Edit to “Solution”

Valter edited
revision #19910 parent #19908 ← older
@@ -14,7 +14,7 @@Solution
1. Horizontal component ($\vec{B}_h$): Directed along the meridians. The cross product $\vec{v} \times \vec{B}_h$ yields a force directed strictly vertically. It causes a redistribution of charges between the roof and the bottom of the fuselage, creating a potential difference $U_{\text{vert}} = v B_{h\perp} H$, where $H$ is the height of the fuselage.
2. Vertical component ($\vec{B}_v$): Pierces the airplane from top to bottom (or bottom to top). The cross product $\vec{v} \times \vec{B}_v$ yields a force directed strictly sideways, along the wingspan. This creates a potential difference $U_{\text{wings}} = v B_v L$, where $L$ is the wingspan.
−Important note: Along the longitudinal axis of the airplane (from tail to nose), the Lorentz force has no projection, since this direction coincides with the velocity vector $\vec{v}$. The potential difference between the nose and the tail is strictly zero. (It could only arise in the hypothetical and absurd case if the airplane flew sideways, along the axis of its own wings).
+Important note: Along the longitudinal axis of the airplane (from tail to nose), the Lorentz force has no projection, since this direction coincides with the velocity vector $\vec{v}$. The potential difference between the nose and the tail is strictly zero. (It could only arise in the hypothetical and absurd case if the airplane flew sideways, along the axis of its own wings, which would be, to say the least, funny).
To determine where the maximum voltage is generated, let us compare $U_{\text{vert}}$ and $U_{\text{wings}}$. The wingspan of passenger airliners ($L \approx 30 \div 80 \text{ m}$) exceeds the fuselage height ($H \approx 5 \div 8 \text{ m}$) by approximately an order of magnitude. Furthermore, in mid-latitudes, the vertical component of the magnetic field $\vec{B}_v$ is comparable to the horizontal one $\vec{B}_h$ or even exceeds it.
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