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Magnetic Effects of Current and Magnetism - Magnetism and Matter (Earth’s Magnetic Field)

Grade 12ICSEPhysics

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

🔑Concepts

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Earth behaves like a magnetic dipole with its magnetic south pole located near the geographic North Pole and its magnetic north pole near the geographic South Pole.

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Magnetic Meridian: A vertical plane passing through the magnetic axis of a freely suspended magnet.

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

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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 that the total magnetic field B⃗\vec{B} of the earth makes 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 of Earth's Magnetic Field (BHB_H): The component of the earth's total magnetic field in the horizontal direction. It is given by BH=Bcos⁡δB_H = B \cos \delta.

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Vertical Component of Earth's Magnetic Field (BVB_V): The component of the earth's total magnetic field in the vertical direction. It is given by BV=Bsin⁡δB_V = B \sin \delta.

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Apparent Dip: If a dip circle is placed in a plane inclined at an angle α\alpha to the magnetic meridian, the observed dip δ′\delta' is called the apparent dip, related to true dip by tan⁡δ′=tan⁡δcos⁡α\tan \delta' = \frac{\tan \delta}{\cos \alpha}.

📐Formulae

BH=Bcos⁡δB_H = B \cos \delta

BV=Bsin⁡δB_V = B \sin \delta

B=BH2+BV2B = \sqrt{B_H^2 + B_V^2}

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

tan⁡δ′=tan⁡δcos⁡α\tan \delta' = \frac{\tan \delta}{\cos \alpha}

cot⁡2δ=cot⁡2δ1+cot⁡2δ2\cot^2 \delta = \cot^2 \delta_1 + \cot^2 \delta_2

💡Examples

Problem 1:

At a certain location, the horizontal component of the Earth's magnetic field is 0.3×10−4 T0.3 \times 10^{-4} \text{ T} and the angle of dip is 60∘60^\circ. Calculate the total magnetic field of the Earth at that location.

Solution:

Given: BH=0.3×10−4 TB_H = 0.3 \times 10^{-4} \text{ T}, δ=60∘\delta = 60^\circ. We know that BH=Bcos⁡δB_H = B \cos \delta. Therefore, B=BHcos⁡δ=0.3×10−4cos⁡60∘=0.3×10−40.5=0.6×10−4 TB = \frac{B_H}{\cos \delta} = \frac{0.3 \times 10^{-4}}{\cos 60^\circ} = \frac{0.3 \times 10^{-4}}{0.5} = 0.6 \times 10^{-4} \text{ T}.

Explanation:

The relationship between the horizontal component and the total field is used here. By dividing the horizontal component by the cosine of the angle of dip, we find the magnitude of the total magnetic field vector.

Problem 2:

A dip circle shows an apparent dip of 45∘45^\circ in a plane which is at an angle of 30∘30^\circ with the magnetic meridian. Determine the true dip δ\delta at that place.

Solution:

Given: Apparent dip δ′=45∘\delta' = 45^\circ, angle with meridian α=30∘\alpha = 30^\circ. Using the formula tan⁡δ=tan⁡δ′cos⁡α\tan \delta = \tan \delta' \cos \alpha, we get tan⁡δ=tan⁡45∘×cos⁡30∘=1×32=0.866\tan \delta = \tan 45^\circ \times \cos 30^\circ = 1 \times \frac{\sqrt{3}}{2} = 0.866. Thus, δ=tan⁡−1(0.866)≈40.9∘\delta = \tan^{-1}(0.866) \approx 40.9^\circ.

Explanation:

When the dip circle is not in the magnetic meridian, the horizontal component effectively becomes BHcos⁡αB_H \cos \alpha, leading to a larger observed (apparent) dip. The formula tan⁡δ′=tan⁡δcos⁡α\tan \delta' = \frac{\tan \delta}{\cos \alpha} is rearranged to solve for the true dip.

Magnetism and Matter (Earth’s Magnetic Field) Revision - Class 12 Physics ICSE