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Zorluk: ZorMagnetism and Earth's Magnetic Field

A dip needle placed within the magnetic meridian at a location on Earth's surface records an angle of dip of 60∘60^\circ. If the vertical component of Earth's magnetic field at this location is 3.46×10−5 T3.46 \times 10^{-5}\text{ T}, what is the horizontal component of Earth's magnetic field? (Take tan⁡60∘=1.73\tan 60^\circ = 1.73, sin⁡60∘=0.87\sin 60^\circ = 0.87, cos⁡60∘=0.50\cos 60^\circ = 0.50)

  1. 2.0×10−5 T2.0 \times 10^{-5}\text{ T}Cevap
  2. B
    6.0×10−5 T6.0 \times 10^{-5}\text{ T}
  3. C
    1.73×10−5 T1.73 \times 10^{-5}\text{ T}
  4. D
    4.0×10−5 T4.0 \times 10^{-5}\text{ T}

Cevap

The horizontal component of Earth's magnetic field is 2.0×10−5 T2.0 \times 10^{-5}\text{ T}.
The horizontal component BhB_h and vertical component BvB_v of Earth's magnetic field are related by tan⁡θ=BvBh\tan \theta = \frac{B_v}{B_h}, where θ\theta is the inclination or dip angle. Substituting Bv=3.46×10−5 TB_v = 3.46 \times 10^{-5}\text{ T} and tan⁡60∘=1.73\tan 60^\circ = 1.73 yields Bh=3.46×10−51.73=2.0×10−5 TB_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}.

Adım Adım Çözüm

1
Identify the relationship between the vertical component (BvB_v), horizontal component (BhB_h), and dip angle (θ\theta).
tan⁡θ=BvBh\tan \theta = \frac{B_v}{B_h}
The angle of dip θ\theta is defined by the direction of Earth's total magnetic field relative to the horizontal plane.
2
Rearrange the equation to express BhB_h in terms of BvB_v and tan⁡θ\tan \theta.
Bh=Bvtan⁡θB_h = \frac{B_v}{\tan \theta}
We need to solve for the horizontal component BhB_h.
3
Substitute the given values Bv=3.46×10−5 TB_v = 3.46 \times 10^{-5}\text{ T} and tan⁡60∘=1.73\tan 60^\circ = 1.73 into the equation.
Bh=3.46×10−51.73=2.0×10−5 TB_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}
Performing the numerical division yields the correct horizontal field intensity.

Anahtar Kavram

Resolution of Earth's Magnetic Field Components
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Konular

Magnetism and Earth's Magnetic Field

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