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| An airplane can be considered as a solid metal conductor moving in the Earth's magnetic field. It contains a multitude of free charges (electrons). During the motion, a Lorentz force acts on each charge $q$ moving with velocity $\vec{v}$ in a magnetic field with induction $\vec{B}$: | | An airplane can be considered as a solid metal conductor moving in the Earth's magnetic field. It contains a multitude of free charges (electrons). During the motion, a Lorentz force acts on each charge $q$ moving with velocity $\vec{v}$ in a magnetic field with induction $\vec{B}$: |
| $$\vec{F}_m = q(\vec{v} \times \vec{B})$$ | | $$\vec{F}_m = q(\vec{v} \times \vec{B})$$ |
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| According to the property of the cross product, the Lorentz force is always perpendicular to the velocity vector of the airplane $\vec{v}$. | | According to the property of the cross product, the Lorentz force is always perpendicular to the velocity vector of the airplane $\vec{v}$. |
| The Earth's magnetic field vector $\vec{B}$ can be decomposed into two independent components: horizontal ($\vec{B}_h$) and vertical ($\vec{B}_v$). | | The Earth's magnetic field vector $\vec{B}$ can be decomposed into two independent components: horizontal ($\vec{B}_h$) and vertical ($\vec{B}_v$). |
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| Let us consider where these components displace the electrons: | | Let us consider where these components displace the electrons: |
| 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. | | 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. | | 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. |
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| 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). |
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| 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. | | 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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| Therefore, for the majority of flights, the product of the values for the wings is significantly larger: | | Therefore, for the majority of flights, the product of the values for the wings is significantly larger: |
| $$v B_v L \gg v B_{h\perp} H$$ | | $$v B_v L \gg v B_{h\perp} H$$ |
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| Exception (Magnetic Equator): At the magnetic equator, the magnetic field lines of the Earth are parallel to its surface ($B_v \approx 0$, while $B_h$ reaches its maximum). If the airplane flies near the magnetic equator strictly to the west or east (the velocity vector is perpendicular to $\vec{B}_h$), then $v B_v L \approx 0$. In this unique case, the maximum voltage will arise between the bottom and the roof of the fuselage ($U_{\text{vert}} > U_{\text{wings}}$). | | Exception (Magnetic Equator): At the magnetic equator, the magnetic field lines of the Earth are parallel to its surface ($B_v \approx 0$, while $B_h$ reaches its maximum). If the airplane flies near the magnetic equator strictly to the west or east (the velocity vector is perpendicular to $\vec{B}_h$), then $v B_v L \approx 0$. In this unique case, the maximum voltage will arise between the bottom and the roof of the fuselage ($U_{\text{vert}} > U_{\text{wings}}$). |
