TS EAMCET · Physics · Magnetic Properties of Matter
An aeroplane is travelling horizontally towards west with a speed of \(540 \mathrm{kmh}^{-1}\). The wing span of the plane is 20 \(\mathrm{m}\). If the horizontal component of the earth's magnetic field at the location is \(2.5 \sqrt{3} \times 10^{-4} \mathrm{~T}\) and the dip angle is \(30^{\circ}\), the potential difference developed between the ends of the wing is
- A \(1 \mathrm{~V}\)
- B \(1.5 \mathrm{~V}\)
- C \(0.75 \mathrm{~V}\)
- D \(0.5 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(0.75 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Speed of aeroplane, \(v=540 \mathrm{~km} / \mathrm{h}\) \(=540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\) Distance between the ends of the wings, \(1=20 \mathrm{~m}\) Magnetic field, \(B=2.5 \sqrt{3} \times 10^{-4} \mathrm{~T}\) Dip angle, \(\theta=30^{\circ}\) The potential…
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