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Difficulty: MediumMagnetism and Earth's Magnetic Field

A magnetic compass needle free to swing in a vertical plane comes to rest at an angle of dip of 30∘30^\circ to the horizontal at a given location. If the magnitude of the Earth's total magnetic field at this point is 5.0×10−5 T5.0 \times 10^{-5}\text{ T}, what is the vertical component of the Earth's magnetic field?

  1. 2.5×10−5 T2.5 \times 10^{-5}\text{ T}Answer
  2. B
    4.33×10−5 T4.33 \times 10^{-5}\text{ T}
  3. C
    1.0×10−4 T1.0 \times 10^{-4}\text{ T}
  4. D
    7.5×10−5 T7.5 \times 10^{-5}\text{ T}

Answer

2.5×10−5 T2.5 \times 10^{-5}\text{ T}
The vertical component of the Earth's magnetic field is found by multiplying the total magnetic field by the sine of the angle of dip: Bv=Bsin⁡θB_v = B \sin\theta. Substituting B=5.0×10−5 TB = 5.0 \times 10^{-5}\text{ T} and θ=30∘\theta = 30^\circ yields 2.5×10−5 T2.5 \times 10^{-5}\text{ T}.

Step-by-Step Solution

1
Identify given parameters and formula for vertical magnetic component
Total magnetic field B=5.0×10−5 TB = 5.0 \times 10^{-5}\text{ T} and angle of dip θ=30∘\theta = 30^\circ. Formula: Bv=Bsin⁡θB_v = B \sin\theta.
The vertical component BvB_v of the Earth's magnetic field is resolved by projecting the total magnetic flux density BB along the vertical axis using the sine of the inclination angle.
2
Substitute values and solve for BvB_v
Bv=5.0×10−5 T×sin⁡(30∘)=5.0×10−5 T×0.5=2.5×10−5 TB_v = 5.0 \times 10^{-5}\text{ T} \times \sin(30^\circ) = 5.0 \times 10^{-5}\text{ T} \times 0.5 = 2.5 \times 10^{-5}\text{ T}.
Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, multiplying gives the vertical component directly.

Key Concept

Resolution of Earth's Magnetic Field Components
Estimated Time:1m 0s
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Topics

Magnetism and Earth's Magnetic Field

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