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A jet plane having a wing-span of $$25$$ m is travelling horizontally towards east with a speed of $$3600$$ km/hour. If the Earth's magnetic field at the location is $$4\times 10^{-4}$$T and the angle of dip is $$30^o$$, then, the potential difference between the ends of the wing is?


A
4 V
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B
5 V
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C
2 V
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D
2.5 V
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Solution

The correct option is C $$5$$ V
$$e=B_v/v$$ But $$\dfrac{B_v}{B_H}=lan 30^o=\dfrac{1}{\sqrt{3}}\Rightarrow B_v=\dfrac{4\times 10^{-4}}{\sqrt{3}}T$$
$$\therefore e=\dfrac{4\times 10^{-4}}{\sqrt{3}}\times 25\times\left(3600\times \dfrac{5}{18}\right)=5.77V$$.

Physics
NCERT
Standard XII

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