An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is \(2.5 \times 10^{-4}\) Wb/m\(^2\) and the angle of dip is \(60^\circ\). The emf induced between the tips of the plane wings will be __________.
Show Hint
- 88.37 mV
- 62.50 mV
- 54.125 mV
- 108.25 mV
The Correct Option is D
Solution and Explanation
Vertical component of earth's field \(B_V = B\sin 60^\circ = 2.5\times10^{-4} \times 0.866 = 2.165\times10^{-4}\) T.
Induced emf \(\varepsilon = B_V\, l\, v = 2.165\times10^{-4} \times 10 \times 50 = 0.10825\) V \(= 108.25\) mV.
Answer: (4) 108.25 mV
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