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Question

Physics Question on Electromagnetic induction

An aeroplane is flying horizontally with a velocity of 360kmh1360\, km\, h^{-1}. The distance between the tips of the wings of the aeroplane is 50m50\, m. The vertical component of the earth's magnetic field is 4×104Wbm24 \times 10^{-4} \, Wb \, m^{-2}. The induced e.m.f. is

A

200 V

B

20 V

C

2 V

D

0.2 V

Answer

2 V

Explanation

Solution

Here,
Velocity of the aeroplane, v=360kmh1v = 360\, km \, h^{-1}
=360×518ms1=100ms1= 360 \times \frac{5}{18} m \, s^{-1} = 100 \, m \, s^{-1}
Distanse between the tips of the wings, l=50ml=50m
Verticai component of earth's magnetic field,
Bv=4×104Wbm2Bv= 4 \times 10^{-4}\, Wb \, m^{-2}
\therefore The induced e.m.f. between the tips of its wings is
ε=Bvlv\varepsilon = B_{v} lv
=(4×104Wbm2)(50m)(100ms1)= \left(4 \times10^{-4} Wb m^{-2}\right)\left(50 m \right)\left( 100 m s^{-1}\right)
=2V= 2 V