TS EAMCET · Physics · Magnetic Properties of Matter
The earth's magnetic field at the geometric poles is \(\sqrt{10} \times 10^5 \mathrm{~T}\). The magnitude of the field at a point on the earth's surface where the radius makes an angle \(\theta\) with the axis of earth's assumed magnetic dipole is \(5 \times 10^5 \mathrm{~T}\). The magnitude of \(\theta\) in degree is :
- A \(30^{\circ}\)
- B \(60^{\circ}\)
- C \(45^{\circ}\)
- D \(75^{\circ}\)
Answer & Solution
Correct Answer
(C) \(45^{\circ}\)
Step-by-step Solution
Detailed explanation
According to the question, In \(\triangle O B C\) \(\because \cos \theta=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{R}}}=\frac{\sqrt{10} \times 10^{-5}}{5 \times 10^{-5}}=\frac{\sqrt{2}}{\sqrt{5}}\) or \(\theta=50.76^{\circ} \approx 45^{\circ}\) Hence, the magnitude of…
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