TS EAMCET · Physics · Magnetic Properties of Matter
At a certain place, the angle of dip is \(60^{\circ}\) and the horizontal component of the earth's magnetic field \(\left(B_H\right)\) is \(0.8 \times 10^{-4} \mathrm{~T}\). The earth's overall magnetic field is
- A \(1.5 \times 10^{-4} \mathrm{~T}\)
- B \(1.6 \times 10^{-3} \mathrm{~T}\)
- C \(1.5 \times 10^{-3} \mathrm{~T}\)
- D \(1.6 \times 10^{-4} \mathrm{~T}\)
Answer & Solution
Correct Answer
(D) \(1.6 \times 10^{-4} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Given, \(\begin{aligned} B_H & =0.8 \times 10^{-4} \mathrm{~T} \\ \theta & =60^{\circ} \\ B_e & =? \end{aligned}\) We know that, \(\begin{aligned} B_H & =B_e \cos \theta \\ 0.8 \times 10^{-4} & =B_e \cos 60^{\circ} \end{aligned}\)…
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