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(A) Speed of the jet plane,$v = 1800\, km/h = 500\, m/s$.Wing span of the jet plane,$l = 25\, m$.Earth's magnetic field strength,$B = 5.0 	imes 10^{-

Vedclass · Accepted Answer

$3.125\, V$

Vedclass · Answer

(A) Speed of the jet plane,$v = 1800\, km/h = 500\, m/s$.Wing span of the jet plane,$l = 25\, m$.Earth's magnetic field strength,$B = 5.0 	imes 10^{-4}\, T$.Angle of dip,$\delta = 30^{\circ}$.The vertical component of the Earth's magnetic field is responsible for the induced $EMF$ across the wings as the plane moves horizontally.$B_v = B \sin \delta = 5 	imes 10^{-4} 	imes \sin 30^{\circ} = 5 	imes 10^{-4} 	imes 0.5 = 2.5 	imes 10^{-4}\, T$.The induced voltage difference between the ends of the wing is given by $e = B_v 	imes l 	imes v$.$e = 2.5 	imes 10^{-4} 	imes 25 	imes 500$.$e = 3.125\, V$.Thus,the voltage difference developed between the ends of the wings is $3.125\, V$.

$A$ jet plane is travelling towards the west at a speed of $1800\, km/h$. What is the voltage difference developed between the ends of the wing having a span of $25\, m$,if the Earth's magnetic field at the location has a magnitude of $5 \times 10^{-4}\, T$ and the dip angle is $30^{\circ}$?

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