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Difficulty: EasyMagnetism and Earth's Magnetic Field

At a specific location on Earth, the total magnetic field intensity is 40 μT40\ \mu\text{T} and the angle of dip is 60∘60^\circ. What is the magnitude of the horizontal component of Earth's magnetic field at this location, in μT\mu\text{T}?

Answer: 20 µT

Answer

The magnitude of the horizontal component of Earth's magnetic field is 20 μT20\ \mu\text{T}.
The horizontal component BhB_h of Earth's magnetic field is derived using Bh=Bcos⁡θB_h = B \cos\theta. Substituting B=40 μTB = 40\ \mu\text{T} and θ=60∘\theta = 60^\circ yields Bh=40×0.5=20 μTB_h = 40 \times 0.5 = 20\ \mu\text{T}.

Step-by-Step Solution

1
Recall the resolving formula for the horizontal component of Earth's magnetic field
Bh=Bcos⁡θB_h = B \cos\theta
The horizontal component is the vector projection of total field BB onto the horizontal plane inclined at angle θ\theta.
2
Evaluate the cosine function for 60∘60^\circ
cos⁡(60∘)=0.5\cos(60^\circ) = 0.5
Standard trigonometric value for 60∘60^\circ.
3
Multiply total magnetic field strength by cos⁡(60∘)\cos(60^\circ)
Bh=40×0.5=20 μTB_h = 40 \times 0.5 = 20\ \mu\text{T}
Calculates the exact horizontal component magnitude.

Key Concept

Components of Earth's Magnetic Field
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Topics

Magnetism and Earth's Magnetic Field

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