Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus D3.1 to D3.6考纲 D3.1 至 D3.6PHYSICS HL
Wire $L = 0.20\ \mathrm{m}$, $I = 4.0\ \mathrm{A}$ east, $B = 0.35\ \mathrm{T}$ up, perpendicular. (a) magnitude of force; (b) direction; (c) effect of making the wire parallel to the field.导线 $L = 0.20\ \mathrm{m}$，$I = 4.0\ \mathrm{A}$ 向东，$B = 0.35\ \mathrm{T}$ 向上，互相垂直。(a) 力的大小；(b) 方向；(c) 把导线转为平行于磁场的影响。
The wire is perpendicular to the field, so $\theta = 90^{\circ}$ and $\sin\theta = 1$. Use $F = BIL\sin\theta$: (M1)
$$ F = (0.35)(4.0)(0.20)(1) = 0.28\ \mathrm{N}. $$(A1)
Fleming's left-hand rule: first finger along $\vec{B}$ (up), second finger along the conventional current (east), thumb gives the force pointing horizontally to the north. (A1)
If the wire lies parallel to $\vec{B}$ then $\theta = 0^{\circ}$, so $\sin\theta = 0$ and $F = BIL\sin\theta = 0$: the force becomes zero. (B1)
导线垂直于磁场，故 $\theta = 90^{\circ}$、$\sin\theta = 1$。用 $F = BIL\sin\theta$：(M1)
$$ F = (0.35)(4.0)(0.20)(1) = 0.28\ \mathrm{N}. $$(A1)
弗莱明左手定则：食指沿 $\vec{B}$（向上），中指沿常规电流（向东），拇指给出力水平指向北方。(A1)
若导线平行于 $\vec{B}$，则 $\theta = 0^{\circ}$，$\sin\theta = 0$，故 $F = BIL\sin\theta = 0$：力变为零。(B1)
Proton $+e$ moves east at $3.0\times10^{6}\ \mathrm{m\,s^{-1}}$ through $B = 0.25\ \mathrm{T}$ up, perpendicular. (a) force magnitude; (b) direction; (c) compare with an electron at the same velocity.质子 $+e$ 以 $3.0\times10^{6}\ \mathrm{m\,s^{-1}}$ 向东穿过向上的 $B = 0.25\ \mathrm{T}$，互相垂直。(a) 力的大小；(b) 方向；(c) 与同速电子比较。
Velocity is perpendicular to the field, so $\theta = 90^{\circ}$. Use $F = qvB\sin\theta$ with $q = e$: (M1)
$$ F = (1.60\times10^{-19})(3.0\times10^{6})(0.25) = 1.2\times10^{-13}\ \mathrm{N}. $$(A1)
For the positive proton, the conventional current is along $\vec{v}$ (east). Fleming's left-hand rule: field up, current east, so the thumb (force) points horizontally to the south. (M1·A1)
The electron has the same charge magnitude, so $F = qvB$ gives the same size, $1.2\times10^{-13}\ \mathrm{N}$. (A1)
Because the electron is negative, its conventional current points opposite to $\vec{v}$, so the force reverses: it points horizontally to the north. (R1)
速度垂直于磁场，故 $\theta = 90^{\circ}$。用 $F = qvB\sin\theta$，$q = e$：(M1)
$$ F = (1.60\times10^{-19})(3.0\times10^{6})(0.25) = 1.2\times10^{-13}\ \mathrm{N}. $$(A1)
对正质子，常规电流沿 $\vec{v}$（向东）。弗莱明左手定则：磁场向上、电流向东，故拇指（力）水平指向南方。(M1·A1)
电子电荷大小相同，故 $F = qvB$ 给出相同大小 $1.2\times10^{-13}\ \mathrm{N}$。(A1)
因电子带负电，其常规电流方向与 $\vec{v}$ 相反，故力方向反转：水平指向北方。(R1)
Proton ($m = 1.67\times10^{-27}\ \mathrm{kg}$, $+e$), $v = 4.0\times10^{6}\ \mathrm{m\,s^{-1}}$, $B = 0.50\ \mathrm{T}$, perpendicular. (a) why circular + radius; (b) period; (c) effect of doubling the speed on the period.质子（$m = 1.67\times10^{-27}\ \mathrm{kg}$，$+e$），$v = 4.0\times10^{6}\ \mathrm{m\,s^{-1}}$，$B = 0.50\ \mathrm{T}$，垂直。(a) 为何为圆 + 半径；(b) 周期；(c) 速率加倍对周期的影响。
The magnetic force $qvB$ stays perpendicular to $\vec{v}$ and is constant in size (the speed is constant), which is exactly the condition for uniform circular motion. (R1)
Set the magnetic force equal to the centripetal force, $qvB = mv^{2}/r$, so $r = mv/(qB)$: (M1)
$$ r = \frac{(1.67\times10^{-27})(4.0\times10^{6})}{(1.60\times10^{-19})(0.50)} \approx 0.084\ \mathrm{m}. $$(A1)
Use $T = 2\pi m/(qB)$: (M1)
$$ T = \frac{2\pi (1.67\times10^{-27})}{(1.60\times10^{-19})(0.50)} \approx 1.3\times10^{-7}\ \mathrm{s}. $$(A1)
$T = 2\pi m/(qB)$ contains no $v$, so the period is unchanged; only the radius would double. (B1)
磁力 $qvB$ 始终垂直于 $\vec{v}$ 且大小恒定（速率不变），这正是匀速圆周运动的条件。(R1)
令磁力等于向心力 $qvB = mv^{2}/r$，得 $r = mv/(qB)$：(M1)
$$ r = \frac{(1.67\times10^{-27})(4.0\times10^{6})}{(1.60\times10^{-19})(0.50)} \approx 0.084\ \mathrm{m}. $$(A1)
用 $T = 2\pi m/(qB)$：(M1)
$$ T = \frac{2\pi (1.67\times10^{-27})}{(1.60\times10^{-19})(0.50)} \approx 1.3\times10^{-7}\ \mathrm{s}. $$(A1)
$T = 2\pi m/(qB)$ 不含 $v$，故周期不变；只有半径会加倍。(B1)
Electron ($m = 9.11\times10^{-31}\ \mathrm{kg}$, $-e$) accelerated from rest through $V = 800\ \mathrm{V}$, plates $0.040\ \mathrm{m}$ apart. (a) field strength; (b) kinetic energy gained; (c) final speed.电子（$m = 9.11\times10^{-31}\ \mathrm{kg}$，$-e$）从静止经 $V = 800\ \mathrm{V}$ 加速，板距 $0.040\ \mathrm{m}$。(a) 场强；(b) 获得的动能；(c) 末速率。
For a uniform field between parallel plates, $E = V/d$: (M1)
$$ E = \frac{800}{0.040} = 2.0\times10^{4}\ \mathrm{V\,m^{-1}}. $$(A1)
The work done by the field on the charge becomes kinetic energy: $E_k = qV$ with $q = e$. (M1)
$$ E_k = (1.60\times10^{-19})(800) = 1.28\times10^{-16}\ \mathrm{J}. $$(A1)
From $E_k = \tfrac{1}{2}mv^{2}$, $v = \sqrt{2E_k/m}$: (M1)
$$ v = \sqrt{\frac{2(1.28\times10^{-16})}{9.11\times10^{-31}}} \approx 1.7\times10^{7}\ \mathrm{m\,s^{-1}}. $$(A1)
平行板间匀强场满足 $E = V/d$：(M1)
$$ E = \frac{800}{0.040} = 2.0\times10^{4}\ \mathrm{V\,m^{-1}}. $$(A1)
电场对电荷做的功转为动能：$E_k = qV$，$q = e$。(M1)
$$ E_k = (1.60\times10^{-19})(800) = 1.28\times10^{-16}\ \mathrm{J}. $$(A1)
由 $E_k = \tfrac{1}{2}mv^{2}$，$v = \sqrt{2E_k/m}$：(M1)
$$ v = \sqrt{\frac{2(1.28\times10^{-16})}{9.11\times10^{-31}}} \approx 1.7\times10^{7}\ \mathrm{m\,s^{-1}}. $$(A1)
Two parallel wires $0.050\ \mathrm{m}$ apart carry $8.0\ \mathrm{A}$ and $12\ \mathrm{A}$ in the same direction. (a) $B$ from the $8.0\ \mathrm{A}$ wire at the other; (b) force per unit length; (c) attract or repel with reason; (d) effect of doubling the separation.两条相距 $0.050\ \mathrm{m}$ 的平行导线同向载流 $8.0\ \mathrm{A}$ 与 $12\ \mathrm{A}$。(a) $8.0\ \mathrm{A}$ 导线在另一根处的 $B$；(b) 单位长度的力；(c) 相吸还是相斥及理由；(d) 间距加倍的影响。
A long straight wire produces $B = \mu_0 I_1/(2\pi r)$ at distance $r$: (M1)
$$ B = \frac{(4\pi\times10^{-7})(8.0)}{2\pi (0.050)} = \frac{(2\times10^{-7})(8.0)}{0.050} = 3.2\times10^{-5}\ \mathrm{T}. $$(A1)
The second wire (current $I_2$) feels a motor-effect force in this field, $F/L = BI_2 = \mu_0 I_1 I_2/(2\pi r)$. (M1)
$$ \frac{F}{L} = \frac{(4\pi\times10^{-7})(8.0)(12)}{2\pi (0.050)} = \frac{(2\times10^{-7})(96)}{0.050}. $$(M1 for substitution)
$$ \frac{F}{L} = 3.8\times10^{-4}\ \mathrm{N\,m^{-1}}. $$(A1)
Same-direction currents attract. (A1)
Each wire sits in the field of the other; applying Fleming's left-hand rule to either wire gives a force pointing toward the other wire, so the pair is pulled together. (R1)
$F/L \propto 1/r$, so doubling $r$ halves the force per unit length. (B1)
长直导线在距离 $r$ 处产生 $B = \mu_0 I_1/(2\pi r)$：(M1)
$$ B = \frac{(4\pi\times10^{-7})(8.0)}{2\pi (0.050)} = \frac{(2\times10^{-7})(8.0)}{0.050} = 3.2\times10^{-5}\ \mathrm{T}. $$(A1)
第二根导线（电流 $I_2$）在此磁场中受电动机效应力，$F/L = BI_2 = \mu_0 I_1 I_2/(2\pi r)$。(M1)
$$ \frac{F}{L} = \frac{(4\pi\times10^{-7})(8.0)(12)}{2\pi (0.050)} = \frac{(2\times10^{-7})(96)}{0.050}. $$(代入得 M1)
$$ \frac{F}{L} = 3.8\times10^{-4}\ \mathrm{N\,m^{-1}}. $$(A1)
同向电流相吸。(A1)
每根导线都处在另一根的磁场中；对任一根用弗莱明左手定则，力都指向对方导线，故两线被拉拢。(R1)
$F/L \propto 1/r$，故 $r$ 加倍则单位长度的力减半。(B1)
Singly charged ions ($+e$) in a uniform $B$; $r$ tabulated against $p = mv$. (a) show $r$ vs $p$ is linear through the origin and state the gradient; (b) gradient; (c) determine $B$; (d) percentage uncertainty in $r$ at $p = 32.0\times10^{-22}$.单位正电荷离子（$+e$）在匀强 $B$ 中；$r$ 对 $p = mv$ 列表。(a) 证明 $r$ 对 $p$ 为过原点直线并说明斜率；(b) 斜率；(c) 求 $B$；(d) $p = 32.0\times10^{-22}$ 处 $r$ 的百分比不确定度。
Start from $r = mv/(qB)$ and recognise that $mv = p$: (M1)
$$ r = \frac{p}{qB} = \left(\frac{1}{qB}\right) p. $$This has the form $r = (\text{gradient})\times p$ with no intercept, so a plot of $r$ against $p$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $1/(qB)$. (A1)
Use two well-separated points, $(8.0\times10^{-22},\,0.010)$ and $(32.0\times10^{-22},\,0.040)$ in SI units: (M1)
$$ \text{gradient} = \frac{0.040 - 0.010}{(32.0 - 8.0)\times10^{-22}} = \frac{0.030}{24.0\times10^{-22}} = 1.25\times10^{19}. $$(A1)
The gradient equals $1/(qB)$, so $B = 1/(q\times\text{gradient})$ with $q = e$: (M1)
$$ B = \frac{1}{(1.60\times10^{-19})(1.25\times10^{19})}. $$(M1 for substitution)
$$ B = \frac{1}{2.0} = 0.50\ \mathrm{T}. $$(A1)
At that point $r = 4.0\ \mathrm{cm}$ with absolute uncertainty $\pm 0.1\ \mathrm{cm}$: (M1)
$$ \frac{0.1}{4.0}\times100\% = 2.5\%. $$(A1)
从 $r = mv/(qB)$ 出发，认出 $mv = p$：(M1)
$$ r = \frac{p}{qB} = \left(\frac{1}{qB}\right) p. $$此式形如 $r = (\text{斜率})\times p$，无截距，故 $r$ 对 $p$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $1/(qB)$。(A1)
用相距较远的两点 $(8.0\times10^{-22},\,0.010)$ 与 $(32.0\times10^{-22},\,0.040)$（SI 单位）：(M1)
$$ \text{斜率} = \frac{0.040 - 0.010}{(32.0 - 8.0)\times10^{-22}} = \frac{0.030}{24.0\times10^{-22}} = 1.25\times10^{19}. $$(A1)
斜率等于 $1/(qB)$，故 $B = 1/(q\times\text{斜率})$，$q = e$：(M1)
$$ B = \frac{1}{(1.60\times10^{-19})(1.25\times10^{19})}. $$(代入得 M1)
$$ B = \frac{1}{2.0} = 0.50\ \mathrm{T}. $$(A1)
该点 $r = 4.0\ \mathrm{cm}$，绝对不确定度 $\pm 0.1\ \mathrm{cm}$：(M1)
$$ \frac{0.1}{4.0}\times100\% = 2.5\%. $$(A1)
Velocity selector, crossed $\vec{E}$ and $\vec{B}$; $E = (2.4\pm0.1)\times10^{4}\ \mathrm{V\,m^{-1}}$, $B = 0.30\pm0.01\ \mathrm{T}$. (a) show $v = E/B$ and why it is charge/mass independent; (b) selected speed; (c) percentage and absolute uncertainty in $v$; (d) deflection direction for a faster particle.速度选择器，正交 $\vec{E}$ 与 $\vec{B}$；$E = (2.4\pm0.1)\times10^{4}\ \mathrm{V\,m^{-1}}$，$B = 0.30\pm0.01\ \mathrm{T}$。(a) 证明 $v = E/B$ 及其与电荷、质量无关；(b) 被选速率；(c) $v$ 的百分比与绝对不确定度；(d) 更快粒子的偏转方向。
The electric force $F_E = qE$ and the magnetic force $F_B = qvB$ act in opposite directions. For no deflection the resultant is zero, so the forces balance: $qE = qvB$. (M1)
The charge $q$ cancels, leaving $v = E/B$. (A1)
Because $q$ has cancelled and no mass ever entered, the selected speed depends only on the field ratio, not on the charge or mass of the particle. (R1)
Substitute into $v = E/B$: (M1)
$$ v = \frac{2.4\times10^{4}}{0.30} = 8.0\times10^{4}\ \mathrm{m\,s^{-1}}. $$(A1)
For a quotient, percentage uncertainties add: (M1)
$$ \frac{\Delta v}{v} = \frac{\Delta E}{E} + \frac{\Delta B}{B} = \frac{0.1}{2.4} + \frac{0.01}{0.30}. $$ $$ \frac{\Delta v}{v} = 4.2\% + 3.3\% = 7.5\%. $$(M1·A1)
So $\Delta v = 0.075\times(8.0\times10^{4}) = 0.6\times10^{4}\ \mathrm{m\,s^{-1}}$, giving $v = (8.0\pm0.6)\times10^{4}\ \mathrm{m\,s^{-1}}$. (A1)
The electric force $qE$ is independent of speed, but the magnetic force $qvB$ grows with $v$. (M1)
For a particle faster than the selected speed, $qvB > qE$, so the magnetic force wins. (A1)
The particle is therefore deflected toward the side on which the magnetic force acts. (R1)
电场力 $F_E = qE$ 与磁场力 $F_B = qvB$ 方向相反。不偏转时合力为零，故两力平衡：$qE = qvB$。(M1)
电荷 $q$ 被消去，得 $v = E/B$。(A1)
由于 $q$ 已消去且质量从未进入，被选速率仅取决于场强之比，与粒子的电荷或质量无关。(R1)
代入 $v = E/B$：(M1)
$$ v = \frac{2.4\times10^{4}}{0.30} = 8.0\times10^{4}\ \mathrm{m\,s^{-1}}. $$(A1)
对于商，百分比不确定度相加：(M1)
$$ \frac{\Delta v}{v} = \frac{\Delta E}{E} + \frac{\Delta B}{B} = \frac{0.1}{2.4} + \frac{0.01}{0.30}. $$ $$ \frac{\Delta v}{v} = 4.2\% + 3.3\% = 7.5\%. $$(M1·A1)
故 $\Delta v = 0.075\times(8.0\times10^{4}) = 0.6\times10^{4}\ \mathrm{m\,s^{-1}}$，即 $v = (8.0\pm0.6)\times10^{4}\ \mathrm{m\,s^{-1}}$。(A1)
电场力 $qE$ 与速率无关，但磁场力 $qvB$ 随 $v$ 增大。(M1)
对比被选速率更快的粒子，$qvB > qE$，磁力占优。(A1)
因此粒子向磁场力作用的一侧偏转。(R1)
Mass spectrometer; selector $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$, $B = 0.20\ \mathrm{T}$; analyser $B' = 0.50\ \mathrm{T}$; ions $+e$; $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$. (a) selected speed; (b) purpose of the selector; (c) radius for neon-20; (d) detector separation neon-20 vs neon-22; (e) effect of increasing $B'$.质谱仪；选择器 $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$、$B = 0.20\ \mathrm{T}$；分析器 $B' = 0.50\ \mathrm{T}$；离子 $+e$；$1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$。(a) 被选速率；(b) 选择器作用；(c) 氖-20 半径；(d) 氖-20 与氖-22 探测器分离；(e) $B'$ 增大的影响。
The velocity selector passes only $v = E/B$: (M1)
$$ v = \frac{3.0\times10^{4}}{0.20} = 1.5\times10^{5}\ \mathrm{m\,s^{-1}}. $$(A1)
It transmits only ions with one definite speed $v = E/B$, regardless of their charge or mass. (B1)
This guarantees every ion entering the analyser has the same known $v$, so the later radius $r = mv/(qB')$ depends only on mass; without it, ions of different speeds of the same mass would land at different radii and blur the result. (R1)
Mass $m = 20\times(1.66\times10^{-27}) = 3.32\times10^{-26}\ \mathrm{kg}$. (M1)
Use $r = mv/(qB')$: (M1)
$$ r_{20} = \frac{(3.32\times10^{-26})(1.5\times10^{5})}{(1.60\times10^{-19})(0.50)} \approx 0.062\ \mathrm{m}. $$(A1)
For neon-22, $m = 22\times(1.66\times10^{-27})$, giving $r_{22} \approx 0.0685\ \mathrm{m}$ from the same formula. (M1)
Each ion completes a semicircle, so the detector spot lies a diameter $2r$ from the entry slit. The separation of the two spots is the difference of diameters: (M1)
$$ \Delta x = 2(r_{22} - r_{20}) = 2(0.0685 - 0.0623) \approx 0.012\ \mathrm{m}. $$(A1)
Since $r = mv/(qB') \propto 1/B'$, a larger $B'$ shrinks every radius. (A1)
The gap $\Delta x = 2v(m_{22}-m_{20})/(qB')$ also scales as $1/B'$, so increasing $B'$ reduces the separation on the detector. (R1)
速度选择器只放行 $v = E/B$：(M1)
$$ v = \frac{3.0\times10^{4}}{0.20} = 1.5\times10^{5}\ \mathrm{m\,s^{-1}}. $$(A1)
它只放行速率为定值 $v = E/B$ 的离子，无论其电荷或质量。(B1)
这保证进入分析器的每个离子都有相同的已知 $v$，于是后续半径 $r = mv/(qB')$ 只依赖质量；若无选择器，同质量但不同速率的离子会落在不同半径，使结果模糊。(R1)
质量 $m = 20\times(1.66\times10^{-27}) = 3.32\times10^{-26}\ \mathrm{kg}$。(M1)
用 $r = mv/(qB')$：(M1)
$$ r_{20} = \frac{(3.32\times10^{-26})(1.5\times10^{5})}{(1.60\times10^{-19})(0.50)} \approx 0.062\ \mathrm{m}. $$(A1)
对氖-22，$m = 22\times(1.66\times10^{-27})$，由同一公式得 $r_{22} \approx 0.0685\ \mathrm{m}$。(M1)
每个离子走半圆，故探测器落点距入口狭缝为直径 $2r$。两落点的分离为直径之差：(M1)
$$ \Delta x = 2(r_{22} - r_{20}) = 2(0.0685 - 0.0623) \approx 0.012\ \mathrm{m}. $$(A1)
由于 $r = mv/(qB') \propto 1/B'$，更大的 $B'$ 使每个半径减小。(A1)
间隔 $\Delta x = 2v(m_{22}-m_{20})/(qB')$ 同样按 $1/B'$ 变化，故增大 $B'$ 会减小探测器上的分离。(R1)
Cyclotron protons ($m = 1.67\times10^{-27}\ \mathrm{kg}$, $+e$), $B = 0.80\ \mathrm{T}$; outer-edge radius $0.30\ \mathrm{m}$. (a) show $v = qBr/m$; (b) speed; (c) kinetic energy; (d) period; (e) why a fixed-frequency voltage works.回旋加速器质子（$m = 1.67\times10^{-27}\ \mathrm{kg}$，$+e$），$B = 0.80\ \mathrm{T}$；最外缘半径 $0.30\ \mathrm{m}$。(a) 证明 $v = qBr/m$；(b) 速率；(c) 动能；(d) 周期；(e) 为何固定频率电压有效。
The magnetic force provides the centripetal force, $qvB = mv^{2}/r$. (M1)
Cancel one factor of $v$ and rearrange: $qB = mv/r$, so $v = qBr/m$. (A1)
Substitute the data: (M1)
$$ v = \frac{(1.60\times10^{-19})(0.80)(0.30)}{1.67\times10^{-27}} \approx 2.3\times10^{7}\ \mathrm{m\,s^{-1}}. $$(A1)
Use $E_k = \tfrac{1}{2}mv^{2}$: (M1)
$$ E_k = \tfrac{1}{2}(1.67\times10^{-27})(2.30\times10^{7})^{2} \approx 4.4\times10^{-13}\ \mathrm{J}. $$(A1)
Use $T = 2\pi m/(qB)$: (M1)
$$ T = \frac{2\pi (1.67\times10^{-27})}{(1.60\times10^{-19})(0.80)} \approx 8.2\times10^{-8}\ \mathrm{s}. $$(A1)
The period $T = 2\pi m/(qB)$ contains no $v$ and no $r$, so it is the same on every loop even as the proton speeds up and spirals outward. (A1)
Therefore the accelerating voltage can be reversed at one fixed frequency $f = 1/T$ and will always be in step with the protons crossing the gap. (R1)
磁力提供向心力，$qvB = mv^{2}/r$。(M1)
约去一个 $v$ 并重排：$qB = mv/r$，故 $v = qBr/m$。(A1)
代入数据：(M1)
$$ v = \frac{(1.60\times10^{-19})(0.80)(0.30)}{1.67\times10^{-27}} \approx 2.3\times10^{7}\ \mathrm{m\,s^{-1}}. $$(A1)
用 $E_k = \tfrac{1}{2}mv^{2}$：(M1)
$$ E_k = \tfrac{1}{2}(1.67\times10^{-27})(2.30\times10^{7})^{2} \approx 4.4\times10^{-13}\ \mathrm{J}. $$(A1)
用 $T = 2\pi m/(qB)$：(M1)
$$ T = \frac{2\pi (1.67\times10^{-27})}{(1.60\times10^{-19})(0.80)} \approx 8.2\times10^{-8}\ \mathrm{s}. $$(A1)
周期 $T = 2\pi m/(qB)$ 不含 $v$ 与 $r$，故即使质子加速并向外螺旋，每圈用时都相同。(A1)
因此加速电压可按单一固定频率 $f = 1/T$ 反向，并始终与穿越缝隙的质子同步。(R1)
Part 1: two parallel wires, each $6.0\ \mathrm{A}$ same direction, $0.040\ \mathrm{m}$ apart. Part 2: electron ($m = 9.11\times10^{-31}\ \mathrm{kg}$, $-e$) enters at $2.0\times10^{7}\ \mathrm{m\,s^{-1}}$ between plates $0.080\ \mathrm{m}$ long with $E = 1.2\times10^{4}\ \mathrm{V\,m^{-1}}$. (a) force per length and attract/repel; (b) acceleration; (c) deflection and path shape.第一部分：两平行导线各 $6.0\ \mathrm{A}$ 同向，相距 $0.040\ \mathrm{m}$。第二部分：电子（$m = 9.11\times10^{-31}\ \mathrm{kg}$，$-e$）以 $2.0\times10^{7}\ \mathrm{m\,s^{-1}}$ 进入板长 $0.080\ \mathrm{m}$、$E = 1.2\times10^{4}\ \mathrm{V\,m^{-1}}$ 的两板之间。(a) 单位长度的力与相吸/相斥；(b) 加速度；(c) 偏转与轨迹形状。
Use $F/L = \mu_0 I_1 I_2/(2\pi r)$ with $I_1 = I_2 = 6.0\ \mathrm{A}$, $r = 0.040\ \mathrm{m}$: (M1)
$$ \frac{F}{L} = \frac{(4\pi\times10^{-7})(6.0)(6.0)}{2\pi (0.040)} = \frac{(2\times10^{-7})(36)}{0.040} = 1.8\times10^{-4}\ \mathrm{N\,m^{-1}}. $$(A1)
The currents are in the same direction, so the wires attract. (A1)
The electric force gives $a = qE/m$ with $q = e$: (M1)
$$ a = \frac{(1.60\times10^{-19})(1.2\times10^{4})}{9.11\times10^{-31}} \approx 2.1\times10^{15}\ \mathrm{m\,s^{-2}}. $$(A1)
The horizontal motion is uniform, so the time inside the plates is $t = \ell/v_0 = 0.080/(2.0\times10^{7}) = 4.0\times10^{-9}\ \mathrm{s}$. (M1)
The vertical deflection is $y = \tfrac{1}{2}at^{2}$:
$$ y = \tfrac{1}{2}(2.11\times10^{15})(4.0\times10^{-9})^{2} \approx 0.017\ \mathrm{m}. $$(A1)
With uniform velocity across the field and uniform acceleration along it, the path inside the plates is a parabola. (B1)
用 $F/L = \mu_0 I_1 I_2/(2\pi r)$，$I_1 = I_2 = 6.0\ \mathrm{A}$、$r = 0.040\ \mathrm{m}$：(M1)
$$ \frac{F}{L} = \frac{(4\pi\times10^{-7})(6.0)(6.0)}{2\pi (0.040)} = \frac{(2\times10^{-7})(36)}{0.040} = 1.8\times10^{-4}\ \mathrm{N\,m^{-1}}. $$(A1)
两电流同向，故两导线相吸。(A1)
电场力给出 $a = qE/m$，$q = e$：(M1)
$$ a = \frac{(1.60\times10^{-19})(1.2\times10^{4})}{9.11\times10^{-31}} \approx 2.1\times10^{15}\ \mathrm{m\,s^{-2}}. $$(A1)
水平运动匀速，故板内时间 $t = \ell/v_0 = 0.080/(2.0\times10^{7}) = 4.0\times10^{-9}\ \mathrm{s}$。(M1)
竖直偏转为 $y = \tfrac{1}{2}at^{2}$：
$$ y = \tfrac{1}{2}(2.11\times10^{15})(4.0\times10^{-9})^{2} \approx 0.017\ \mathrm{m}. $$(A1)
沿垂直方向匀速、沿场方向匀加速，故板内轨迹为抛物线。(B1)