"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$-$4 Wb/m2 and the angle of dip is 60$^\circ$. The emf induced between the tips of the plane wings will be __________."

Question

Accepted Answer

"$\varepsilon$_ind = (B_v) LV and B_v = B_Total sin60^o

$\therefore$ $\varepsilon$_ind = (2.5 $\times$ 10^$-$4)(sin 60^o) $\times$ 10 $\times$ 180 $\times$ ${5 \over {18}}$

= 108.25 mV "

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^$-$4 Wb/m² and the angle of dip is 60$^\circ$. The emf induced between the tips of the plane wings will be __________.

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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$-$4 Wb/m2 and the angle of dip is 60$^\circ$. The emf induced between the tips of the plane wings will be __________.

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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^$-$4 Wb/m² and the angle of dip is 60$^\circ$. The emf induced between the tips of the plane wings will be __________.