An aeroplane, with its wings spread $10 \mathrm{~m}$, is flying at a speed of $180 \mathrm{~km} / \mathrm{h}$ in a horizontal direction. The total intensity of earth's field at that part is $2.5 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}$ and the angle of dip is $60^{\circ}$. The emf induced between the tips of the plane wings will be :-
Correct Option: 1
$\epsilon=[\overrightarrow{\mathrm{B}} \overrightarrow{\mathrm{V}} \overrightarrow{\mathrm{L}}]=\mathrm{BVL} \sin \theta$
$=\left(2.5 \times 10^{-4} \mathrm{~T}\right)\left(180 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\right)(10 \mathrm{~m}) \sin 60^{\circ}$
$=108.25 \times 10^{-3} \mathrm{~V}$
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