PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. For every direction answer, state the hand rule you use and whether the charge is positive or negative. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。每个方向类答案都要写明所用的手定则以及电荷的正负。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1force on a wire $F=BIL\sin\theta$载流导线受力 $F=BIL\sin\theta$[4 marks]
A straight wire of length $0.20\ \mathrm{m}$ carries a current of $4.0\ \mathrm{A}$ in a uniform magnetic field of flux density $0.35\ \mathrm{T}$. The current flows horizontally toward the east and the field points vertically upward, so the wire is perpendicular to the field.一根长 $0.20\ \mathrm{m}$ 的直导线在磁感应强度为 $0.35\ \mathrm{T}$ 的匀强磁场中载流 $4.0\ \mathrm{A}$。电流水平指向正东，磁场竖直向上，故导线垂直于磁场。
(a)Calculate the magnitude of the force on the wire.计算导线受力的大小。[2]
(b)Using Fleming's left-hand rule, state the direction of the force on the wire.用弗莱明左手定则写出导线受力的方向。[1]
(c)State what would happen to the magnitude of the force if the wire were rotated to lie parallel to the field, and justify your answer.说明若把导线转到与磁场平行，受力大小会如何变化，并说明理由。[1]
Q2MEDIUMPaper 1force on a moving charge (Lorentz)运动电荷受力（洛伦兹力）[6 marks]
A proton (charge $+e$) moves due east at $3.0\times10^{6}\ \mathrm{m\,s^{-1}}$ through a uniform magnetic field of $0.25\ \mathrm{T}$ directed vertically upward, so that the velocity is perpendicular to the field.一个质子（电荷 $+e$）以 $3.0\times10^{6}\ \mathrm{m\,s^{-1}}$ 向正东穿过竖直向上、大小为 $0.25\ \mathrm{T}$ 的匀强磁场，速度垂直于磁场。
(a)Calculate the magnitude of the magnetic force on the proton.计算质子受到的磁力大小。[2]
(b)Using Fleming's left-hand rule, state the direction of the force on the proton.用弗莱明左手定则写出质子受力的方向。[2]
(c)An electron is now sent through the same field with the same velocity. State, with a reason, how the magnitude and direction of the force compare with those on the proton.现在让一个电子以相同速度穿过同一磁场。说明并解释其受力的大小与方向相比质子如何。[2]
A proton ($m = 1.67\times10^{-27}\ \mathrm{kg}$, charge $+e$) enters a uniform magnetic field of $0.50\ \mathrm{T}$ at $4.0\times10^{6}\ \mathrm{m\,s^{-1}}$, moving perpendicular to the field.一个质子（$m = 1.67\times10^{-27}\ \mathrm{kg}$，电荷 $+e$）以 $4.0\times10^{6}\ \mathrm{m\,s^{-1}}$ 垂直于磁场进入大小为 $0.50\ \mathrm{T}$ 的匀强磁场。
(a)Explain why the proton follows a circular path, and calculate the radius of that path.解释质子为何做圆周运动，并计算该圆轨道的半径。[3]
(b)Calculate the period of the orbit.计算轨道的周期。[2]
(c)State how the period would change if the proton's speed were doubled, and justify your answer.说明若质子速率加倍，周期将如何变化，并说明理由。[1]
Q4HARDPaper 1charge in an electric field $F=qE$, $W=qV$电场中的电荷 $F=qE$、$W=qV$[6 marks]
An electron ($m = 9.11\times10^{-31}\ \mathrm{kg}$, charge $-e$) is accelerated from rest through a potential difference of $800\ \mathrm{V}$ between two parallel plates separated by $0.040\ \mathrm{m}$.一个电子（$m = 9.11\times10^{-31}\ \mathrm{kg}$，电荷 $-e$）在相距 $0.040\ \mathrm{m}$ 的两平行板之间，从静止经 $800\ \mathrm{V}$ 的电势差加速。
(a)Calculate the magnitude of the electric field strength between the plates.计算两板之间电场强度的大小。[2]
(b)Calculate the kinetic energy gained by the electron.计算电子获得的动能。[2]
(c)Hence calculate the final speed of the electron.由此计算电子的末速率。[2]
Two long straight parallel wires are $0.050\ \mathrm{m}$ apart. One carries a current of $8.0\ \mathrm{A}$ and the other a current of $12\ \mathrm{A}$, both in the same direction. Take $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$.两条相距 $0.050\ \mathrm{m}$ 的长直平行导线，一根载流 $8.0\ \mathrm{A}$，另一根载流 $12\ \mathrm{A}$，两者方向相同。取 $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$。
(a)Calculate the magnetic flux density produced by the $8.0\ \mathrm{A}$ wire at the position of the $12\ \mathrm{A}$ wire.计算 $8.0\ \mathrm{A}$ 导线在 $12\ \mathrm{A}$ 导线所在位置产生的磁感应强度。[2]
(b)Calculate the force per unit length between the two wires.计算两导线之间单位长度的力。[3]
(c)State whether the wires attract or repel, and use a hand-rule argument to justify your answer.说明两导线是相吸还是相斥，并用手定则论证。[2]
(d)State how the force per unit length would change if the separation were doubled.说明若间距加倍，单位长度的力将如何变化。[1]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward careful reading of gradients, correct linearisation, and proper handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查对斜率的细致读取、正确的线性化以及对不确定度的妥善处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q6HARDPaper 1Bradius vs momentum linear graph半径对动量的线性图[10 marks]
Singly charged ions (charge $+e$) are fired perpendicular into a uniform magnetic field and follow circular arcs. For ions of several different momenta $p = mv$, the orbit radius $r$ is measured and tabulated against $p$:带单位正电荷（电荷 $+e$）的离子垂直射入匀强磁场，沿圆弧运动。对若干不同动量 $p = mv$ 的离子测量轨道半径 $r$，并将 $r$ 对 $p$ 列表：
$p\ /\ 10^{-22}\ \mathrm{kg\,m\,s^{-1}}$
$8.0$
$16.0$
$24.0$
$32.0$
$r\ /\ \mathrm{cm}$
$1.0$
$2.0$
$3.0$
$4.0$
(a)Starting from $r = mv/(qB)$, show that a graph of $r$ against $p$ should be a straight line through the origin, and state what the gradient represents.从 $r = mv/(qB)$ 出发，证明 $r$ 对 $p$ 的图应为过原点的直线，并说明斜率代表什么。[3]
(b)Calculate the gradient of the line.计算该直线的斜率。[2]
(c)Hence determine the magnetic flux density $B$ of the field.由此求磁场的磁感应强度 $B$。[3]
(d)Each radius is measured with an absolute uncertainty of $\pm 0.1\ \mathrm{cm}$. For the point at $p = 32.0\times10^{-22}\ \mathrm{kg\,m\,s^{-1}}$, calculate the percentage uncertainty in $r$.每个半径的绝对不确定度为 $\pm 0.1\ \mathrm{cm}$。对 $p = 32.0\times10^{-22}\ \mathrm{kg\,m\,s^{-1}}$ 处的点，计算 $r$ 的百分比不确定度。[2]
In a velocity selector a uniform electric field and a uniform magnetic field are arranged perpendicular to each other and to the beam. A charged particle passes straight through undeflected when the two forces balance. The fields are measured as $E = (2.4 \pm 0.1)\times10^{4}\ \mathrm{V\,m^{-1}}$ and $B = 0.30 \pm 0.01\ \mathrm{T}$.在速度选择器中，匀强电场与匀强磁场相互垂直，并都垂直于束流。当两个力平衡时，带电粒子不偏转直线通过。两场测得 $E = (2.4 \pm 0.1)\times10^{4}\ \mathrm{V\,m^{-1}}$、$B = 0.30 \pm 0.01\ \mathrm{T}$。
(a)Show that the speed of the particles that pass straight through is given by $v = E/B$, and explain why this speed does not depend on the charge or mass of the particle.证明直线通过的粒子速率由 $v = E/B$ 给出，并解释为何此速率与粒子的电荷或质量无关。[3]
(b)Calculate the selected speed $v$.计算被选速率 $v$。[2]
(c)Calculate the percentage uncertainty in $v$ and hence state $v$ with its absolute uncertainty.计算 $v$ 的百分比不确定度，并由此写出带绝对不确定度的 $v$。[4]
(d)A particle enters with a speed greater than $v$. State, with a reason, the direction in which it is deflected (toward the electric-force side or the magnetic-force side).一个粒子以大于 $v$ 的速率进入。说明并解释它偏向哪一侧（电场力一侧还是磁场力一侧）。[3]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem with a clear diagram and labelled field directions. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先画清晰示意图并标注场的方向。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
A mass spectrometer separates singly ionised atoms (charge $+e$). Ions first pass through a velocity selector with $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$ and $B = 0.20\ \mathrm{T}$ crossed perpendicular. The emerging ions then enter a region of pure magnetic field $B' = 0.50\ \mathrm{T}$, where they follow semicircular arcs and strike a detector. Take $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$.一台质谱仪分离单电离原子（电荷 $+e$）。离子先通过速度选择器，其中相互垂直的 $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$ 与 $B = 0.20\ \mathrm{T}$。射出的离子随后进入纯磁场区域 $B' = 0.50\ \mathrm{T}$，沿半圆弧运动并打到探测器上。取 $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$。
(a)Calculate the speed of the ions that pass through the velocity selector.计算通过速度选择器的离子速率。[2]
(b)Explain the purpose of the velocity selector in front of the magnetic field, in terms of how it improves the measurement.从如何改善测量的角度，解释磁场前置速度选择器的作用。[2]
(c)A neon-20 ion has mass $20\ \mathrm{u}$. Calculate the radius of its semicircular path in the field $B'$.一个氖-20 离子质量为 $20\ \mathrm{u}$。计算它在磁场 $B'$ 中半圆轨道的半径。[3]
(d)A neon-22 ion (mass $22\ \mathrm{u}$) travels through the same instrument. Calculate the distance on the detector between the landing points of the neon-20 and neon-22 ions.一个氖-22 离子（质量 $22\ \mathrm{u}$）通过同一仪器。计算氖-20 与氖-22 离子在探测器上落点之间的距离。[3]
(e)State, with a reason, how the separation on the detector would change if $B'$ were increased.说明并解释若 $B'$ 增大，探测器上的分离会如何变化。[2]
In a cyclotron, protons ($m = 1.67\times10^{-27}\ \mathrm{kg}$, charge $+e$) spiral in a uniform magnetic field of $0.80\ \mathrm{T}$. At its outer edge a proton travels in a circle of radius $0.30\ \mathrm{m}$.在回旋加速器中，质子（$m = 1.67\times10^{-27}\ \mathrm{kg}$，电荷 $+e$）在 $0.80\ \mathrm{T}$ 的匀强磁场中螺旋运动。在最外缘，质子沿半径 $0.30\ \mathrm{m}$ 的圆运动。
(a)By equating the magnetic force to the centripetal force, show that the speed of a charged particle on a circular path is $v = qBr/m$.通过令磁力等于向心力，证明做圆周运动的带电粒子速率为 $v = qBr/m$。[2]
(b)Calculate the speed of the proton at the outer edge.计算最外缘处质子的速率。[2]
(c)Calculate the kinetic energy of the proton at the outer edge, in joules.计算最外缘处质子的动能（单位焦耳）。[2]
(d)Calculate the period of the proton's circular motion.计算质子圆周运动的周期。[2]
(e)Explain why a cyclotron can use an accelerating voltage of fixed frequency, even though the protons speed up on each loop.解释为何回旋加速器可以使用固定频率的加速电压，尽管质子每圈都在加速。[2]
Part 1: Two long parallel wires, each carrying $6.0\ \mathrm{A}$ in the same direction, are separated by $0.040\ \mathrm{m}$. Part 2: an electron ($m = 9.11\times10^{-31}\ \mathrm{kg}$, charge $-e$) enters mid-way between two parallel deflecting plates moving horizontally at $2.0\times10^{7}\ \mathrm{m\,s^{-1}}$. The plates are $0.080\ \mathrm{m}$ long and produce a uniform field of $1.2\times10^{4}\ \mathrm{V\,m^{-1}}$ perpendicular to the entry velocity.第一部分：两条长平行导线各载流 $6.0\ \mathrm{A}$、方向相同，相距 $0.040\ \mathrm{m}$。第二部分：一个电子（$m = 9.11\times10^{-31}\ \mathrm{kg}$，电荷 $-e$）从两偏转板正中以 $2.0\times10^{7}\ \mathrm{m\,s^{-1}}$ 水平进入。板长 $0.080\ \mathrm{m}$，产生垂直于入射速度的 $1.2\times10^{4}\ \mathrm{V\,m^{-1}}$ 匀强电场。
(a)For the wires, calculate the force per unit length and state whether they attract or repel.对两导线，计算单位长度的力，并说明相吸还是相斥。[3]
(b)For the electron, calculate the acceleration produced by the electric field.对电子，计算电场产生的加速度。[2]
(c)Calculate the vertical deflection of the electron as it leaves the plates, and state the shape of its path inside the field.计算电子离开偏转板时的竖直偏转，并说明它在场内的轨迹形状。[3]