Question

# A jet plane is traveling towards the west at speed of $1800\mathrm{km}/\mathrm{h}$. What is the voltage difference developed between the ends of the wing having a span of $25\mathrm{m}$, if the Earth's magnetic field at the location has a magnitude of $5×{10}^{-4}\mathrm{T}$ and the dip angle is $30°$?

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Solution

## Step 1: Given data and diagram,Speed of jet $\mathrm{v}=1800\mathrm{Km}{\mathrm{h}}^{-1}$Length of wing span $\mathrm{l}=25\mathrm{m}$Magnetic field $\mathrm{B}=5×{10}^{-4}\mathrm{T}$Dip angle $\mathrm{\delta }=30°$ Step 2: Formula UsedThe vertical component of Earth's magnetic field is given by the formula ${B}_{v}=B\mathrm{sin}\left(\delta \right)$. The induced voltage can be given by the formula $E={B}_{v}vl$.Step 3: Calculate the voltage difference developed between the ends of the wing:The velocity of the jet plane, $v=1800\mathrm{km}/\mathrm{h}$. Convert it into $m/s$ by multiplying by $\frac{1000}{3600}$.$⇒v=\frac{1800×1000}{3600}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}⇒v=\frac{1800000}{3600}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}⇒v=500\mathrm{m}/\mathrm{s}$Calculate the vertical component of Earth's magnetic field (here, the vertical component is considered because only the vertical component of magnetic field is normal to both the wings and the direction of motion)$⇒{B}_{v}=\left(5×{10}^{-4}\right)\mathrm{sin}\left(30°\right)\mathrm{T}\phantom{\rule{0ex}{0ex}}⇒{B}_{v}=\left(5×{10}^{-4}\right)×\frac{1}{2}\mathrm{T}\phantom{\rule{0ex}{0ex}}⇒{B}_{v}=2.5×{10}^{-4}\mathrm{T}$Calculate the induced voltage$⇒E=\left(2.5×{10}^{-4}\right)×500×25\mathrm{V}\phantom{\rule{0ex}{0ex}}⇒E=3.125\mathrm{V}$Hence, the voltage difference developed between the ends of the wing is $3.125\mathrm{V}$.

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