Unit D.3: Motion in Electromagnetic Fields单元 D.3:电磁场中的运动
The dynamics half of Theme D "Fields". How charges and currents respond to magnetic and electric fields: the motor-effect force on a current-carrying conductor, the magnetic force on a single moving charge, the circular orbits charges trace in a uniform magnetic field, the parabolic deflection of charges in a uniform electric field, the force between two parallel currents, and the crossed-field velocity selector behind mass spectrometers and the cathode-ray tube. These rules underpin every electromagnetic device on the syllabus, and they feed straight into D.4 Induction.主题 D"场"的动力学部分。讲电荷与电流如何响应磁场与电场:载流导线上的电动机效应力、单个运动电荷受到的磁力、电荷在匀强磁场中描出的圆轨道、电荷在匀强电场中的抛物线偏转、两条平行电流之间的力,以及质谱仪与阴极射线管背后的正交场速度选择器。这些规则支撑了大纲上的每一个电磁器件,并直接通往 D.4 电磁感应。
How to use this guide本指南使用说明
D.3 is the unit where direction matters as much as magnitude. Almost every mark is split between getting the size of a force right (plug into the data-booklet formula) and getting its direction right (apply a hand rule cleanly). Train the two skills in parallel: for each formula, also rehearse the matching hand rule, and always ask "is the charge positive or negative?" before reading off a direction.D.3 是方向与大小同等重要的单元。几乎每一分都被拆成两半:算对力的大小(代入数据手册公式)以及判对力的方向(干净地用手定则)。两项技能并行训练:每记一个公式,同时复习对应的手定则;读方向前永远先问"电荷是正还是负?"。
Memorise four formulas: $F = BIL\sin\theta$ (wire), $F = qvB\sin\theta$ (charge), $r = \dfrac{mv}{qB}$ (orbit radius), $v = \dfrac{E}{B}$ (velocity selector). For direction, use Fleming's left-hand rule (thumb = force, first finger = field, second finger = conventional current). A magnetic force never does work, so it changes direction, not speed.
背熟四个公式:$F = BIL\sin\theta$(导线)、$F = qvB\sin\theta$(电荷)、$r = \dfrac{mv}{qB}$(轨道半径)、$v = \dfrac{E}{B}$(速度选择器)。方向用弗莱明左手定则(拇指=力,食指=场,中指=常规电流)。磁力永不做功,只改变方向、不改变速率。
Be able to derive $r = mv/(qB)$ by setting the magnetic force equal to the centripetal force, and show the period $T = 2\pi m/(qB)$ is independent of speed. State clearly that for a negative charge the force reverses relative to a positive charge in the same $\vec{v}$ and $\vec{B}$. Explain crossed fields by force balance $qE = qvB$, and connect the velocity selector to the mass spectrometer and CRT.
能通过令磁力等于向心力推出 $r = mv/(qB)$,并证明周期 $T = 2\pi m/(qB)$ 与速率无关。明确指出:在相同 $\vec{v}$、$\vec{B}$ 下,负电荷受力方向与正电荷相反。用力平衡 $qE = qvB$ 解释正交场,并把速度选择器与质谱仪、阴极射线管联系起来。
Force on a Current-Carrying Conductor载流导线在磁场中受力 D.3 SL+HL
- $\theta$ is measured between the current direction and $\vec{B}$. Force is maximum when $\theta = 90^\circ$ (wire perpendicular to field) and zero when $\theta = 0^\circ$ (wire parallel to field).
- Data booklet:
F = BIL sin θ. Units: $B$ in $\mathrm{T}$, $I$ in $\mathrm{A}$, $L$ in $\mathrm{m}$, giving $F$ in $\mathrm{N}$.
- $\theta$ 是电流方向与 $\vec{B}$ 之间的夹角。$\theta = 90^\circ$(导线垂直于磁场)时力最大,$\theta = 0^\circ$(导线平行于磁场)时力为零。
- 数据手册:
F = BIL sin θ。单位:$B$ 取 $\mathrm{T}$、$I$ 取 $\mathrm{A}$、$L$ 取 $\mathrm{m}$,得 $F$ 单位 $\mathrm{N}$。
A straight wire of length $0.25\ \mathrm{m}$ carries a current of $3.0\ \mathrm{A}$ and lies at $30^\circ$ to a uniform magnetic field of flux density $0.40\ \mathrm{T}$. Find the magnitude of the force on the wire.长度 $0.25\ \mathrm{m}$ 的直导线载流 $3.0\ \mathrm{A}$,与磁感应强度 $0.40\ \mathrm{T}$ 的匀强磁场成 $30^\circ$ 角。求导线受力的大小。
Identify. Use $F = BIL\sin\theta$, the data-booklet motor-effect law. Here $B = 0.40\ \mathrm{T}$, $I = 3.0\ \mathrm{A}$, $L = 0.25\ \mathrm{m}$, $\theta = 30^\circ$.
识别。用数据手册的电动机效应公式 $F = BIL\sin\theta$。这里 $B = 0.40\ \mathrm{T}$、$I = 3.0\ \mathrm{A}$、$L = 0.25\ \mathrm{m}$、$\theta = 30^\circ$。
Set up. Substitute, with $\sin 30^\circ = 0.50$.
列式。代入,$\sin 30^\circ = 0.50$。
$$ F = (0.40)(3.0)(0.25)(0.50). $$Execute.
计算。
$$ F = 0.15\ \mathrm{N}. $$Evaluate. The $\sin 30^\circ = 0.5$ factor halves the force compared with a perpendicular wire ($0.30\ \mathrm{N}$). Forgetting the angle factor is the most common error here.
评估。$\sin 30^\circ = 0.5$ 使力比垂直时($0.30\ \mathrm{N}$)减半。漏掉角度因子是此处最常见的错误。
Going deeper: why $F = BIL\sin\theta$ comes from the force on moving charges深入:$F = BIL\sin\theta$ 为何源自运动电荷受力
A current is a stream of charge carriers, each with charge $q$ drifting at speed $v_d$. In a wire of length $L$ and cross-section carrying $n$ carriers per unit volume, the current is $I = nAqv_d$. Each carrier feels $qv_dB\sin\theta$; summing over all carriers in the length $L$ gives the wire force.
电流是一束载流子,每个带电荷 $q$、以漂移速率 $v_d$ 运动。长度 $L$、单位体积载流子数为 $n$ 的导线中,电流 $I = nAqv_d$。每个载流子受力 $qv_dB\sin\theta$;对长度 $L$ 内全部载流子求和得到导线所受力。
$$ F = (nAL)\,(q v_d B \sin\theta) = (nAqv_d)\,L B \sin\theta = B I L \sin\theta. $$So the macroscopic wire law and the microscopic single-charge law $F = qvB\sin\theta$ are the same physics seen at two scales. This is why D3.1 and D3.2 share one hand rule.
所以宏观导线公式与微观单电荷公式 $F = qvB\sin\theta$ 是同一物理在两个尺度上的表现。这正是 D3.1 与 D3.2 共用同一手定则的原因。
Magnetic Force on a Single Moving Charge单个运动电荷的磁力(洛伦兹力) D.3 SL+HL
- Data booklet:
F = qvB sin θ. Maximum when $\vec{v} \perp \vec{B}$; zero when $\vec{v} \parallel \vec{B}$. - The force is always perpendicular to $\vec{v}$, so a magnetic field does no work: it changes the direction of motion, never the speed or kinetic energy.
- 数据手册:
F = qvB sin θ。$\vec{v} \perp \vec{B}$ 时最大;$\vec{v} \parallel \vec{B}$ 时为零。 - 力始终垂直于 $\vec{v}$,故磁场不做功:只改变运动方向,绝不改变速率或动能。
A proton moves east at $2.0\times10^{6}\ \mathrm{m\,s^{-1}}$ through a $0.30\ \mathrm{T}$ field pointing vertically upward ($\vec{v} \perp \vec{B}$). The proton charge is $1.6\times10^{-19}\ \mathrm{C}$. Find the magnitude of the force, and state how it would change for an electron moving the same way.质子以 $2.0\times10^{6}\ \mathrm{m\,s^{-1}}$ 向东穿过竖直向上的 $0.30\ \mathrm{T}$ 磁场($\vec{v} \perp \vec{B}$)。质子电荷 $1.6\times10^{-19}\ \mathrm{C}$。求力的大小,并说明若换成同向运动的电子会如何变化。
Identify. Use $F = qvB\sin\theta$ with $\theta = 90^\circ$, so $\sin\theta = 1$.
识别。用 $F = qvB\sin\theta$,$\theta = 90^\circ$,$\sin\theta = 1$。
$$ F = (1.6\times10^{-19})(2.0\times10^{6})(0.30). $$Execute.
计算。
$$ F = 9.6\times10^{-14}\ \mathrm{N}. $$Direction. By Fleming's left-hand rule (current east = $\vec{v}$ east for the positive proton, field up), the force points horizontally to the south.
方向。由弗莱明左手定则(正质子的电流向东=$\vec{v}$ 向东,磁场向上),力水平指向南方。
Electron. An electron has the same speed and field, so the magnitude $9.6\times10^{-14}\ \mathrm{N}$ is identical (its charge magnitude is the same). But because conventional current is opposite to its velocity, the force points horizontally to the north — exactly reversed.
电子。电子速率与磁场相同,故大小同为 $9.6\times10^{-14}\ \mathrm{N}$(电荷大小相同)。但其常规电流与速度相反,故力水平指向北方——方向恰好相反。
Evaluate. Sign of charge flips the direction but never the magnitude. This sign rule is the heart of D.3 direction questions.
评估。电荷符号只改变方向、绝不改变大小。这条符号规则是 D.3 方向题的核心。
Going deeper: the full Lorentz force and why $\vec{B}$ does no work深入:完整洛伦兹力与磁场为何不做功
The complete force on a charge in combined fields is $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$. The magnetic part $q\vec{v}\times\vec{B}$ is, by the cross product, perpendicular to $\vec{v}$. Power delivered is $P = \vec{F}\cdot\vec{v}$; for the magnetic term this is $q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$ because a vector perpendicular to $\vec{v}$ has zero dot product with $\vec{v}$.
电荷在复合场中受到的完整力为 $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$。其磁分量 $q\vec{v}\times\vec{B}$ 由叉积可知垂直于 $\vec{v}$。功率 $P = \vec{F}\cdot\vec{v}$;磁项给出 $q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$,因为垂直于 $\vec{v}$ 的矢量与 $\vec{v}$ 的标量积为零。
Hence the kinetic energy $\tfrac12 mv^2$ is constant under a purely magnetic force — the speed is fixed and only the direction turns. Electric fields, by contrast, can do work and change kinetic energy (D3.4).
因此在纯磁力作用下动能 $\tfrac12 mv^2$ 守恒——速率不变,只有方向转动。相反,电场可以做功并改变动能(D3.4)。
Circular Motion of a Charged Particle带电粒子的圆周运动 D.3 SL+HL
- $r$ grows with momentum $mv$ and shrinks with stronger $B$.
- The period $T$ is independent of speed and radius — faster particles trace bigger circles but take the same time per loop. This is the cyclotron principle.
- $r$ 随动量 $mv$ 增大,随 $B$ 增强而减小。
- 周期 $T$ 与速率和半径无关——快粒子描出更大的圆,但每圈用时相同。这正是回旋加速器原理。
A proton ($m = 1.67\times10^{-27}\ \mathrm{kg}$, $q = 1.6\times10^{-19}\ \mathrm{C}$) moves at $5.0\times10^{6}\ \mathrm{m\,s^{-1}}$ perpendicular to a $0.50\ \mathrm{T}$ field. Find the radius of its circular path and the period of its orbit.质子($m = 1.67\times10^{-27}\ \mathrm{kg}$,$q = 1.6\times10^{-19}\ \mathrm{C}$)以 $5.0\times10^{6}\ \mathrm{m\,s^{-1}}$ 垂直于 $0.50\ \mathrm{T}$ 磁场运动。求其圆轨道半径与周期。
Identify. Perpendicular entry gives circular motion; use $r = mv/(qB)$ and $T = 2\pi m/(qB)$.
识别。垂直入射产生圆周运动;用 $r = mv/(qB)$ 与 $T = 2\pi m/(qB)$。
Radius.
半径。
$$ r = \frac{(1.67\times10^{-27})(5.0\times10^{6})}{(1.6\times10^{-19})(0.50)} \approx 0.10\ \mathrm{m}. $$Period.
周期。
$$ T = \frac{2\pi (1.67\times10^{-27})}{(1.6\times10^{-19})(0.50)} \approx 1.3\times10^{-7}\ \mathrm{s}. $$Evaluate. If the proton were sped up, $r$ would grow proportionally but $T$ would not change at all — the speed cancels out of the period. That speed-independence is the key conceptual point examiners probe.
评估。若质子加快,$r$ 按比例增大,而 $T$ 完全不变——速率在周期中被抵消。这种"与速率无关"正是考官重点考查的概念。
Going deeper: deriving the speed-independent period深入:推导与速率无关的周期
Start from $r = mv/(qB)$ and the circumference relation $T = 2\pi r / v$. Substitute $r$:
从 $r = mv/(qB)$ 与周长关系 $T = 2\pi r / v$ 出发,代入 $r$:
$$ T = \frac{2\pi}{v}\cdot\frac{mv}{qB} = \frac{2\pi m}{qB}. $$The speed $v$ cancels, so $T$ depends only on the mass-to-charge ratio $m/q$ and the field $B$. A faster particle covers a longer circumference at a proportionally higher speed, so the time per revolution is unchanged. This is exactly why a cyclotron can use a fixed-frequency accelerating voltage as the particle spirals outward.
速率 $v$ 被消去,故 $T$ 仅依赖荷质比 $m/q$ 与磁场 $B$。快粒子在按比例更高的速率下走更长的周长,故每圈用时不变。这正是回旋加速器能在粒子向外螺旋时使用固定频率加速电压的原因。
Charged Particle in a Uniform Electric Field匀强电场中的带电粒子 D.3 SL+HL
E = V/d).
Parabolic deflection. A charge entering perpendicular to the field (along $x$ at speed $v_0$) behaves like a projectile: uniform velocity along $x$, uniform acceleration along $y$. The path is a parabola, $y = \tfrac12 a (x/v_0)^2$.
Work and energy. Moving a charge $q$ through a potential difference $V$ does work
$$ W = qV. $$
If accelerated from rest, this work becomes kinetic energy: $qV = \tfrac12 m v^{2}$ — the key energy-gain relation (data booklet W = qV).
E = V/d)。
抛物线偏转。垂直于电场进入(沿 $x$ 方向、速率 $v_0$)的电荷如同抛体:沿 $x$ 匀速、沿 $y$ 匀加速。轨迹为抛物线,$y = \tfrac12 a (x/v_0)^2$。
功与能。使电荷 $q$ 通过电势差 $V$ 所做的功为
$$ W = qV. $$
若从静止加速,此功转化为动能:$qV = \tfrac12 m v^{2}$——关键的能量增益关系(数据手册 W = qV)。
An electron ($m = 9.11\times10^{-31}\ \mathrm{kg}$, $q = 1.6\times10^{-19}\ \mathrm{C}$) is accelerated from rest through a potential difference of $500\ \mathrm{V}$. Find the kinetic energy gained and the final speed.电子($m = 9.11\times10^{-31}\ \mathrm{kg}$,$q = 1.6\times10^{-19}\ \mathrm{C}$)从静止经 $500\ \mathrm{V}$ 电势差加速。求获得的动能与末速率。
Identify. Energy gained equals work done: $W = qV$ (data booklet). Then set $W = \tfrac12 m v^{2}$ to find speed.
识别。获得的能量等于做的功:$W = qV$(数据手册)。再令 $W = \tfrac12 m v^{2}$ 求速率。
Kinetic energy.
动能。
$$ E_k = qV = (1.6\times10^{-19})(500) = 8.0\times10^{-17}\ \mathrm{J}. $$Final speed. From $\tfrac12 m v^{2} = E_k$:
末速率。由 $\tfrac12 m v^{2} = E_k$:
$$ v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2(8.0\times10^{-17})}{9.11\times10^{-31}}} \approx 1.3\times10^{7}\ \mathrm{m\,s^{-1}}. $$Evaluate. The accelerating field does work and raises kinetic energy — in sharp contrast to a magnetic field, which never changes speed. The energy $qV$ is also called $500\ \mathrm{eV}$ in electronvolts.
评估。加速电场做功并提高动能——与永不改变速率的磁场形成鲜明对比。能量 $qV$ 以电子伏表示即 $500\ \mathrm{eV}$。
Going deeper: the deflection inside parallel plates深入:平行板内的偏转
A charge enters horizontally at $v_0$ along $x$, between plates of length $\ell$ with field $E = V/d$ pointing along $y$. The acceleration is $a = qE/m = qV/(md)$, purely vertical. The horizontal motion is unaccelerated, so the time inside is $t = \ell / v_0$. The vertical deflection on exit is
电荷以 $v_0$ 沿 $x$ 水平进入长度为 $\ell$ 的两板之间,场 $E = V/d$ 沿 $y$ 方向。加速度 $a = qE/m = qV/(md)$,纯竖直。水平运动无加速,故板内时间 $t = \ell / v_0$。出口处竖直偏转为
$$ y = \frac{1}{2} a t^{2} = \frac{qV\ell^{2}}{2 m d v_0^{2}}. $$This is precisely the projectile decomposition from kinematics, with $qE/m$ playing the role of $g$. Increasing $V$ or $\ell$, or lowering $v_0$, increases the deflection — exactly how a cathode-ray tube steers its beam.
这正是运动学中的抛体分解,其中 $qE/m$ 扮演 $g$ 的角色。增大 $V$ 或 $\ell$、或减小 $v_0$ 都使偏转增大——这正是阴极射线管偏转电子束的方式。
Forces Between Two Parallel Current-Carrying Wires两条平行载流导线间的力 D.3 SL · HL深化
- Currents in the same direction → wires attract.
- Currents in opposite directions → wires repel.
- 电流同向 → 两线相吸。
- 电流反向 → 两线相斥。
Two long parallel wires $0.10\ \mathrm{m}$ apart carry currents of $10\ \mathrm{A}$ and $15\ \mathrm{A}$ in the same direction. Find the force per unit length between them and state whether they attract or repel. Take $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$.两条相距 $0.10\ \mathrm{m}$ 的长平行导线,同向载流 $10\ \mathrm{A}$ 与 $15\ \mathrm{A}$。求它们之间单位长度的力,并判断相吸还是相斥。取 $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$。
Identify. Same-direction currents attract. Magnitude per length from $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$.
识别。同向电流相吸。每米大小由 $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$ 给出。
Execute.
计算。
$$ \frac{F}{L} = \frac{(4\pi\times10^{-7})(10)(15)}{2\pi (0.10)} = \frac{(2\times10^{-7})(150)}{0.10} = 3.0\times10^{-4}\ \mathrm{N\,m^{-1}}. $$Direction. Currents are parallel and in the same direction, so the wires are pulled together: they attract.
方向。电流平行且同向,故两线被拉拢:相吸。
Evaluate. The shortcut $\mu_0/(2\pi) = 2\times10^{-7}$ keeps the arithmetic clean. Reversing one current would flip the sign to repulsion without changing the magnitude.
评估。用速算 $\mu_0/(2\pi) = 2\times10^{-7}$ 可使运算简洁。把其中一个电流反向只改变符号为排斥,大小不变。
Going deeper: why same-direction currents attract深入:同向电流为何相吸
Wire 1 (current up the page) produces a field that, at the location of wire 2 to its right, points into the page (right-hand grip rule). Wire 2 also carries current up the page. Apply Fleming's left-hand rule to wire 2: current up, field into the page, force points to the left — back toward wire 1. By symmetry wire 1 is pushed toward wire 2. Hence same-direction currents attract.
导线 1(电流向上)在其右侧导线 2 处产生指向纸面内的磁场(右手握螺旋定则)。导线 2 也载向上的电流。对导线 2 用弗莱明左手定则:电流向上、磁场入纸面,力指向左——朝向导线 1。由对称性,导线 1 被推向导线 2。故同向电流相吸。
Reverse one current and every direction flips, so the forces point apart: opposite-direction currents repel. This force-per-length expression historically defined the ampere.
把一个电流反向,所有方向随之翻转,故力相互背离:反向电流相斥。这一单位长度受力表达式历史上曾用于定义安培。
Combined Fields: the Velocity Selector复合场:速度选择器 D.3 SL · HL深化
- Electric force $F_E = qE$ (independent of speed).
- Magnetic force $F_B = qvB$ (grows with speed).
- 电场力 $F_E = qE$(与速率无关)。
- 磁场力 $F_B = qvB$(随速率增大)。
In a velocity selector the electric field is $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$ and the perpendicular magnetic field is $B = 0.20\ \mathrm{T}$. Find the speed of the ions that pass through undeflected.速度选择器中电场 $E = 3.0\times10^{4}\ \mathrm{V\,m^{-1}}$,与之垂直的磁场 $B = 0.20\ \mathrm{T}$。求不偏转通过的离子速率。
Identify. Undeflected means the electric and magnetic forces balance: $qE = qvB$, so $v = E/B$.
识别。不偏转意味着电场力与磁场力平衡:$qE = qvB$,故 $v = E/B$。
Execute.
计算。
$$ v = \frac{E}{B} = \frac{3.0\times10^{4}}{0.20} = 1.5\times10^{5}\ \mathrm{m\,s^{-1}}. $$Evaluate. The charge $q$ cancels, so this speed is selected regardless of the ion's charge or mass. Ions faster than this are deflected by the dominant magnetic force; slower ions are deflected by the dominant electric force.
评估。电荷 $q$ 被消去,故无论离子电荷或质量如何,都筛选出此速率。比它快的离子被占优的磁力偏转;比它慢的被占优的电力偏转。
Going deeper: how a mass spectrometer uses the selector深入:质谱仪如何利用选择器
A mass spectrometer chains two of this unit's ideas. First, a velocity selector passes only ions with $v = E/B$, guaranteeing every surviving ion has the same known speed. Those ions then enter a second region of pure magnetic field $B'$, where they follow circular arcs of radius
质谱仪串联了本单元两个思想。首先,速度选择器只放行 $v = E/B$ 的离子,保证每个通过的离子都有相同的已知速率。这些离子随后进入第二个纯磁场区域 $B'$,沿半径为下式的圆弧运动
$$ r = \frac{mv}{qB'}. $$Because $v$, $q$ and $B'$ are fixed and known, $r$ is directly proportional to mass $m$. Heavier ions land farther out on the detector, so position maps directly onto mass. The velocity selector (D3.6) and the gyroradius formula (D3.3) work together to make the measurement clean.
由于 $v$、$q$ 与 $B'$ 固定且已知,$r$ 与质量 $m$ 成正比。较重的离子落在探测器更外侧,故位置直接对应质量。速度选择器(D3.6)与回旋半径公式(D3.3)协同使测量变得清晰。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Use Fleming's left-hand rule with conventional current. For an electron or any negative charge, point the "current" finger opposite to the velocity.
- 用弗莱明左手定则时取常规电流方向。对电子或任何负电荷,"电流"指向与速度相反。
- Ask "positive or negative?" before reading off any direction. Markschemes credit the correct sense; a flipped sign loses the direction mark.
- 读方向前先问"正还是负?"。评分按正确指向给分;方向弄反会丢方向分。
- Both $F = BIL\sin\theta$ and $F = qvB\sin\theta$ carry the angle factor. Maximum at $90^\circ$, zero at $0^\circ$. Check whether the question gives a perpendicular geometry (then $\sin\theta = 1$).
- $F = BIL\sin\theta$ 与 $F = qvB\sin\theta$ 都带角度因子。$90^\circ$ 时最大,$0^\circ$ 时为零。先确认题目是否为垂直几何(则 $\sin\theta = 1$)。
- A magnetic field does no work. Never let it change kinetic energy or speed — it only bends the path.
- 磁场不做功。绝不让它改变动能或速率——它只弯曲路径。
- Derive $r = mv/(qB)$ by equating $qvB$ to $mv^2/r$. State this step explicitly to earn the method mark.
- 令 $qvB = mv^2/r$ 推出 $r = mv/(qB)$。明确写出此步以拿方法分。
- Remember $T = 2\pi m/(qB)$ has no $v$. "Increase the speed" changes $r$ but not $T$ — a favourite trap.
- 记住 $T = 2\pi m/(qB)$ 不含 $v$。"增大速率"改变 $r$ 但不改变 $T$——常见陷阱。
- Electric field: parabola, does work, changes kinetic energy via $W = qV$. Force is independent of speed.
- 电场:抛物线,做功,经 $W = qV$ 改变动能。力与速率无关。
- Magnetic field: circle, does no work, constant speed. Force grows with speed via $qvB$. In the velocity selector these two compete to give $v = E/B$.
- 磁场:圆,不做功,速率恒定。力随速率经 $qvB$ 增大。速度选择器中两者竞争给出 $v = E/B$。
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Unit D.3 Practice Quiz单元 D.3 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对(不看笔记、一次过)才打勾。
- Compute the force on a current-carrying wire with $F = BIL\sin\theta$, including the angle factor用 $F = BIL\sin\theta$(含角度因子)求载流导线受力
- Apply Fleming's left-hand rule to find the direction of the force on a wire or charge用弗莱明左手定则判定导线或电荷受力方向
- Compute the magnetic force on a moving charge with $F = qvB\sin\theta$用 $F = qvB\sin\theta$ 求运动电荷的磁力
- Reverse the force direction correctly for a negative charge (electron) versus a positive charge正确判定负电荷(电子)相对正电荷受力方向相反
- Explain why a magnetic field does no work and changes direction, not speed解释磁场为何不做功,只改变方向不改变速率
- Derive and use $r = mv/(qB)$ by equating magnetic and centripetal force通过令磁力等于向心力推导并使用 $r = mv/(qB)$
- State that the period $T = 2\pi m/(qB)$ is independent of speed and use it说明周期 $T = 2\pi m/(qB)$ 与速率无关并应用
- Identify a parabolic path in a uniform electric field and use $W = qV = \tfrac12 mv^2$识别匀强电场中的抛物线轨迹并使用 $W = qV = \tfrac12 mv^2$
- Find the energy and final speed of a charge accelerated through a potential difference求电荷经电势差加速后的能量与末速率
- State whether two parallel currents attract or repel from their relative directions由两平行电流的相对方向判定相吸或相斥
- HL Compute the force per unit length between parallel wires with $F/L = \mu_0 I_1 I_2/(2\pi r)$用 $F/L = \mu_0 I_1 I_2/(2\pi r)$ 求平行导线单位长度受力
- Apply the velocity selector $v = E/B$ and link it to the mass spectrometer and CRT应用速度选择器 $v = E/B$ 并联系质谱仪与阴极射线管
IB Paper-Style PracticeIB 试卷风格练习
D.3 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_D3_*.html with the bilingual built-in pattern.
D.3 配套 Practice 与 Solutions 在排期,上线后位于 Practice Questions/Unit_D3_*.html。