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Magnetism and Matter - Earth’s Magnetic Field

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Earth's magnetic field behaves as if a powerful bar magnet is placed at its center, with its south pole pointing towards the geographic North.

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Magnetic Declination (θ\theta): The angle between the geographic meridian and the magnetic meridian at a point on Earth's surface.

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Magnetic Inclination or Dip (δ\delta): The angle made by the total magnetic field of the Earth (BeB_e) with the horizontal direction in the magnetic meridian. At the magnetic equator, δ=0∘\delta = 0^\circ; at the magnetic poles, δ=90∘\delta = 90^\circ.

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Horizontal Component (BHB_H): The component of the total magnetic field of Earth (BeB_e) in the horizontal direction. This component is measured using a vibration magnetometer.

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Magnetic Meridian: An imaginary vertical plane passing through the magnetic axis of a freely suspended magnet at a location.

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Geographic Meridian: A vertical plane passing through the geographic North and South poles at a location.

📐Formulae

BH=Becos⁡δB_H = B_e \cos \delta

BV=Besin⁡δB_V = B_e \sin \delta

tan⁡δ=BVBH\tan \delta = \frac{B_V}{B_H}

Be=BH2+BV2B_e = \sqrt{B_H^2 + B_V^2}

tan⁡δ′=tan⁡δcos⁡β (Apparent Dip formula, where β is the angle with the magnetic meridian)\tan \delta' = \frac{\tan \delta}{\cos \beta} \text{ (Apparent Dip formula, where } \beta \text{ is the angle with the magnetic meridian)}

💡Examples

Problem 1:

In the magnetic meridian of a certain place, the horizontal component of the Earth's magnetic field is 0.26 G0.26 \, G and the dip angle is 60∘60^\circ. What is the magnetic field of the Earth at this location?

Solution:

Given BH=0.26 GB_H = 0.26 \, G and δ=60∘\delta = 60^\circ. We use the formula BH=Becos⁡δB_H = B_e \cos \delta. Be=BHcos⁡δ=0.26cos⁡60∘B_e = \frac{B_H}{\cos \delta} = \frac{0.26}{\cos 60^\circ} Be=0.260.5=0.52 GB_e = \frac{0.26}{0.5} = 0.52 \, G

Explanation:

The total magnetic field BeB_e is related to its horizontal component BHB_H via the cosine of the angle of dip. Dividing the horizontal component by cos⁡δ\cos \delta gives the magnitude of the total vector.

Problem 2:

At a certain location, the horizontal component of Earth's magnetic field is equal to the vertical component. Find the angle of dip at that place.

Solution:

Given BH=BVB_H = B_V. Using the relation tan⁡δ=BVBH\tan \delta = \frac{B_V}{B_H}, we get: tan⁡δ=BHBH=1\tan \delta = \frac{B_H}{B_H} = 1 δ=tan⁡−1(1)=45∘\delta = \tan^{-1}(1) = 45^\circ

Explanation:

When the vertical and horizontal components are equal in magnitude, the resultant magnetic field vector bisects the angle between the horizontal and vertical axes, resulting in a dip angle of 45∘45^\circ.