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An airplane with a 20 wingspread is flyi...

An airplane with a 20 wingspread is flying at 250m/s straight south parallel to the earth's surface. The earths magnetic field has a horizontal component of `2xx10^(-5)Wb//m^(2)` and the dip angle is `60^(@)`. Calculate the induced e.m.f. between the plane tips.

Text Solution

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The magnitude of e.m.f induced across the tips of the wing of the air plane is given by `epsilon` = Blv `" "`…(1)
Now, here B = B, of earth.
Given `B_(H) = 2 xx 10^(5) wb//m^(2)` and the angle of dip `delta = 60^(@)` .
`therefore B_(v) = B_(H) tan 60^(@)`
`rArr B_(v) 2 xx 10^(-5) xx sqrt(3) xx 250 xx 20 ` yolt
`rArr epsilon ` = 0.173 volt.
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