📚 AS Physics Unit 4 June 2019 Question Paper — Formula Derivation | AS物理单元4 2019年6月试卷公式推导
In the June 2019 Edexcel AS Physics Unit 4 (WPH14/01) paper, a classic derivation question tests students’ grasp of circular motion and magnetic fields. Usually, the task is to derive the radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field. This derivation weaves together Newton’s second law and the magnetic Lorentz force, making it a must-know. This article steps through the derivation in full detail, offers physical insights, and points out common errors.
2019年6月爱德思AS物理第四单元（WPH14/01）试卷含有一道经典推导题，考查学生对圆周运动与磁场的掌握。常见任务是推导带电粒子在垂直于匀强磁场方向运动时的圆周轨道半径。该推导将牛顿第二定律与磁洛伦兹力结合，是必会内容。本文将逐步详拆这一推导，提供物理洞察，并指出常见错误。
1. Setting the Scene: Particle in a Magnetic Field | 设定场景：磁场中的粒子
Imagine a charged particle with charge q moving at speed v in a uniform magnetic field of flux density B. Its velocity is perpendicular to the field lines. The magnetic force acting on it is always at right angles to both v and B, resulting in a circular path at constant speed.
设想一个电荷量为 q 的带电粒子以速率 v 在磁通量密度为 B 的匀强磁场中运动，速度方向垂直于磁场线。它受到的磁力始终与 v 和 B 垂直，从而形成恒定速率的圆周轨迹。
This setup is typical in mass spectrometers and particle accelerators. In the 2019 Unit 4 question, candidates must build the expression for radius r from fundamental principles.
这种配置在质谱仪和粒子加速器中十分典型。在2019年第四单元考题中，考生须从基本原理构建半径 r 的表达式。
2. Forces Involved: Magnetic Lorentz Force | 涉及的力：洛伦兹磁力
The magnitude of the magnetic force on a moving charge is given by F = Bqv sinθ, where θ is the angle between velocity and magnetic field. When the particle enters perpendicularly, sin90° = 1, so the force simplifies to F = Bqv. The direction is determined by Fleming’s left-hand rule for a positive charge or the Lorentz right-hand rule.
运动电荷所受磁力大小为 F = Bqv sinθ，θ 为速度与磁场间的夹角。垂直入射时 sin90° = 1，力简化为 F = Bqv。方向由弗莱明左手定则（正电荷）或洛伦兹右手定则确定。
Fmagnetic = B q v
This magnetic force acts as the centripetal force required to keep the particle on its circular track. As it has no component along the direction of motion, the speed stays unchanged.
该磁力充当维持粒子圆周运动的向心力。由于它在运动方向上无分量，速率保持不变。
3. Centripetal Force Requirement for Circular Motion | 圆周运动所需的向心力
For an object of mass m travelling in a circle of radius r at speed v, the inward centripetal force needed is F = mv²/r. This net force is supplied entirely by the magnetic force in our scenario.
质量为 m 的物体以速率 v 沿半径为 r 的圆周运动时，所需向心力为 F = mv²/r。在此情景下，该净向心力完全由磁力提供。
Fcentripetal = m v2 / r
Equating the two forces gives the basic relationship. Note that the particle’s velocity changes direction continuously while its magnitude remains fixed.
令两力相等即可得到基本关系。注意粒子速度大小恒定，方向不断变化。
4. Equating Forces: qvB = mv²/r | 力平衡：qvB = mv²/r
Set the magnetic force equal to the centripetal force: Bqv = mv²/r. Cancel one power of v (v ≠ 0) to obtain Bq = mv/r. This step leads directly to the radius formula.
令磁力等于向心力：Bqv = mv²/r。约去一个 v（v ≠ 0）得 Bq = mv/r。这一步直接指向半径公式。
B q v = m v2 / r → B q = m v / r
From here, rearrangement gives the desired expression for the circular path’s radius.
由此整理即可得到圆周路径的半径表达式。
5. Solving for Radius r = mv/(Bq) | 求解半径 r = mv/(Bq)
Solving for r yields r = (m v) / (B q). This compact formula states that the radius is directly proportional to the particle’s momentum mv and inversely proportional to both the magnetic flux density and the charge.
求解 r 得 r = (m v) / (B q)。这一简洁公式表明，半径与粒子动量 mv 成正比，与磁通量密度及电荷量成反比。
r = m v / (B q)
When the problem asks for an expression in terms of kinetic energy, replace v using v = √(2Ek/m), giving r = √(2mEk) / (B q). This variant is often tested.
若题目要求用动能表示，可用 v = √(2Ek/m) 替换，得 r = √(2mEk) / (B q)。该变体常被考查。
6. Understanding the Dependence on Mass, Velocity, Charge, and Field | 理解质量、速度、电荷与磁场的依赖关系
The proportionalities r ∝ m, r ∝ v, r ∝ 1/B, and r ∝ 1/q offer useful qualitative predictions. Heavier or faster particles sweep out larger circles, whereas a stronger field or larger charge tightens the curve.
比例关系 r ∝ m、r ∝ v、r ∝ 1/B 与 r ∝ 1/q 提供了有用的定性判断。粒子越重或越快，圆半径越大；磁场越强或电荷越多，轨道越紧。
In a cloud chamber, for example, electrons produce tight spirals because of their small mass, while protons or alpha particles create broader tracks. This demonstrates the derived dependence.
例如在云室中，电子因质量小而出现紧螺旋，质子或α粒子则产生较宽的径迹，这正是上述依赖关系的体现。
7. Deriving the Period of Circular Motion | 推导圆周运动周期
The period T is the time for one full revolution: T = circumference / speed = 2πr / v. Substitute the radius expression r = mv/(Bq) to get T = 2π(mv/(Bq)) / v = 2πm / (Bq). Importantly, T does not depend on speed.
周期 T 是完成一圈的时间：T = 周长 / 速率 = 2πr / v。代入半径表达式 r = mv/(Bq)，得 T = 2π(mv/(Bq)) / v = 2πm / (Bq)。关键之处在于，T 与速度无关。
T = 2π m / (B q)
This means all particles with the same charge-to-mass ratio take identical time for a full circle, regardless of their speed (provided v ≪ c). This is the fundamental idea behind the cyclotron.
这意味着荷质比相同的所有粒子完成一整圈所需时间相同，与其速率无关（只要 v ≪ c）。这正是回旋加速器的基本原理。
8. Deriving the Frequency (Cyclotron Frequency) | 推导频率（回旋频率）
Frequency f is the reciprocal of period: f = 1/T = (B q) / (2π m). This is universally known as the cyclotron frequency. The angular frequency follows as ω = 2πf = (B q) / m.
频率 f 是周期的倒数：f = 1/T = (B q) / (2π m)，通称回旋频率。角频率 ω = 2πf = (B q) / m。
f = B q / (2π m) ; ω = B q / m
In a cyclotron, the alternating electric field must oscillate at this precise frequency to synchronise with the particle’s motion and boost its energy each half-turn.
在回旋加速器中，交变电场必须以此频率振荡，才能与粒子运动同步，在每半圈为其加速。
9. Checking Homogeneity of Units | 检查单位的一致性
A quick unit check confirms the radius expression is dimensionally sound. Using SI: [m] = kg, [v] = m/s, [B] = T = N/(A·m) = kg/(A·s²), [q] = A·s. Then [r] = kg·(m/s) / ( (kg/(A·s²))·(A·s) ) = (kg·m/s) / (kg/s) = m, which is correct.
快速单位检验可确认半径表达式量纲正确。在国际单位制下：[m]=kg, [v]=m/s, [B]=T = N/(A·m) = kg/(A·s²), [q]=A·s。于是 [r]=kg·(m/s) / ( (kg/(A·s²))·(A·s) ) = (kg·m/s)/(kg/s)=m，无误。
Dimensional analysis is a powerful tool to spot algebraic mistakes, and examiners often expect candidates to show it.
量纲分析是发现代数错误的利器，阅卷人常期望考生展示这一步骤。
10. Worked Example and Application | 实例计算与应用
Example: A proton (mass 1.67×10-27 kg, charge 1.60×10-19 C) moves at 2.0×106 m/s perpendicular to a 0.50 T magnetic field. Determine the radius of its path.
例题：一个质子（质量 1.67×10-27 kg，电荷 1.60×10<
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