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Difficulty: MediumMagnetism and Earth's Magnetic Field

At a geographical research station, a bar magnet placed along the magnetic meridian neutralizes the horizontal component of the Earth's magnetic field at its neutral points. If the measured horizontal component of the Earth's magnetic field at this site is 1.8×10−5 T1.8 \times 10^{-5}\text{ T} and the angle of dip (inclination) is 60∘60^\circ, what is the magnitude of the Earth's total magnetic field intensity?

  1. 3.6×10−5 T3.6 \times 10^{-5}\text{ T}Answer
  2. B
    0.9×10−5 T0.9 \times 10^{-5}\text{ T}
  3. C
    1.56×10−5 T1.56 \times 10^{-5}\text{ T}
  4. D
    3.12×10−5 T3.12 \times 10^{-5}\text{ T}

Answer

The magnitude of the Earth's total magnetic field intensity is 3.6×10−5 T3.6 \times 10^{-5}\text{ T}.
The horizontal component BhB_h of the Earth's magnetic field is related to the total magnetic intensity BB and the angle of dip θ\theta by Bh=Bcos⁡θB_h = B \cos\theta. Dividing 1.8×10−5 T1.8 \times 10^{-5}\text{ T} by cos⁡60∘=0.5\cos 60^\circ = 0.5 gives 3.6×10−5 T3.6 \times 10^{-5}\text{ T}.

Step-by-Step Solution

1
Identify given parameters and formula relating field components.
Horizontal component Bh=1.8×10−5 TB_h = 1.8 \times 10^{-5}\text{ T}, Angle of dip θ=60∘\theta = 60^\circ. Formula: Bh=Bcos⁡θB_h = B \cos\theta.
The horizontal component of the Earth's magnetic field is the projection of the total field onto the horizontal plane.
2
Rearrange formula to solve for the total magnetic field intensity BB.
B=Bhcos⁡θB = \frac{B_h}{\cos\theta}.
To isolate total magnetic field intensity BB from the given horizontal component.
3
Substitute numerical values and evaluate.
B=1.8×10−5 Tcos⁡60∘=1.8×10−50.5=3.6×10−5 TB = \frac{1.8 \times 10^{-5}\text{ T}}{\cos 60^\circ} = \frac{1.8 \times 10^{-5}}{0.5} = 3.6 \times 10^{-5}\text{ T}.
cos⁡60∘=0.5\cos 60^\circ = 0.5, yielding a simple exact calculation.

Key Concept

Resolution of Earth's Magnetic Field Components
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Topics

Magnetism and Earth's Magnetic Field

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