you tilt the dip circle. The apparent dip is different from the true dip unless the dip circle is in the magnetic meridian. The relation is $\cot^2\delta = \cot^2\delta_1 + \cot^2\delta_2$ for two perpendicular planes. This appeared in JEE Main 2023.

B_H = B\cos\delta, \quad B_V = B\sin\delta, \quad \tan\delta = \frac{B_V}{B_H}

B = \sqrt{B_H^2 + B_V^2}

Apparent dip in two perpendicular planes: $\cot^2\delta = \cot^2\delta_1 + \cot^2\delta_2$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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