JEE PYQ: Electromagnetic Induction Question 5
Question 5 - 2021 (26 Feb Shift 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/m$^2$ and the angle of dip is $60°$. The emf induced between the tips of the plane wings will be:
(1) 88.37 mV
(2) 62.50 mV
(3) 54.125 mV
(4) 108.25 mV
Show Answer
Answer: (4)
Solution
The vertical component $B_V = B\sin 60° = \frac{\sqrt{3}}{2}B$. The induced EMF $= B_V \times l \times v = \frac{\sqrt{3}}{2} \times 2.5 \times 10^{-4} \times 10 \times 50 = 108.25 \times 10^{-3}$ V $= 108.25$ mV.
BETA
