📚 IB Physics: Analysing Trajectories of Charged Particles in Magnetic Fields | IB物理：带电粒子在磁场中的运动轨迹分析
When a charged particle enters a magnetic field, its path can curve into circles, helixes, or even remain a straight line, depending on the angle between its velocity and the magnetic field. This behaviour is central to IB Physics, connecting electromagnetism with circular motion and energy conservation. In this article, we will analyse the possible trajectories, derive the key equations, and apply them to exam-style situations.
当带电粒子进入磁场时，其运动轨迹可以是圆弧、螺旋线，甚至保持直线，具体取决于速度与磁场之间的夹角。这一行为是IB物理的核心内容，将电磁学与圆周运动、能量守恒紧密联系起来。本文将系统分析可能的轨迹，推导关键方程，并应用于考试常见情境。
1. The Lorentz Force and Its Direction | 洛伦兹力及其方向
A charged particle moving in a magnetic field experiences the Lorentz force. In vector form, the force is given by the cross product of velocity v and magnetic field B, multiplied by the charge q.
在磁场中运动的带电粒子会受到洛伦兹力。矢量形式下，力等于电荷量 q 乘以速度 v 与磁场 B 的叉积。
F = q v × B
For the magnitude of the force, we use the angle θ between the velocity vector and the magnetic field vector.
力的大小使用速度矢量与磁场矢量之间的夹角 θ 来计算。
F = qvB sin θ
To find the direction of the force on a positive charge, use the right-hand rule: point your fingers in the direction of v, curl them toward B, and your thumb points along F. For a negative charge, reverse the direction of the thumb.
要判断正电荷受力方向，请使用右手定则：右手手指指向 v 的方向，再向 B 的方向弯曲，拇指所指即为正电荷受力方向。对于负电荷，则将拇指方向取反。
2. No Work Done and Constant Speed | 力不做功与速率恒定
Because the magnetic force is always perpendicular to the displacement of the particle, the work done by the magnetic force is zero at every instant.
由于磁场力始终垂直于粒子的位移，因此磁场力在任意时刻做功均为零。
W = F d cos 90° = 0
If no other forces do work, the kinetic energy of the particle stays constant. The magnetic field can change the direction of the velocity, but it cannot change the speed of the particle.
如果没有其他力做功，粒子的动能保持不变。磁场可以改变速度的方向，但不能改变粒子速度的大小。
This is a powerful exam concept: even when the trajectory curves into a circle, the particle never speeds up or slows down due to the magnetic force alone.
这是一个重要的考试概念：即使轨迹弯曲成圆，仅由磁场力作用时，粒子也永远不会加速或减速。
3. Circular Motion for Perpendicular Entry | 垂直射入时的圆周运动
= 1. The magnetic force provides the centripetal force needed for uniform circular motion.
当速度与磁场恰好垂直时，θ = 90°，sin θ = 1。此时磁场力提供匀速圆周运动所需的向心力。
qvB = mv² / r
Solving for the radius of the circular path gives the most important equation for this topic.
解出圆周运动的半径，可以得到本主题最重要的方程。
r = mv / (qB)
The period of the circular motion, T, is the circumference divided by the speed.
圆周运动的周期 T 等于周长除以速率。
T = 2πr / v = 2πm / (qB)
A striking result is that the period does not depend on the speed or the radius. A faster particle moves in a larger circle but completes one revolution in exactly the same time as a slower particle with the same mass and charge.
一个显著结论是：周期与速度、半径无关。速度更快的粒子在更大的圆上运动，但与质量、电荷相同的较慢粒子完成一圈所需的时间完全相同。
4. Helical Motion for Oblique Entry | 斜射入时的螺旋运动
If the velocity makes an angle θ with the magnetic field, only the component perpendicular to the field causes circular motion. The component parallel to the field is unaffected by the magnetic force.
如果速度与磁场成夹角 θ，只有垂直于磁场的速度分量产生圆周运动；平行于磁场的速度分量不受磁场力影响。
v⊥ = v sin θ, v∥ = v cos θ
The radius of the helical path is determined by the perpendicular component of velocity.
螺旋轨迹的半径由速度的垂直分量决定。
r = m v sin θ / (qB)
As the particle circles around the field line, it also drifts along the field direction. After one period T, the distance travelled parallel to the field is called the pitch p.
粒子一边绕磁感线旋转，一边沿磁场方向漂移。经过一个周期 T 后，沿磁场方向前进的距离称为螺距 p。
p = v∥ T = 2πm v cos θ / (qB)
Helical motion explains many natural phenomena, such as charged particles spiralling along Earth's magnetic field lines near the polar regions.
螺旋运动可以解释许多自然现象，例如带电粒子在地球极区附近沿磁感线螺旋运动。
5. Comparing Protons, Electrons and Alpha Particles | 质子、电子与α粒子的轨迹比较
. Different particles therefore follow very different paths. The table below compares the radius of a proton, an electron and an alpha particle, both for equal speed and for equal kinetic energy, relative to a proton.
在相同磁场中，半径 r = mv/(qB) 取决于 m/q 的比值。因此不同粒子的路径差异很大。下表比较了质子、电子和α粒子相对于质子的半径，分别在速度相同和动能相同的条件下。
| Particle | 粒子 | Charge | 电荷 | Mass relative to proton | 相对质子的质量 | r for same speed | 相同速度时的半径 | r for same kinetic energy | 相同动能时的半径 |
|---|---|---|---|---|
| Proton | 质子 | +e | 1 | 1 | 1 |
| Electron | 电子 | −e | 1/1836 | 1/1836, opposite direction | 方向相反 | 1/42.8, opposite direction | 方向相反 |
| Alpha particle | α粒子 | +2e | 4 | 2 | 1 |
For equal speed, r is proportional to m/q, so the alpha particle has twice the radius of a proton. For equal kinetic energy, r is proportional to √m, so a proton and an alpha particle happen to have the same radius if q = 2e and m = 4m_p.
在速度相同时，r 与 m/q 成正比，因此α粒子的半径是质子的两倍。在动能相同时，r 与 √m/q 成正比，因此当 q = 2e、m = 4m_p 时，质子和α粒子的半径恰好相同。
6. Worked Example: Circular Trajectory | 例题精讲：圆周轨迹
with a speed of 2.0 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.50 T. Given mp 10⁻²⁷ kg and e = 1.60 × 10⁻¹⁹ C, find the radius of the path and the period of revolution.
一个质子以 2.0 × 10⁶ m/s 的速度垂直于 0.50 T 的匀强磁场运动。已知 mp = 1.67 × 10⁻²⁷ kg，e = 1.60 × 10⁻¹⁹ C，求轨迹半径和运动周期。
The radius is found using r = mv/(qB).
半径由 r = mv/(qB) 求出。
r = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.50) = 0.04175 m ≈ 4.2 cm
The period is found using T = 2πm/(qB).
周期由 T = 2πm/(qB) 求出。
T = (2π × 1
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