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Magnetism and Matter · Class 12 · CBSE · Science Physics

The horizontal component of Earth's magnetic field at a place is 3.6 times 10^{-5}text{ T} and the angle of dip is 60^circ. The magnitude of the Eart…

Practice question from Magnetism and Matter (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

medium · MCQ · Q-BA3DF1

The horizontal component of Earth's magnetic field at a place is 3.6×10−5 T3.6 \times 10^{-5}\text{ T} and the angle of dip is 60∘60^\circ. The magnitude of the Earth's magnetic field at this place is:

  1. A.1.8 \times 10^{-5}\text{ T}
  2. B.3.6 \times 10^{-5}\text{ T}
  3. C.7.2 \times 10^{-5}\text{ T}Correct answer
  4. D.7.2 \times 10^{-4}\text{ T}

Solution

The horizontal component is BH=Bcos⁡δB_H = B \cos\delta, where BB is total Earth's magnetic field and δ\delta is the angle of dip. Given BH=3.6×10−5 TB_H = 3.6 \times 10^{-5}\text{ T} and δ=60∘\delta = 60^\circ. Therefore, B=BHcos⁡(60∘)=3.6×10−51/2=7.2×10−5 TB = \frac{B_H}{\cos(60^\circ)} = \frac{3.6 \times 10^{-5}}{1/2} = 7.2 \times 10^{-5}\text{ T}.

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