An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km h$$^{-1}$$ in a horizontal direction. The total intensity of earth's field at that part is $$2.5 \times 10^{-4}$$ Wb m$$^{-2}$$ and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

The aeroplane flies horizontally with its wings spread $$l = 10$$ m at speed $$v = 180$$ km/h $$= 50$$ m/s. The total intensity of earth's magnetic field is $$B = 2.5 \times 10^{-4}$$ Wb/m$$^2$$ and the angle of dip is $$\delta = 60°$$.

When the plane flies horizontally, the EMF is induced due to the vertical component of the earth's magnetic field, which is $$B_v = B\sin\delta = 2.5 \times 10^{-4} \times \sin 60° = 2.5 \times 10^{-4} \times \frac{\sqrt{3}}{2}$$.

The induced EMF between the wing tips is $$\varepsilon = B_v \times l \times v = 2.5 \times 10^{-4} \times \frac{\sqrt{3}}{2} \times 10 \times 50$$.

Calculating: $$\varepsilon = 2.5 \times 10^{-4} \times 0.866 \times 500 = 2.5 \times 10^{-4} \times 433 = 0.10825$$ V $$= 108.25$$ mV.