JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 2)
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 __________.
88.37 mV
62.50 mV
54.125 mV
108.25 mV
Explanation
$$\varepsilon $$ind = (Bv) LV and Bv = BTotal sin60o
$$ \therefore $$ $$\varepsilon $$ind = (2.5 $$\times$$ 10$$-$$4)(sin 60o) $$\times$$ 10 $$\times$$ 180 $$\times$$ $${5 \over {18}}$$
= 108.25 mV
$$ \therefore $$ $$\varepsilon $$ind = (2.5 $$\times$$ 10$$-$$4)(sin 60o) $$\times$$ 10 $$\times$$ 180 $$\times$$ $${5 \over {18}}$$
= 108.25 mV
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