An aircraft with a wingspan of 40 m flies with a speed of \( 1080\,\mathrm{km\,h^{-1}} \) in the eastward direction at a constant altitude in a region where the vertical component of the magnetic field is \( 1.75\times10^{-5}\,\mathrm{T} \). The emf that develops between the tips of the wings is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.21 V
A straight conductor moving perpendicular to a magnetic field develops a motional emf \( \varepsilon=Blv \) between its ends. The wings move horizontally. So only the vertical component of the field is perpendicular to both the wing and its velocity. Here \( l \) is the wingspan.
The speed in SI units is:
\( \begin{aligned} v &= 1080\times\frac{1000\,\mathrm{m}}{3600\,\mathrm{s}} \\ &= 300\,\mathrm{m\,s^{-1}} \end{aligned} \)
The emf between the wing tips is:
\( \begin{aligned} \varepsilon &= Blv \\ &= 1.75\times10^{-5}\,\mathrm{T}\times40\,\mathrm{m}\times300\,\mathrm{m\,s^{-1}} \\ &= 0.21\,\mathrm{V} \end{aligned} \)