Examkin

Question

Difficulty: EasyMagnetism and Earth's Magnetic Field

At a point on the Earth's surface, the total magnetic intensity is 4.0×10−5 T4.0 \times 10^{-5}\text{ T} and the angle of dip is 60∘60^\circ. What is the horizontal component of the Earth's magnetic field at this point?

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

Answer

The horizontal component of the Earth's magnetic field is 2.0×10−5 T2.0 \times 10^{-5}\text{ T}.
The horizontal component BhB_h of the Earth's magnetic field is given by resolving the total magnetic intensity BB along the horizontal direction using Bh=Bcos⁡θB_h = B \cos \theta. Substituting B=4.0×10−5 TB = 4.0 \times 10^{-5}\text{ T} and θ=60∘\theta = 60^\circ yields Bh=4.0×10−5×0.5=2.0×10−5 TB_h = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}.

Step-by-Step Solution

1
Identify the given values and formula
Total field B=4.0×10−5 TB = 4.0 \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 Earth's magnetic field is obtained by resolving the total magnetic vector along the horizontal plane.
2
Substitute the given values into the equation
Bh=4.0×10−5 T×cos⁡(60∘)=4.0×10−5×0.5=2.0×10−5 TB_h = 4.0 \times 10^{-5}\text{ T} \times \cos(60^\circ) = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}
Since cos⁡(60∘)=0.5\cos(60^\circ) = 0.5, evaluating the product gives the exact horizontal component.

Key Concept

Resolution of Earth's Magnetic Field Components
Rate this question

Topics

Magnetism and Earth's Magnetic Field

More questions from this topic