Q. An aeroplane is flying horizontally with a velocity of $360\, km\, h^{-1}$. The distance between the tips of the wings of the aeroplane is $50\, m$. The vertical component of the earth's magnetic field is $4 \times 10^{-4} \, Wb \, m^{-2}$. The induced e.m.f. is
Solution:
Here,
Velocity of the aeroplane, $v = 360\, km \, h^{-1}$
$ = 360 \times \frac{5}{18} m \, s^{-1} = 100 \, m \, s^{-1}$
Distanse between the tips of the wings, $l=50m$
Verticai component of earth's magnetic field,
$Bv= 4 \times 10^{-4}\, Wb \, m^{-2}$
$ \therefore $ The induced e.m.f. between the tips of its wings is
$\varepsilon = B_{v} lv $
$ = \left(4 \times10^{-4} Wb m^{-2}\right)\left(50 m \right)\left( 100 m s^{-1}\right)$
$ = 2 V$
