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The wing span of an aeroplane is 40m. The plane is flying horizontally due north at 360 km/h. What is the potential difference developed between the wing-tips if the horizontla component of the Earth's magnetic field B_(h) = 3.2 xx 10^(-5)T and the angle of dip at the place is 60^(@) ? |
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Answer» Solution :Data: l=40m, V=360 km/H `=360 xx 5/18= 100` m/s, `B_(h) = 3.2 xx 10^(-5) T, delta = 60^(@)` The area swept out by the wing per UNIT time= lv. The magnetic lines of induction perpendicular to this area are those of the vertical component `B_(v)` of the Earth's magnetic field. `B_(v) = B_(h) tan delta` where `delta` is the angle of DIP. `therefore` The magnetic FLUX cut by the wing per unit time `=(dPhi_(m))/(dt) = B_(v)(lv) = B_(h)tan delta(lv)` `=(3.2 xx 10^(-5))(tan 60^(@))(40)(100)` `=(12.8 xx 10^(-2))(1.732) = 0.2217 Wb//s` `therefore` The induced emf, `E=-(dPhi_(m))/(dt)` (nummerically) `=0.2217` V |
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