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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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An aeroplane in which the distance between the tips of using is $50\;m$ is flying horizontally with speed of $360\;km/hr$ over a place where the vertical component of earth's magnetic field is $2.0 \times 10^{-4}\;Weber /m^{2}$ The potential difference between the tips of wings will be

$(a)\;0.1\;V \\ (b)\;1.0\;V \\(c)\;0.2\;V \\(d)\;2.0\;V $

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Induced Emf
$e=B_v vl$
Velocity of plane $= \large\frac{360 \times 1000}{3600}$$=100\;m/s$
$\therefore e= 2 \times 10^{-4} \times 100 \times 50$
$\qquad= 1V$
Hence b is the correct answer.
answered Mar 21, 2014 by meena.p

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