Правка разделов «Statement», «Solution», «Answer»

Alexphysics правка от 23 июн. 2026 г., 23:58

правка #19134 предыдущая #19133 ← раньше

en/14.4.25.md +7 −5

		@@ -1,7 +1,11 @@
			### Statement

		−	$14.4.25~~.$ [Insert the problem statement]~~
		+	$14.4.25$

		+	Estimate at what minimum energy electrons located at an altitude of h =
		+	1000 km will be able to reach the Earth’s surface in the equator region, if the
		+	Earth’s magnetic field induction B = 30 µT?
		+
			### 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:

			$K = \sqrt{9.0^2 + 0.511^2} - 0.511
			\approx \sqrt{81 + 0.261} - 0.511
		@@ -45,12 +49,10 @@Solution
			\approx 9.014 - 0.511
			\approx 8.5\ \text{MeV}$

		−	Result
		+	$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$

		−	\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]
		+	$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$

	@@ -1,7 +1,11 @@
	### Statement		### Statement

	$14.4.25~~.$ [Insert the problem statement]~~		$14.4.25$

			Estimate at what minimum energy electrons located at an altitude of h =
			1000 km will be able to reach the Earth’s surface in the equator region, if the
			Earth’s magnetic field induction B = 30 µT?

	### Solution		### Solution

	Physical criterion		Physical criterion

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

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

	Here p is the relativistic momentum of the electron		Here p is the relativistic momentum of the electron

	Calculation of the momentum		Calculation of the momentum

	Substituting the values		Substituting the values

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

	We multiply by c to work in energy units:		We multiply by c to work in energy units:

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

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

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

	Kinetic energy		Kinetic energy

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

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

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

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

	Substituting the values:		Substituting the values:

	$K = \sqrt{9.0^2 + 0.511^2} - 0.511		$K = \sqrt{9.0^2 + 0.511^2} - 0.511
	\approx \sqrt{81 + 0.261} - 0.511		\approx \sqrt{81 + 0.261} - 0.511
	@@ -45,12 +49,10 @@Solution
	\approx 9.014 - 0.511		\approx 9.014 - 0.511
	\approx 8.5\ \text{MeV}$		\approx 8.5\ \text{MeV}$

	Result		$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$

	\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.		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		#### Answer

	[Insert a concise answer or boxed result]		$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$