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Zorluk: Çok zorMagnetism and Earth's Magnetic Field

A dip circle is set up in a vertical plane that is inclined at an angle of 60∘60^\circ to the magnetic meridian. The needle comes to rest at an apparent angle of dip of 45∘45^\circ. If the actual horizontal component of the Earth's magnetic field in the magnetic meridian is 4.0×10−5 T4.0 \times 10^{-5}\text{ T}, what is the vertical component of the Earth's magnetic field at that location?

  1. A
    4.0×10−5 T4.0 \times 10^{-5}\text{ T}
  2. 2.0×10−5 T2.0 \times 10^{-5}\text{ T}Cevap
  3. C
    3.5×10−5 T3.5 \times 10^{-5}\text{ T}
  4. D
    8.0×10−5 T8.0 \times 10^{-5}\text{ T}

Cevap

The vertical component of the Earth's magnetic field is 2.0×10−5 T2.0 \times 10^{-5}\text{ T}.
In a vertical plane inclined at an angle α\alpha to the magnetic meridian, the vertical component BVB_V remains constant while the effective horizontal component becomes BH′=BHcos⁡αB_H' = B_H \cos \alpha. Substituting BH=4.0×10−5 TB_H = 4.0 \times 10^{-5}\text{ T} and α=60∘\alpha = 60^\circ yields BH′=2.0×10−5 TB_H' = 2.0 \times 10^{-5}\text{ T}. Using the formula for apparent dip tan⁡θ′=BV/BH′\tan \theta' = B_V / B_H' with θ′=45∘\theta' = 45^\circ gives tan⁡45∘=1\tan 45^\circ = 1, which confirms BV=BH′=2.0×10−5 TB_V = B_H' = 2.0 \times 10^{-5}\text{ T}.

Adım Adım Çözüm

1
Determine the effective horizontal component of the magnetic field in the plane of inclination
BH′=BHcos⁡α=(4.0×10−5 T)×cos⁡60∘=2.0×10−5 TB_H' = B_H \cos \alpha = (4.0 \times 10^{-5}\text{ T}) \times \cos 60^\circ = 2.0 \times 10^{-5}\text{ T}
When a dip circle is rotated by an angle α\alpha away from the magnetic meridian, the horizontal component acting along the plane of the dip circle is reduced to BHcos⁡αB_H \cos \alpha.
2
Relate the apparent angle of dip to the vertical component and effective horizontal component
tan⁡θ′=BVBH′\tan \theta' = \frac{B_V}{B_H'}
The vertical component BVB_V remains unchanged regardless of the vertical plane's orientation.
3
Substitute the given values to solve for BVB_V
BV=BH′tan⁡45∘=(2.0×10−5 T)×1=2.0×10−5 TB_V = B_H' \tan 45^\circ = (2.0 \times 10^{-5}\text{ T}) \times 1 = 2.0 \times 10^{-5}\text{ T}
Since tan⁡45∘=1\tan 45^\circ = 1, the vertical component is equal to the resolved horizontal component.

Anahtar Kavram

Apparent Dip Angle and Resolution of Earth's Magnetic Field Components
Tahmini Süre:2m 0s
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Magnetism and Earth's Magnetic Field

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