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en/14.4.29.md
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| + | ### Statement | ||
| + | |||
| + | $14.4.29.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + | Transformation to a system without an electric field | ||
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| + | 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: | ||
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| + | $\mathbf{v}_d = \frac{\mathbf{E} \times \mathbf{B}}{B^2}, \qquad v_d = \frac{E}{B} = kc$ | ||
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| + | 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 $\beta_1 c$ | ||
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| + | Relativistic velocity composition | ||
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| + | 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: | ||
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| + | Maximum velocity (when $\beta_1 $and k point in the same direction): | ||
| + | |||
| + | $v_{\max} = \frac{\beta_1 c + kc}{1 + \beta_1 k} = c\,\frac{\beta_1 + k}{1 + \beta_1 k}$ | ||
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| + | Minimum velocity (when $\beta_1$ and k point in opposite directions): | ||
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| + | $v_{\min} = \frac{\beta_1 c - kc}{1 - \beta_1 k} = c\,\frac{\beta_1 - k}{1 - \beta_1 k}$ | ||
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| + | The statement tells us that this minimum velocity is precisely $\beta c$ Therefore: | ||
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| + | $\beta c = c\,\frac{\beta_1 - k}{1 - \beta_1 k} \quad\Rightarrow\quad \beta = \frac{\beta_1 - k}{1 - \beta_1 k}$ | ||
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| + | Expression for the maximum velocity in terms of$ \beta and k$ | ||
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| + | From the previous relation we can solve for $\beta_1$ | ||
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| + | $\beta_1 = \frac{\beta + k}{1 + \beta k}$. | ||
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| + | Substituting this expression into the formula for the maximum velocity we obtain: | ||
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| + | $v_{\max} = c\,\frac{\dfrac{\beta + k}{1 + \beta k} + k}{1 + \dfrac{\beta + k}{1 + \beta k}\,k} | ||
| + | = c\,\frac{(\beta + k) + k(1 + \beta k)}{(1 + \beta k) + (\beta + k)k}$ | ||
| + | |||
| + | Simplifying the numerator and denominator: | ||
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| + | $v_{\max} = c\,\frac{\beta + k + k + \beta k^2}{1 + \beta k + \beta k + k^2} | ||
| + | = c\,\frac{\beta(1 + k^2) + 2k}{1 + k^2 + 2\beta k}$ | ||
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| + | Final result | ||
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| + | 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: | ||
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| + | $\boxed{v_{\max} = c\,\frac{2k + (1 + k^2)\beta}{1 + k^2 + 2k\beta}}$ | ||
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| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.4.29.$ [Insert the problem statement] | |||
| ### Solution | |||
| Transformation to a system without an electric field | |||
| 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: | |||
| $v_{\max} = c\,\frac{\beta + k + k + \beta k^2}{1 + \beta k + \beta k + k^2} | |||
| = c\,\frac{\beta(1 + k^2) + 2k}{1 + k^2 + 2\beta k}$ | |||
| Final result | |||
| 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] | |||