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Difficulty: MediumMagnetism and Earth's Magnetic Field

At a geographical survey location, the horizontal component of the Earth's magnetic field is measured as 3.0×10−5 T3.0 \times 10^{-5}\text{ T}. If the angle of dip at this location is 60∘60^\circ, what is the magnitude of the vertical component of the Earth's magnetic field?

  1. A
    1.7×10−5 T1.7 \times 10^{-5}\text{ T}
  2. 5.2×10−5 T5.2 \times 10^{-5}\text{ T}Answer
  3. C
    1.5×10−5 T1.5 \times 10^{-5}\text{ T}
  4. D
    6.0×10−5 T6.0 \times 10^{-5}\text{ T}

Answer

5.2×10−5 T5.2 \times 10^{-5}\text{ T}
The vertical component (BvB_v) and horizontal component (BhB_h) of Earth's magnetic field are related by the angle of dip (θ\theta) through tan⁡θ=BvBh\tan\theta = \frac{B_v}{B_h}. Substituting Bh=3.0×10−5 TB_h = 3.0 \times 10^{-5}\text{ T} and θ=60∘\theta = 60^\circ yields Bv=3.0×10−5×1.732=5.2×10−5 TB_v = 3.0 \times 10^{-5} \times 1.732 = 5.2 \times 10^{-5}\text{ T}.

Step-by-Step Solution

1
Identify the relationship between 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 dip angle θ\theta represents the inclination of the total magnetic field relative to the horizontal plane.
2
Rearrange the equation to solve for the vertical component BvB_v.
Bv=Bh×tan⁡θB_v = B_h \times \tan\theta
Multiplying both sides by BhB_h isolates BvB_v.
3
Substitute the given values into the formula and calculate.
Bv=3.0×10−5 T×tan⁡60∘=3.0×10−5 T×1.732=5.196×10−5 T≈5.2×10−5 TB_v = 3.0 \times 10^{-5}\text{ T} \times \tan 60^\circ = 3.0 \times 10^{-5}\text{ T} \times 1.732 = 5.196 \times 10^{-5}\text{ T} \approx 5.2 \times 10^{-5}\text{ T}
Using tan⁡60∘≈1.732\tan 60^\circ \approx 1.732 gives the vertical magnetic field magnitude.

Key Concept

Components of Earth's Magnetic Field and Angle of Dip
Estimated Time:1m 15s
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Topics

Magnetism and Earth's Magnetic Field

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