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  • An air plane with 20m wing spread is flying at 250 m/s straight along south parallel to the earth surface. The earth’s magnetic field has a horizontal component of<img alt="\mathrm {2 \times

An air plane with 20m wing spread is flying at 250 m/s straight along south parallel to the earth surface. The earth’s magnetic field has a horizontal component of\mathrm {2 \times 10^{-5} Wb/m^2} and the dip angle is \mathrm {60^o} The induced emf.between the wing tips.

Option: 1

5 \sqrt{3} \times 10^{-3} \mathrm{~V}


Option: 2

3 \sqrt{3} \times 10^{-3} \mathrm{~V}


Option: 3

4 \sqrt{3} \times 10^{-3} \mathrm{~V}


Option: 4

\mathrm{zero}


Answers (1)

best_answer

The magnitude of e.m.f. induced across the tips of the wing of the air plane is given by
\varepsilon=\mathrm{B} \ell \mathrm{V} . . . (1)

Now, here \mathrm{B = Bv} of earth.

Given \mathrm{B}_{\mathrm{H}}=2 \times 10^{-5} \mathrm{wb} / \mathrm{m}^2 and the angle of dip \delta=60^{\circ} \text {. }

\mathrm{\begin{aligned} & \therefore B_V=B_H \tan 60^{\circ} . \\ & \Rightarrow B_V=2 \times 10^{-5} \times \sqrt{3} \mathrm{~Wb} / \mathrm{m}^2 \end{aligned}} . . . (2)

From (1) and (2) we obtain,

\mathrm{\begin{aligned} &\varepsilon=2 \times 10^{-5} \times \sqrt{3} \times 250 \text { volt }\\ &\Rightarrow \varepsilon=5 \sqrt{3} \times 10^{-3} \text { Volt. } \end{aligned}}

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jitender.kumar

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