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

Alexphysics правка от 24 июн. 2026 г., 00:39

правка #19139 предыдущая #19138 ← раньше позже →

en/14.4.29.md +12 −7

		@@ -1,13 +1,22 @@
			### Statement

		−	$14.4.29.$ ~~[Insert the problem statement]~~
		+	$14.4.29.$

		+	What is the maximum velocity of a charged particle in the crossed electric and
		+	magnetic fields−→E and−→B (−→E ⊥−→B ), if minimal cost the velocity is equal to βc?
		+	Eβ
		+
			### Solution

			Transformation to a system without an electric field

		−	~~If $E < cB (that is, k < 1)$, th~~ere exists an inertial reference frame moving with drift velocity ~~$\mathbf{v}_d$ relative to the laboratory, in which the electric field vanishes and only an effective magnetic field remains. The velocity of that frame is precisely the electric drift velocity:~~
		+	If $E < cB $ (that is, $k < 1$),
		+	there exists an inertial reference frame moving with drift velocity

		+	$\mathbf{v}_d$
		+
		+	relative to the laboratory, in which the electric field vanishes and only an effective magnetic field remains. The velocity of that frame is precisely the electric drift velocity:
		+
			$\mathbf{v}_d = \frac{\mathbf{E} \times \mathbf{B}}{B^2}, \qquad v_d = \frac{E}{B} = kc$

			In that privileged system, the particle feels no electric force and moves only under the magnetic field, describing a uniform circular motion with a constant speed that we will call $\beta_1 c$

			Relativistic velocity composition

			Upon returning to the laboratory system, the velocity of the particle is obtained by combining the circular velocity in the moving system $(\beta_1 c)$ with the drift velocity of the system itself$ (v_d = kc)$ Since both motions are collinear at certain instants (parallel or antiparallel), the relativistic addition formula gives the extreme values of the observed velocity:

			Maximum velocity (when $\beta_1 $and k point in the same direction):

			$v_{\max} = \frac{\beta_1 c + kc}{1 + \beta_1 k} = c\,\frac{\beta_1 + k}{1 + \beta_1 k}$

			Minimum velocity (when $\beta_1$ and k point in opposite directions):

			$v_{\min} = \frac{\beta_1 c - kc}{1 - \beta_1 k} = c\,\frac{\beta_1 - k}{1 - \beta_1 k}$

			The statement tells us that this minimum velocity is precisely $\beta c$ Therefore:

			$\beta c = c\,\frac{\beta_1 - k}{1 - \beta_1 k} \quad\Rightarrow\quad \beta = \frac{\beta_1 - k}{1 - \beta_1 k}$

			Expression for the maximum velocity in terms of$ \beta and k$

			From the previous relation we can solve for $\beta_1$

			$\beta_1 = \frac{\beta + k}{1 + \beta k}$.

			Substituting this expression into the formula for the maximum velocity we obtain:

			$v_{\max} = c\,\frac{\dfrac{\beta + k}{1 + \beta k} + k}{1 + \dfrac{\beta + k}{1 + \beta k}\,k}
			= c\,\frac{(\beta + k) + k(1 + \beta k)}{(1 + \beta k) + (\beta + k)k}$

			Simplifying the numerator and denominator:

			$v_{\max} = c\,\frac{\beta + k + k + \beta k^2}{1 + \beta k + \beta k + k^2}
		@@ -45,12 +54,8 @@Solution
			= c\,\frac{\beta(1 + k^2) + 2k}{1 + k^2 + 2\beta k}$


		−	Final result
		+	#### Answer

			The maximum velocity of the particle in crossed fields, expressed in terms of the minimum velocity \beta c and the parameter k = E/(cB), is:

			$\boxed{v_{\max} = c\,\frac{2k + (1 + k^2)\beta}{1 + k^2 + 2k\beta}}$
		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]