📚 Combined Electric and Magnetic Fields | 电场与磁场的结合
In Edexcel A-Level Physics, the study of combined electric and magnetic fields brings together two fundamental areas of electromagnetism. When charged particles move through regions where both an electric field and a magnetic field are present, their motion is governed by the Lorentz force law, giving rise to a range of important practical devices such as velocity selectors, mass spectrometers, and cyclotrons. Understanding the vector nature of these fields and how they interact with moving charges is essential for analysing particle trajectories, energy changes, and force equilibrium conditions.
在Edexcel A-Level物理课程中，电场与磁场的结合研究将电磁学的两个基本领域联系在了一起。当带电粒子在同时存在电场和磁场的区域中运动时，其运动由洛伦兹力定律支配，由此衍生出一系列重要的实用装置，如速度选择器、质谱仪和回旋加速器。理解这些场的矢量特性以及它们与运动电荷的相互作用，对于分析粒子轨迹、能量变化和力的平衡条件至关重要。
1. The Lorentz Force - Vector Formulation | 洛伦兹力——矢量形式
s given by the Lorentz force equation: F = q(E + v × B). The electric force qE acts parallel or antiparallel to the field direction, while the magnetic force qv × B is perpendicular to both the velocity vector and the magnetic field vector.
电荷 q 在电场 E 和磁场（严格来说是磁通密度）B 中以速度 v 运动时，所受到的总电磁力由洛伦兹力方程给出：F = q(E + v × B)。电场力 qE 与电场方向平行或反平行，而磁场力 qv × B 同时垂直于速度矢量和磁场矢量。
The direction of the magnetic component is found using Fleming' a negative charge, the force direction is reversed. The magnitude of the magnetic force is |q| the angle between v and B. When v is parallel to B, sin θ = 0 and no magnetic force acts.
磁场分量的方向对于正电荷可用弗莱明左手定则判断；对于负电荷，力的方向相反。磁场力的大小为 |q|vB sin θ，其中 θ 是 v 与 B 之间的夹角。当 v 平行于 B 时，sin θ = 0，磁场力为零。
| Quantity | Symbol | Unit |
|---|---|---|
| Electric field strength | E | N C⁻¹ or V m⁻¹ |
| Magnetic flux density | B | T (tesla) |
| Charge | q | C |
| Velocity | v | m s⁻¹ |
2. Motion in Uniform Electric Fields | 在均匀电场中的运动
arabolic trajectory analogous to projectile motion under gravity. The acceleration a = qE/m is constant, so kinematic equations apply. For an electron accelerated from rest through a potential difference V, the kinetic energy gained is:
当带电粒子垂直于初速度方向进入均匀电场时，会受到恒定的电场力 qE，其轨迹呈抛物线，类似于重力作用下的抛体运动。加速度 a = qE/m 是恒定的，因此可应用运动学方程。对于从静止加速穿过电势差 V 的电子，获得的动能为：
½mv² = qV → v = √2qV/m
If the particle enters the field parallel to the field lines, it simply accelerates or decelerates along a straight line depending on the sign of the charge and field direction.
若粒子平行于电场线进入电场，则只会沿着直线加速或减速，具体取决于电荷符号和电场方向。
3. Motion in Uniform Magnetic Fields – Circular Paths | 在均匀磁场中的运动——圆周路径
its velocity. This provides the centripetal force required for circular motion: qvB = mv²/r, giving the radius of the path r = mv/(qB). The period of revolution T = 2πr/v = 2πm/(qB) is independent of speed, a fact exploited in the cyclotron.
垂直于均匀磁场运动的带电粒子受到大小恒定为 qvB 且始终与速度垂直的力。这提供了圆周运动所需的向心力：qvB = mv²/r，从而得出路径半径 r = mv/(qB)。旋转周期 T = 2πr/v = 2πm/(qB) 与速度无关，回旋加速器正是利用了这一事实。
If the velocity has a component parallel to B, the particle follows a helical path because the parallel component remains unchanged while the perpendicular component causes circular motion.
如果速度具有平行于 B 的分量，则粒子会沿螺旋路径运动，因为平行分量保持不变，而垂直分量引起圆周运动。
4. Velocity Selector – Balanced Fields | 速度选择器——平衡场
rtically and B horizontally into the page, while the charged particles travel horizontally. The electric force qE is upward (for a positive charge), and the magnetic force qvB is downward. When these are equal in magnitude, qE = qvB, the net force is zero, so particles pass through undeflected.
速度选择器利用相互垂直的电场和磁场，使运动电荷所受的力可以相互抵消。通常，E 指向垂直方向，B 指向纸面内水平方向，而带电粒子沿水平方向运动。电场力 qE 向上（对正电荷），磁场力 qvB 向下。当二者大小相等，即 qE = qvB 时，合外力为零，粒子无偏转地通过。
ct speed are selected; faster or slower particles experience a net force and are deflected away from the exit slit.
该条件给出了特定的速度 v = E/B。只有具有该精确速度的粒子才能被选出；速度更快或更慢的粒子会感受到净力，从而偏离出口狭缝。
5. Thomson's e/m Experiment – Historical Context | 汤姆孙的 e/m 实验——历史背景
J.J. Thomson' electron beam and then adding a magnetic field to cancel the deflection, he determined v = E/B. Removing the electric field allowed the magnetic field alone to bend the beam into a circular arc of radius r, giving e/m = E/(B²r).
J.J. 汤姆孙于 1897 年测量电子荷质比 (e/m) 的实验利用了组合场。他首先施加已知电场使电子束偏转，再加入磁场来抵消偏转，从而确定 v = E/B。移除电场后，仅靠磁场使电子束弯曲成半径为 r 的圆弧，得出 e/m = E/(B²r)。
Thomson' C kg⁻¹, establishing that electrons are fundamental particles with a specific charge vastly greater than that of any ion, thus proving the existence of subatomic particles.
汤姆孙的测量值约为 1.76 × 10¹¹ C kg⁻¹，这表明电子是一种基本粒子，其比电荷远大于任何离子，从而证明了亚原子粒子的存在。
6. Mass Spectrometry – Separation by Mass-to-Charge Ratio | 质谱分析法——按质荷比分离
ectrometer uses a velocity selector (E and B perpendicular) to ensure all ions entering the main magnetic field B₀ have the same speed v = E/B. In the uniform field B₀, ions move in semicircular paths with radius r = mv/(qB₀).
质谱仪利用组合电场和磁场，根据离子的质荷比 (m/q) 进行分离。班布里奇质谱仪使用速度选择器（E 与 B 垂直）确保所有进入主磁场 B₀ 的离子具有相同的速度 v = E/B。在均匀磁场 B₀ 中，离子沿半径 r = mv/(qB₀) 的半圆路径运动。
₀B r/E. By measuring r (or the position on a photographic plate or detector), different isotopes or ion species can be identified and their relative abundances determined.
因此，质荷比由半径确定：m/q = B₀r/v = B₀B r/E。通过测量 r（或照相底板/探测器上的位置），可以鉴别不同的同位素或离子种类，并测定其相对丰度。
7. Cyclotron – Acceleration with Constant Period | 回旋加速器——恒定周期加速
d in a uniform magnetic field B. The magnetic field forces particles into a circular path inside each dee; each time the particle crosses the gap between dees, the electric field from the alternating voltage gives it a boost. Since T = 2πm/(qB) is independent of speed, the frequency f = qB/(2πm) of the alternating voltage must match the cyclotron frequency for synchronous acceleration.
回旋加速器利用位于均匀磁场 B 中的两个中空 D 形电极（D 形盒）之间的高频交变电压来加速带电粒子。磁场迫使粒子在每个 D 形盒内做圆周运动；粒子每次穿过盒间缝隙时，交变电压产生的电场会使其加速。由于 T = 2πm/(qB) 与速度无关，交变电压的频率 f = qB/(2πm) 必须与回旋频率一致，以实现同步加速。
²B²R²/(2m). However, at relativistic speeds, mass increases and T changes, limiting the cyclotron to non-relativistic energies unless a synchrocyclotron is used.
获得的动能与 D 形盒的最终半径 R 有关：KE = q²B²R²/(2m)。不过，在接近相对论性速度时，质量增大、周期 T 改变，从而限制了回旋加速器只适用于非相对论性能量，除非使用同步回旋加速器。
8. Hall Effect – Combined Fields in a Conductor | 霍尔效应——导体中的组合场
ce a magnetic force qvₐB (vₐ is drift velocity) that pushes them to one side, creating a transverse electric field. Equilibrium is reached when the electric force qEₕ equals the magnetic force: qEₕ = qvₐB, so the Hall voltage Vₕ = Eₕw = vₐBw, where w is the width of the conductor.
当载流导体或半导体置于垂直磁场中时，会产生霍尔效应。载流子受到磁场力 qvₐB（vₐ 为漂移速度）的作用，被推向一侧，从而建立起横向电场。当电场力 qEₕ 与磁场力大小相等，即 qEₕ = qvₐB 时达到平衡，因此霍尔电压 Vₕ = Eₕw = vₐBw，其中 w 为导体宽度。
icient is defined as Rₕ = 1/(nq), and measuring Vₕ allows determination of carrier density and type (positive or negative).
利用电流 I = nqAvₐ（n 为载流子浓度，A 为横截面积），可以推导出 Vₕ = BI/(nqt)，其中 t 为厚度。霍尔系数定义为 Rₕ = 1/(nq)，通过测量 Vₕ 可确定载流子浓度及其类型（正或负）。
9. Force Equilibrium Problems – Solving Strategies | 力平衡问题——解题策略
ply the Lorentz force law component by component. If the particle moves in a straight line with constant velocity, the net force must be zero, which often gives v = E/B for perpendicular fields. For curved paths, the net force provides the centripetal force: |q|vB - |q|E = mv²/r, accounting for direction.
在电场力和磁场力共同作用的问题中，第一步是画出显示所有力矢量的受力图。然后逐分量应用洛伦兹力定律。若粒子以恒定速度沿直线运动，则合外力必为零，对于垂直场这通常给出 v = E/B。对于曲线路径，合外力提供向心力：|q|vB - |q|E = mv²/r，并注意方向。
Common pitfalls include forgetting to consider the sign of the charge, misapplying the right-hand/left-hand rules, and confusing electric and magnetic force directions. Always check that E and B are truly perpendicular and note whether the fields are uniform.
常见的易错点包括忽略电荷符号、错误使用右手/左手定则，以及混淆电场力和磁场力的方向。务必检查 E 与 B 是否真正垂直，并注意场是否均匀。
10. Worked Example – Velocity Selector and Mass Spec Calculation | 典型例题——速度选择器与质谱计算
eld B₀ = 0.30 T where they move in semicircles of radius 12 cm. Determine the mass of one ion.
一束带单个正电荷的离子通过一个速度选择器，其中 E = 2.5 × 10⁴ V m⁻¹，B = 0.15 T。然后离子进入均匀磁场 B₀ = 0.30 T，在其中做半径为 12 cm 的半圆运动。求单个离子的质量。
Step 1: Selected speed v = E/B = 2.5 × 10⁴ / 0.15 = 1.67 × 10⁵ m s⁻¹. Step 2: In field B₀, radius r = mv/(qB₀), so m = qB₀r/v. 1.6 × 10⁻¹⁹ C, r = 0.12 m: m = (1.6 × 10⁻¹⁹)(0.30)(0.12) / (1.67 × 10⁵) ≈ 3.45 × 10⁻²⁶ kg.
步骤1：选择速度 v = E/B = 2.5 × 10⁴ / 0.15 = 1.67 × 10⁵ m s⁻¹。 步骤2：在磁场 B₀ 中，半径 r = mv/(qB₀)，因此 m = qB₀r/v。 代入 q = +1.6 × 10⁻¹⁹ C，r = 0.12 m：m = (1.6 × 10⁻¹⁹)(0.30)(0.12) / (1.67 × 10⁵) ≈ 3.45 × 10⁻²⁶ kg。
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