An aircraft with a wingspan of $40 m$ flies with a speed of $1080 k m / h r$ in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earth's magnetic field $1.75 \times 10^{- 5} T$ . Then the emf developed between the tips of the wings is

0.5 V

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0.34 V

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0.21 V

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2.1 V

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Solution

The correct option is D $0.21 V$
Given : $B = 1.75 \times 10^{- 5} T$ $l = 40 m$
Speed of aircraft $v = 1080$ km/h $= 1080 \times \frac{5}{18} = 300$ $m / s$
Emf induced $E = B v l$
$∴$ $E = (1.75 \times 10^{- 5}) (300) (40) = 0.21$ volts

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An aircraft with a wingspan of 40 m flies with a speed of 1080 km / hr in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earth’s magnetic field $1.75 \times 10^{- 5} T$ . Then the emf developed between the tips of the wings is

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An aircraft with a wingspan of 40 m flies with a speed of 1080 km/hr in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earth's magnetic field 1.75×10−5T. Then the emf developed between the tips of the wings is

The correct option is D 0.21 VGiven : B=1.75×10−5T l=40mSpeed of aircraft v=1080 km/h =1080×518=300 m/sEmf induced E=Bvl∴ E=(1.75×10−5)(300)(40)=0.21 volts

An aircraft with a wingspan of $40 m$ flies with a speed of $1080 k m / h r$ in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earth's magnetic field $1.75 \times 10^{- 5} T$ . Then the emf developed between the tips of the wings is

The correct option is D $0.21 V$
Given : $B = 1.75 \times 10^{- 5} T$ $l = 40 m$
Speed of aircraft $v = 1080$ km/h $= 1080 \times \frac{5}{18} = 300$ $m / s$
Emf induced $E = B v l$
$∴$ $E = (1.75 \times 10^{- 5}) (300) (40) = 0.21$ volts