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NCERT Class XII Chapter
Electromagnetic Induction
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Question : 10 of 17
Marks: +1, -0
A jet plane is travelling towards west at a speed of 1800 km h–1h^{–1}. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–410^{–4} T and the dip angle is 30°.
Solution:  
Earth magnetic field will have two components, BHB_H and BVB_V. It is vertical component which develop induced emf across the wing in N-S direction.
BvB_v = B sin δ
BvB_v = 5 × 10–410^{–4} sin 30°
BvB_v = 2.5 × 10–410^{–4} T
Induced emf ε = BvB_v vl (Since 1800 km h–1h^{–1} = 500 m s–1s^{–1})
ε = 2.5 × 10–410^{–4} × 500 × 25
ε = 3.125 V
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