Question

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

• A. 200 V
• B. 20 V
• C. 2 V
• D. 0.2 V

Answer 1

Answer: (C)

Here,
Velocity of the aeroplane, $v=360km{h}^{-1}$
$=360\times \frac{5}{18}m{s}^{-1}=100m{s}^{-1}$
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
$\epsilon =B_{v}lv$
$=\left(4\times 10^{-4}Wb{m}^{-2}\right)\left(50m\right)\left(100m{s}^{-1}\right)$
$=2V$