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en/10.2.2.md
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| + | ### Statement | ||
| + | |||
| + | $10.2.2.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $10.2.2^\ast$. Estimate the drift velocity of an electron across an inhomogeneous magnetic field, whose induction components are $B_x = 0$, $B_y = 0$, $B_z = B_0(1 + \alpha x)$. The electron velocity is $v$, $v \ll eB_0/(\alpha m_e)$. | ||
| + | |||
| + | <div align="center"> | ||
| + | <img src="https://savchenkosolutions.com/10/10.2.2/statement.png" alt="For problem 10.2.2"> | ||
| + | </div> | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Since the problem asks to <i>estimate</i> the velocity (which is also indicated by the $\approx$ sign in the author's answer), numerical coefficients on the order of unity can be neglected. We will provide two methods: a simplified kinematic estimation (corresponding to the textbook's answer) and a strict analytical derivation. | ||
| + | |||
| + | <b>Method 1. Kinematic estimation</b> | ||
| + | Let's approximately represent the electron's trajectory as two glued semicircles. In the upper part of the trajectory, the field is weaker ($B_1$), and the radius is larger ($R_1$). In the lower part, the field is stronger ($B_2$), and the radius is smaller ($R_2$). | ||
| + | |||
| + | The trajectory radius is $R = \frac{m_e v}{eB}$. The displacement of the electron along the $y$-axis during one full revolution is equal to the difference of the diameters: | ||
| + | $$\Delta y = 2R_2 - 2R_1 = \frac{2m_e v}{e} \left( \frac{1}{B_2} - \frac{1}{B_1} \right) = \frac{2m_e v}{e} \frac{B_1 - B_2}{B_1 B_2}$$ | ||
| + | |||
| + | The distance between the centers of the semicircles is approximately equal to $2R \approx \frac{2m_e v}{eB_0}$. Over this distance, the magnetic field changes by: | ||
| + | $$dB = \alpha B_0 \cdot \Delta x = \alpha B_0 \frac{2m_e v}{eB_0} = \frac{2\alpha m_e v}{e}$$ | ||
| + | |||
| + | Then the field difference is $B_1 - B_2 = -dB$, and the denominator is $B_1 B_2 \approx B_0^2$. The magnitude of the displacement is: | ||
| + | $$|\Delta y| \approx \frac{2m_e v}{e B_0^2} dB = \frac{2m_e v}{e B_0^2} \frac{2\alpha m_e v}{e} = \frac{4\alpha m_e^2 v^2}{e^2 B_0^2}$$ | ||
| + | |||
| + | The time of one revolution is $T \approx \frac{2\pi m_e}{e B_0}$. The estimated drift velocity is: | ||
| + | $$v_d = \frac{|\Delta y|}{T} = \frac{4\alpha m_e^2 v^2 / (e^2 B_0^2)}{2\pi m_e / (e B_0)} = \frac{2}{\pi} \frac{\alpha m_e v^2}{e B_0}$$ | ||
| + | |||
| + | Dropping the factor $\frac{2}{\pi} \approx 0.63$ (since we are only interested in the order of magnitude), we obtain the author's estimation: | ||
| + | $$v_d \approx \frac{\alpha m_e v^2}{e B_0}$$ | ||
| + | |||
| + | --- | ||
| + | |||
| + | <b>Method 2. Strict calculation (via magnetic moment)</b> | ||
| + | For an exact calculation, we use the equivalent magnetic dipole method. | ||
| + | An electron moving in a circular orbit is equivalent to a closed current loop. Since $v \ll eB_0/(\alpha m_e)$, the field variation over the scale of one loop is very small. The radius is $R = \frac{m_e v}{e B_0}$, and the period is $T = \frac{2\pi m_e}{e B_0}$. | ||
| + | The effective current of such a loop is: | ||
| + | $$I = \frac{e}{T} = \frac{e v}{2\pi R}$$ | ||
| + | |||
| + | The magnetic moment of this microcurrent is $\mu = I S$, where the area is $S = \pi R^2$: | ||
| + | $$\mu = \frac{e v}{2\pi R} \cdot \pi R^2 = \frac{e v R}{2} = \frac{m_e v^2}{2 B_0}$$ | ||
| + | |||
| + | The energy of the magnetic dipole in an external magnetic field is $U = \mu B_z$. | ||
| + | An effective force acts on the dipole in an inhomogeneous field, tending to push it into the region of the weaker field: | ||
| + | $$F_x = -\frac{\partial U}{\partial x} = -\mu \frac{\partial B_z}{\partial x} = -\frac{m_e v^2}{2 B_0} \cdot B_0 \alpha = -\frac{m_e v^2 \alpha}{2}$$ | ||
| + | |||
| + | The drift velocity under the action of the transverse force $\vec{F}$ is calculated as $\vec{v}_d = \frac{\vec{F} \times \vec{B}}{q B^2}$. Substituting the magnitude of the force, we get the exact value of the gradient drift velocity: | ||
| + | $$v_d = \frac{F_x}{eB_0} = \frac{1}{2} \frac{m_e v^2 \alpha}{e B_0}$$ | ||
| + | |||
| + | As can be seen, the strict calculation yields an additional coefficient of $1/2$, which is lost in the rough kinematic estimation. The velocity vector is directed to the left (along the $-y$ axis). | ||
| + | |||
| + | <i>Note: The figure in the problem statement mistakenly shows the trajectory of a positively charged particle drifting to the right. A negatively charged electron in this field rotates counterclockwise and drifts to the left.</i> | ||
| + | |||
| + | #### Answer | ||
| + | $$v_d \approx \frac{\alpha m_e v^2}{e B_0}$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $10.2.2.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $10.2.2^\ast$. Estimate the drift velocity of an electron across an inhomogeneous magnetic field, whose induction components are $B_x = 0$, $B_y = 0$, $B_z = B_0(1 + \alpha x)$. The electron velocity is $v$, $v \ll eB_0/(\alpha m_e)$. | |||
| <div align="center"> | |||
| <img src="https://savchenkosolutions.com/10/10.2.2/statement.png" alt="For problem 10.2.2"> | |||
| </div> | |||
| ### Solution | |||
| Since the problem asks to <i>estimate</i> the velocity (which is also indicated by the $\approx$ sign in the author's answer), numerical coefficients on the order of unity can be neglected. We will provide two methods: a simplified kinematic estimation (corresponding to the textbook's answer) and a strict analytical derivation. | |||
| <b>Method 1. Kinematic estimation</b> | |||
| Let's approximately represent the electron's trajectory as two glued semicircles. In the upper part of the trajectory, the field is weaker ($B_1$), and the radius is larger ($R_1$). In the lower part, the field is stronger ($B_2$), and the radius is smaller ($R_2$). | |||
| The trajectory radius is $R = \frac{m_e v}{eB}$. The displacement of the electron along the $y$-axis during one full revolution is equal to the difference of the diameters: | |||
| $$\Delta y = 2R_2 - 2R_1 = \frac{2m_e v}{e} \left( \frac{1}{B_2} - \frac{1}{B_1} \right) = \frac{2m_e v}{e} \frac{B_1 - B_2}{B_1 B_2}$$ | |||
| The distance between the centers of the semicircles is approximately equal to $2R \approx \frac{2m_e v}{eB_0}$. Over this distance, the magnetic field changes by: | |||
| $$dB = \alpha B_0 \cdot \Delta x = \alpha B_0 \frac{2m_e v}{eB_0} = \frac{2\alpha m_e v}{e}$$ | |||
| Then the field difference is $B_1 - B_2 = -dB$, and the denominator is $B_1 B_2 \approx B_0^2$. The magnitude of the displacement is: | |||
| $$|\Delta y| \approx \frac{2m_e v}{e B_0^2} dB = \frac{2m_e v}{e B_0^2} \frac{2\alpha m_e v}{e} = \frac{4\alpha m_e^2 v^2}{e^2 B_0^2}$$ | |||
| The time of one revolution is $T \approx \frac{2\pi m_e}{e B_0}$. The estimated drift velocity is: | |||
| $$v_d = \frac{|\Delta y|}{T} = \frac{4\alpha m_e^2 v^2 / (e^2 B_0^2)}{2\pi m_e / (e B_0)} = \frac{2}{\pi} \frac{\alpha m_e v^2}{e B_0}$$ | |||
| Dropping the factor $\frac{2}{\pi} \approx 0.63$ (since we are only interested in the order of magnitude), we obtain the author's estimation: | |||
| $$v_d \approx \frac{\alpha m_e v^2}{e B_0}$$ | |||
| --- | |||
| <b>Method 2. Strict calculation (via magnetic moment)</b> | |||
| For an exact calculation, we use the equivalent magnetic dipole method. | |||
| An electron moving in a circular orbit is equivalent to a closed current loop. Since $v \ll eB_0/(\alpha m_e)$, the field variation over the scale of one loop is very small. The radius is $R = \frac{m_e v}{e B_0}$, and the period is $T = \frac{2\pi m_e}{e B_0}$. | |||
| The effective current of such a loop is: | |||
| $$I = \frac{e}{T} = \frac{e v}{2\pi R}$$ | |||
| The magnetic moment of this microcurrent is $\mu = I S$, where the area is $S = \pi R^2$: | |||
| $$\mu = \frac{e v}{2\pi R} \cdot \pi R^2 = \frac{e v R}{2} = \frac{m_e v^2}{2 B_0}$$ | |||
| The energy of the magnetic dipole in an external magnetic field is $U = \mu B_z$. | |||
| An effective force acts on the dipole in an inhomogeneous field, tending to push it into the region of the weaker field: | |||
| $$F_x = -\frac{\partial U}{\partial x} = -\mu \frac{\partial B_z}{\partial x} = -\frac{m_e v^2}{2 B_0} \cdot B_0 \alpha = -\frac{m_e v^2 \alpha}{2}$$ | |||
| The drift velocity under the action of the transverse force $\vec{F}$ is calculated as $\vec{v}_d = \frac{\vec{F} \times \vec{B}}{q B^2}$. Substituting the magnitude of the force, we get the exact value of the gradient drift velocity: | |||
| $$v_d = \frac{F_x}{eB_0} = \frac{1}{2} \frac{m_e v^2 \alpha}{e B_0}$$ | |||
| As can be seen, the strict calculation yields an additional coefficient of $1/2$, which is lost in the rough kinematic estimation. The velocity vector is directed to the left (along the $-y$ axis). | |||
| <i>Note: The figure in the problem statement mistakenly shows the trajectory of a positively charged particle drifting to the right. A negatively charged electron in this field rotates counterclockwise and drifts to the left.</i> | |||
| #### Answer | |||
| $$v_d \approx \frac{\alpha m_e v^2}{e B_0}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||