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| $10.1.17.$ [Insert the problem statement] |
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| $10.1.17.$ a) In a flat capacitor of length $l$, the electric field strength is $E$, and the magnetic field induction, directed along $E$, is $B$. At the entrance to the capacitor, there is a radioactive source emitting electrons with different velocities. A thin beam is formed from them, which passes through the capacitor and then hits a photographic plate located at a distance $L \gg l$. What line-trace will the electrons "draw" on the photographic plate if their deviations from a straight-line trajectory are small? |
| $10.1.17.$ a) In a flat capacitor of length $l$, the electric field strength is $E$, and the magnetic field induction, directed along $E$, is $B$. At the entrance to the capacitor, there is a radioactive source emitting electrons with different velocities. A thin beam is formed from them, which passes through the capacitor and then hits a photographic plate located at a distance $L \gg l$. What line-trace will the electrons "draw" on the photographic plate if their deviations from a straight-line trajectory are small? |
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| b) Find the line-trace of the electrons on the photographic plate for $B = 1$ T, $E = 5 \cdot 10^5$ V/m, $l = 5$ cm, $L = 50$ cm. |
| b) Find the line-trace of the electrons on the photographic plate for $B = 1$ T, $E = 5 \cdot 10^5$ V/m, $l = 5$ cm, $L = 50$ cm. |
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| c) At a high electron velocity, its mass noticeably changes; according to the Lorentz formula $m = m_e / \sqrt{1 - \beta^2}$, where $\beta$ is the ratio of the electron velocity to the speed of light, $m_e$ is the electron rest mass. Solve problem 10.1.17a taking into account the effect of the electron mass change. |
| c) At a high electron velocity, its mass noticeably changes; according to the Lorentz formula $m = m_e / \sqrt{1 - \beta^2}$, where $\beta$ is the ratio of the electron velocity to the speed of light, $m_e$ is the electron rest mass. Solve problem 10.1.17a taking into account the effect of the electron mass change. |
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| ### Solution |
| ### Solution |
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| a) Let us introduce a coordinate system as shown in the figure for the problem: the beam flies along the $x$-axis, the fields are directed vertically upwards along the $y$-axis ($\vec{E} = E \hat{j}$, $\vec{B} = B \hat{j}$). The force acting on the electron (with charge $q = -e$) is the Lorentz force: | | a) Let us introduce a coordinate system as shown in the figure for the problem: the beam flies along the $x$-axis, the fields are directed vertically upwards along the $y$-axis ($\vec{E} = E \hat{j}$, $\vec{B} = B \hat{j}$). The force acting on the electron (with charge $q = -e$) is the Lorentz force: |
| $$\vec{F} = -e\vec{E} - e(\vec{v} \times \vec{B})$$ | | $$\vec{F} = -e\vec{E} - e(\vec{v} \times \vec{B})$$ |
| Since the deviations are small, the time of flight inside the capacitor is $t_1 \approx l/v$. | | Since the deviations are small, the time of flight inside the capacitor is $t_1 \approx l/v$. |
| The transverse forces acting on the electron: | | The transverse forces acting on the electron: |
| $F_y = -eE$ (the electric force deflects it downwards, opposite to the $y$-axis) | | $F_y = -eE$ (the electric force deflects it downwards, opposite to the $y$-axis) |
| $F_z = -e(v\hat{i} \times B\hat{j})_z = -evB$ (the magnetic force deflects it opposite to the $z$-axis) | | $F_z = -e(v\hat{i} \times B\hat{j})_z = -evB$ (the magnetic force deflects it opposite to the $z$-axis) |
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| The transverse velocity components upon exiting the capacitor: | | The transverse velocity components upon exiting the capacitor: |
| $$v_z = a_z t_1 = -\frac{evB}{m} \frac{l}{v} = -\frac{eBl}{m}$$ | | $$v_z = a_z t_1 = -\frac{evB}{m} \frac{l}{v} = -\frac{eBl}{m}$$ |
| $$v_y = a_y t_1 = -\frac{eE}{m} \frac{l}{v} = -\frac{eEl}{mv}$$ | | $$v_y = a_y t_1 = -\frac{eE}{m} \frac{l}{v} = -\frac{eEl}{mv}$$ |
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| Since $L \gg l$, we can neglect the small displacement of the electron inside the capacitor itself and consider the displacement only during the free flight to the screen, where the drift time is $t_2 \approx L/v$: | | Since $L \gg l$, we can neglect the small displacement of the electron inside the capacitor itself and consider the displacement only during the free flight to the screen, where the drift time is $t_2 \approx L/v$: |
| $$z \approx v_z t_2 = -\frac{eBlL}{mv}$$ | | $$z \approx v_z t_2 = -\frac{eBlL}{mv}$$ |
| $$y \approx v_y t_2 = -\frac{eElL}{mv^2}$$ | | $$y \approx v_y t_2 = -\frac{eElL}{mv^2}$$ |
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| To find the equation of the trajectory on the screen (in coordinates $z, y$), let us express the velocity $v$ from the equation for the horizontal deflection $z$: | | To find the equation of the trajectory on the screen (in coordinates $z, y$), let us express the velocity $v$ from the equation for the horizontal deflection $z$: |
| $$v = -\frac{eBlL}{mz}$$ | | $$v = -\frac{eBlL}{mz}$$ |
| Let us substitute the obtained expression into the equation for the vertical deflection $y$: | | Let us substitute the obtained expression into the equation for the vertical deflection $y$: |
| $$y = -\frac{eElL}{m} \left( -\frac{mz}{eBlL} \right)^2 = - \frac{mE}{e B^2 l L} z^2$$ | | $$y = -\frac{eElL}{m} \left( -\frac{mz}{eBlL} \right)^2 = - \frac{mE}{e B^2 l L} z^2$$ |
| The trace on the screen is a branch of a parabola (the electrons fall into the $-z, -y$ quadrant). | | The trace on the screen is a branch of a parabola (the electrons fall into the $-z, -y$ quadrant). |
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| b) Let us substitute the numerical values. Using the electron mass $m_e \approx 9.1 \cdot 10^{-31}$ kg and the elementary charge $e \approx 1.6 \cdot 10^{-19}$ C: | | b) Let us substitute the numerical values. Using the electron mass $m_e \approx 9.1 \cdot 10^{-31}$ kg and the elementary charge $e \approx 1.6 \cdot 10^{-19}$ C: |
| $$y = - \frac{9.1 \cdot 10^{-31} \cdot 5 \cdot 10^5}{1.6 \cdot 10^{-19} \cdot 1^2 \cdot 0.05 \cdot 0.5} z^2 \approx -1.14 \cdot 10^{-4} \cdot z^2 \; (\text{m})$$ | | $$y = - \frac{9.1 \cdot 10^{-31} \cdot 5 \cdot 10^5}{1.6 \cdot 10^{-19} \cdot 1^2 \cdot 0.05 \cdot 0.5} z^2 \approx -1.14 \cdot 10^{-4} \cdot z^2 \; (\text{m})$$ |
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| c) For relativistic velocities, it is necessary to use the momentum $\vec{p} = \gamma m_e \vec{v}$. | | c) For relativistic velocities, it is necessary to use the momentum $\vec{p} = \gamma m_e \vec{v}$. |
| The change in the transverse momentum inside the capacitor is determined by the impulse of the acting forces $\Delta \vec{p} = \vec{F} t_1$: | | The change in the transverse momentum inside the capacitor is determined by the impulse of the acting forces $\Delta \vec{p} = \vec{F} t_1$: |
| $$p_z = -evB \frac{l}{v} = -eBl$$ | | $$p_z = -evB \frac{l}{v} = -eBl$$ |
| $$p_y = -eE \frac{l}{v} = -\frac{eEl}{v}$$ | | $$p_y = -eE \frac{l}{v} = -\frac{eEl}{v}$$ |
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| The coordinates on the screen are determined by the velocities $v_z = p_z / (\gamma m_e)$ and $v_y = p_y / (\gamma m_e)$: | | The coordinates on the screen are determined by the velocities $v_z = p_z / (\gamma m_e)$ and $v_y = p_y / (\gamma m_e)$: |
| $$z = v_z \frac{L}{v} = \frac{p_z L}{\gamma m_e v} = -\frac{eBlL}{p}$$ | | $$z = v_z \frac{L}{v} = \frac{p_z L}{\gamma m_e v} = -\frac{eBlL}{p}$$ |
| $$y = v_y \frac{L}{v} = \frac{p_y L}{\gamma m_e v} = - \frac{eElL}{p v}$$ | | $$y = v_y \frac{L}{v} = \frac{p_y L}{\gamma m_e v} = - \frac{eElL}{p v}$$ |
| where $p = \gamma m_e v$ is the total momentum of the electron. | | where $p = \gamma m_e v$ is the total momentum of the electron. |
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| From these equations, we can express the total momentum $p$ and the product $pv$: | | From these equations, we can express the total momentum $p$ and the product $pv$: |
| $$p = -\frac{eBlL}{z} \quad \text{and} \quad pv = - \frac{eElL}{y}$$ | | $$p = -\frac{eBlL}{z} \quad \text{and} \quad pv = - \frac{eElL}{y}$$ |
| Dividing the second expression by the first, we find the velocity: | | Dividing the second expression by the first, we find the velocity: |
| $$v = \frac{pv}{p} = \frac{E z}{B y}$$ | | $$v = \frac{pv}{p} = \frac{E z}{B y}$$ |
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| We use the relativistic relation between momentum and velocity: $p = \frac{m_e v}{\sqrt{1 - (v/c)^2}}$. Let us substitute the expressions for $p$ and $v$ into this relation: | | We use the relativistic relation between momentum and velocity: $p = \frac{m_e v}{\sqrt{1 - (v/c)^2}}$. Let us substitute the expressions for $p$ and $v$ into this relation: |
| $$-\frac{eBlL}{z} = \frac{m_e (Ez / By)}{\sqrt{1 - \left( \frac{Ez}{Bcy} \right)^2}}$$ | | $$-\frac{eBlL}{z} = \frac{m_e (Ez / By)}{\sqrt{1 - \left( \frac{Ez}{Bcy} \right)^2}}$$ |
| Let us square both sides of the equation to eliminate the square root and the minus sign: | | Let us square both sides of the equation to eliminate the square root and the minus sign: |
| $$\frac{e^2 B^2 l^2 L^2}{z^2} \left( 1 - \frac{E^2 z^2}{B^2 c^2 y^2} \right) = \frac{m_e^2 E^2 z^2}{B^2 y^2}$$ | | $$\frac{e^2 B^2 l^2 L^2}{z^2} \left( 1 - \frac{E^2 z^2}{B^2 c^2 y^2} \right) = \frac{m_e^2 E^2 z^2}{B^2 y^2}$$ |
| Expanding the brackets: | | Expanding the brackets: |
| $$\frac{e^2 B^2 l^2 L^2}{z^2} - \frac{e^2 E^2 l^2 L^2}{c^2 y^2} = \frac{m_e^2 E^2 z^2}{B^2 y^2}$$ | | $$\frac{e^2 B^2 l^2 L^2}{z^2} - \frac{e^2 E^2 l^2 L^2}{c^2 y^2} = \frac{m_e^2 E^2 z^2}{B^2 y^2}$$ |
| Let us multiply the entire expression by $z^2 y^2$ and express $y^2$: | | Let us multiply the entire expression by $z^2 y^2$ and express $y^2$: |
| $$e^2 B^2 l^2 L^2 y^2 = \frac{e^2 E^2 l^2 L^2 z^2}{c^2} + \frac{m_e^2 E^2 z^4}{B^2}$$ | | $$e^2 B^2 l^2 L^2 y^2 = \frac{e^2 E^2 l^2 L^2 z^2}{c^2} + \frac{m_e^2 E^2 z^4}{B^2}$$ |
| $$y^2 = \frac{E^2 z^2}{e^2 B^2 l^2 L^2} \left( \frac{e^2 l^2 L^2}{c^2} + \frac{m_e^2 z^2}{B^2} \right)$$ | | $$y^2 = \frac{E^2 z^2}{e^2 B^2 l^2 L^2} \left( \frac{e^2 l^2 L^2}{c^2} + \frac{m_e^2 z^2}{B^2} \right)$$ |
| Taking the square root (and choosing the minus sign, since the deflection occurs in the negative direction of the $y$-axis), we get: | | Taking the square root (and choosing the minus sign, since the deflection occurs in the negative direction of the $y$-axis), we get: |
| $$y = - \frac{m_e E}{e B^2 l L} z \sqrt{z^2 + \left( \frac{eBlL}{m_e c} \right)^2}$$ | | $$y = - \frac{m_e E}{e B^2 l L} z \sqrt{z^2 + \left( \frac{eBlL}{m_e c} \right)^2}$$ |
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| #### Answer | | #### Answer |
| a) $y = - \frac{m_e E}{e B^2 l L} z^2$ | | a) $y = - \frac{m_e E}{e B^2 l L} z^2$ |
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| b) $y = -1.14 \cdot 10^{-4} \cdot z^2$ m | | b) $y = -1.14 \cdot 10^{-4} \cdot z^2$ m |
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| c) $y = - \frac{m_e E}{e B^2 l L} z \sqrt{z^2 + \left( \frac{eBlL}{m_e c} \right)^2}$ | | c) $y = - \frac{m_e E}{e B^2 l L} z \sqrt{z^2 + \left( \frac{eBlL}{m_e c} \right)^2}$ |
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| *Note on the official answer:* | | *Note on the official answer:* |
