$10.2.2^\ast$. 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 $v$, $v \ll eB_0/(\alpha m_e)$.
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<img src="https://savchenkosolutions.com/10/10.2.2/statement.png" alt="For problem 10.2.2">
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### 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.
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<b>Method 1. Kinematic estimation</b>
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<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:
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:
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:
$$v_d \approx\frac{\alpha m_e v^2}{e B_0}$$
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<b>Method 2. Strict calculation (via magnetic moment)</b>
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<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 $S = \pi R^2$:
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:
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).
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<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
$$v_d \approx\frac{\alpha m_e v^2}{e B_0}$$
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#### Answer
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[Insert a concise answer or boxed result]
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### Statement
### 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 $v$, $v \ll eB_0/(\alpha m_e)$.
$10.2.2^\ast$. 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
### 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.
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>
<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 ($R_2$).
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:
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:
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:
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:
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:
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:
$$v_d \approx\frac{\alpha m_e v^2}{e B_0}$$
$$v_d \approx\frac{\alpha m_e v^2}{e B_0}$$
@@ -36,7 +28,7 @@Solution
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<b>Method 2. Strict calculation (via magnetic moment)</b>
<b>Method 2. Strict calculation (via magnetic moment)</b>\
For an exact calculation, we use the equivalent magnetic dipole method.
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 $T = \frac{2\pi m_e}{e B_0}$.
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:
The effective current of such a loop is:
$$I = \frac{e}{T} = \frac{e v}{2\pi R}$$
$$I = \frac{e}{T} = \frac{e v}{2\pi R}$$
The magnetic moment of this microcurrent is $\mu = I S$, where the area is $S = \pi R^2$:
The magnetic moment of this microcurrent is $\mu = I S$, where the area is $S = \pi R^2$:
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:
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:
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).
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>
