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Difficulty: MediumMagnetism and Earth's Magnetic Field

A dip needle placed in the magnetic meridian at a location on the Earth's surface makes an angle of 30∘30^\circ with the horizontal. If the total intensity of the Earth's magnetic field at this location is 6.0×10−5 T6.0 \times 10^{-5}\text{ T}, what is the vertical component of the Earth's magnetic field?

  1. 3.0×10−5 T3.0 \times 10^{-5}\text{ T}Answer
  2. B
    5.2×10−5 T5.2 \times 10^{-5}\text{ T}
  3. C
    6.0×10−5 T6.0 \times 10^{-5}\text{ T}
  4. D
    1.2×10−4 T1.2 \times 10^{-4}\text{ T}

Answer

The vertical component of the Earth's magnetic field is 3.0×10−5 T3.0 \times 10^{-5}\text{ T}.
The vertical component of the Earth's magnetic field is given by Bv=Bsin⁡θB_v = B \sin \theta. Substituting B=6.0×10−5 TB = 6.0 \times 10^{-5}\text{ T} and θ=30∘\theta = 30^\circ yields Bv=6.0×10−5×0.5=3.0×10−5 TB_v = 6.0 \times 10^{-5} \times 0.5 = 3.0 \times 10^{-5}\text{ T}.

Step-by-Step Solution

1
Identify the given physical quantities from the problem statement.
Total intensity B=6.0×10−5 TB = 6.0 \times 10^{-5}\text{ T} and inclination angle θ=30∘\theta = 30^\circ.
These parameters are required to compute the resolved component.
2
Recall the formula for resolving the vertical component of Earth's magnetic field.
Bv=Bsin⁡θB_v = B \sin \theta
The vertical component corresponds to the vertical side of the right triangle formed by total field BB and angle of dip θ\theta.
3
Substitute the given values into the formula and calculate.
Bv=(6.0×10−5 T)×sin⁡(30∘)=(6.0×10−5)×0.5=3.0×10−5 TB_v = (6.0 \times 10^{-5}\text{ T}) \times \sin(30^\circ) = (6.0 \times 10^{-5}) \times 0.5 = 3.0 \times 10^{-5}\text{ T}.
Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, multiplying 6.0×10−56.0 \times 10^{-5} by 0.50.5 yields 3.0×10−5 T3.0 \times 10^{-5}\text{ T}.

Key Concept

Resolution of Earth's total magnetic field into horizontal and vertical components
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Topics

Magnetism and Earth's Magnetic Field

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