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Difficulty: EasyMagnetism and Earth's Magnetic Field

At a specific location on the Earth's surface, the horizontal component of the Earth's magnetic field is 3.2×10−5 T3.2 \times 10^{-5}\text{ T} and the vertical component is also 3.2×10−5 T3.2 \times 10^{-5}\text{ T}. What is the angle of dip at this location?

  1. 45∘45^\circAnswer
  2. B
    0∘0^\circ
  3. C
    90∘90^\circ
  4. D
    30∘30^\circ

Answer

The angle of dip at this location is 45∘45^\circ.
The angle of dip θ\theta is defined by tan⁡θ=BvBh\tan\theta = \frac{B_v}{B_h}. Since the vertical and horizontal components of Earth's magnetic field are equal, their ratio is equal to 1. The angle whose tangent is 1 is 45∘45^\circ.

Step-by-Step Solution

1
Identify the relationship between the horizontal component (BhB_h), vertical component (BvB_v), and angle of dip (θ\theta).
tan⁡θ=BvBh\tan\theta = \frac{B_v}{B_h}
The angle of dip θ\theta is the angle made by the Earth's total magnetic field vector with the horizontal.
2
Substitute the given values into the formula.
tan⁡θ=3.2×10−5 T3.2×10−5 T=1\tan\theta = \frac{3.2 \times 10^{-5}\text{ T}}{3.2 \times 10^{-5}\text{ T}} = 1
Both BvB_v and BhB_h are given as 3.2×10−5 T3.2 \times 10^{-5}\text{ T}.
3
Calculate the angle θ\theta.
θ=arctan⁡(1)=45∘\theta = \arctan(1) = 45^\circ
The inverse tangent of 1 is 45∘45^\circ.

Key Concept

Earth's Magnetic Field Components and Angle of Dip
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Topics

Magnetism and Earth's Magnetic Field

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