The distance between the ends of the wings of an airplane is $50 m$ . It is flying in a horizontal plane at a speed of $360 K m / h o u r$ . The vertical component of earth's magnetic field at that place is $2.0 \times 10^{- 4} W b / m^{2}$ , then the potential difference induced between the ends of the wings will be

0.1 v o l t

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1.0 v o l t

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0.2 v o l t

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0.01 v o l t

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Solution

The correct option is D $1.0 v o l t$
$e m f = B l v$
$= 2 \times 10^{- 4} \times 50 \times 360 \times \frac{1000}{3600} = 1 V$

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The distance between the ends of the wings of an airplane is 50m. It is flying in a horizontal plane at a speed of 360Km/hour. The vertical component of earth's magnetic field at that place is 2.0×10−4Wb/m2, then the potential difference induced between the ends of the wings will be

The correct option is D 1.0voltemf=Blv=2×10−4×50×360×10003600=1V

The correct option is D $1.0 v o l t$
$e m f = B l v$
$= 2 \times 10^{- 4} \times 50 \times 360 \times \frac{1000}{3600} = 1 V$