Новое решение

Alexphysics правка от 23 июн. 2026 г., 23:53

правка #19133 позже →

en/14.4.25.md +56 −0

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		+	### Statement
		+
		+	$14.4.25.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Physical criterion
		+
		+	At the magnetic equator, the field $\mathbf{B}$ is horizontal.
		+	An electron moving toward Earth experiences a perpendicular magnetic force that curves its trajectory.
		+	For it not to become trapped in a spiral, the Larmor radius must be at least of the order of the height h:
		+
		+	$r_L = \frac{p}{eB} \gtrsim h \quad\Rightarrow\quad p \approx e B h$.
		+
		+	Here p is the relativistic momentum of the electron
		+
		+	Calculation of the momentum
		+
		+	Substituting the values
		+
		+	$p = (1.6\times 10^{-19})(3.0\times 10^{-5})(10^6) = 4.8\times 10^{-18}\ \text{kg·m/s}$
		+
		+	We multiply by c to work in energy units:
		+
		+	$pc = 4.8\times 10^{-18} \times 3.0\times 10^8 = 1.44\times 10^{-9}\ \text{J}$
		+
		+	Convert to MeV $(1\ \text{MeV} = 1.6\times 10^{-13}\ \text{J})$
		+
		+	$pc = \frac{1.44\times 10^{-9}}{1.6\times 10^{-13}}\ \text{MeV} \approx 9.0\ \text{MeV}$
		+
		+	Kinetic energy
		+
		+	The electron is clearly ultra‑relativistic $(pc \gg m_e c^2)$ The exact relation between momentum and total energy is:
		+
		+	$E_{\text{total}} = \sqrt{(pc)^2 + (m_e c^2)^2}$
		+
		+	and the kinetic energy is obtained by subtracting the rest mass:
		+
		+	$K = E_{\text{total}} - m_e c^2$
		+
		+	Substituting the values:
		+
		+	$K = \sqrt{9.0^2 + 0.511^2} - 0.511
		+	\approx \sqrt{81 + 0.261} - 0.511
		+	\approx 9.014 - 0.511
		+	\approx 8.5\ \text{MeV}$
		+
		+	Result
		+
		+	\boxed{K_{\min} \approx 8.5\ \text{MeV}}.
		+
		+	Electrons need a kinetic energy of the order of several MeV to cross the equatorial magnetic field and reach the ground. Less energetic electrons become trapped in the radiation belts.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,56 @@
			### Statement

			$14.4.25.$ [Insert the problem statement]

			### Solution

			Physical criterion

			At the magnetic equator, the field $\mathbf{B}$ is horizontal.
			An electron moving toward Earth experiences a perpendicular magnetic force that curves its trajectory.
			For it not to become trapped in a spiral, the Larmor radius must be at least of the order of the height h:

			$r_L = \frac{p}{eB} \gtrsim h \quad\Rightarrow\quad p \approx e B h$.

			Here p is the relativistic momentum of the electron

			Calculation of the momentum

			Substituting the values

			$p = (1.6\times 10^{-19})(3.0\times 10^{-5})(10^6) = 4.8\times 10^{-18}\ \text{kg·m/s}$

			We multiply by c to work in energy units:

			$pc = 4.8\times 10^{-18} \times 3.0\times 10^8 = 1.44\times 10^{-9}\ \text{J}$

			Convert to MeV $(1\ \text{MeV} = 1.6\times 10^{-13}\ \text{J})$

			$pc = \frac{1.44\times 10^{-9}}{1.6\times 10^{-13}}\ \text{MeV} \approx 9.0\ \text{MeV}$

			Kinetic energy

			The electron is clearly ultra‑relativistic $(pc \gg m_e c^2)$ The exact relation between momentum and total energy is:

			$E_{\text{total}} = \sqrt{(pc)^2 + (m_e c^2)^2}$

			and the kinetic energy is obtained by subtracting the rest mass:

			$K = E_{\text{total}} - m_e c^2$

			Substituting the values:

			$K = \sqrt{9.0^2 + 0.511^2} - 0.511
			\approx \sqrt{81 + 0.261} - 0.511
			\approx 9.014 - 0.511
			\approx 8.5\ \text{MeV}$

			Result

			\boxed{K_{\min} \approx 8.5\ \text{MeV}}.

			Electrons need a kinetic energy of the order of several MeV to cross the equatorial magnetic field and reach the ground. Less energetic electrons become trapped in the radiation belts.

			#### Answer

			[Insert a concise answer or boxed result]