IB Physics HL · 鼎睿学苑

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。

IB Physics · Theme D.3 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL mix6 个核心专题 · SL + HL 混合

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?"。

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If you are cramming如果你在临阵磨枪

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。

If you are going for a 7如果你目标是 7 分

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。

HL flagHL 标记说明 D.3 is taken by both SL and HL students. The core force laws and circular-motion treatment are common; the quantitative force-between-parallel-wires expression and the deeper crossed-field / mass-spectrometer analysis are flagged HL where they extend beyond the SL requirement. SL students may treat the HL-flagged derivations as enrichment.D.3 由 SL 与 HL 学生共同学习。核心受力定律与圆周运动处理为共同内容;平行载流导线间力的定量表达式,以及更深入的正交场 / 质谱仪分析,超出 SL 要求处标记 HL。SL 学生可把带 HL 标记的推导视为拓展。

Force on a Current-Carrying Conductor载流导线在磁场中受力 D.3 SL+HL

The motor-effect force. A wire of length $L$ carrying current $I$ at angle $\theta$ to a magnetic field of flux density $B$ (the magnetic field strength, units tesla $\mathrm{T}$) feels a force $$ F = B I L \sin\theta. $$
  • $\theta$ 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 $\mathrm{N}$.
Direction — Fleming's left-hand rule. Left hand: thumb = thrust (force), First finger = Field ($\vec{B}$), seCond finger = Current ($I$, conventional). The three are mutually perpendicular.
电动机效应力(安培力)。长度 $L$、载流 $I$ 的导线与磁感应强度(magnetic field strength,磁场强度,单位特斯拉 $\mathrm{T}$)成 $\theta$ 角时,受到的力为 $$ F = B I L \sin\theta. $$
  • $\theta$ 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 $\mathrm{N}$。
方向——弗莱明左手定则。左手:拇指=推力(力),食指=磁场($\vec{B}$),中指=电流($I$,常规方向)。三者两两垂直。
Worked Example D3.1 (force on a wire)D3.1 例题(导线受力)

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。

A $0.50\ \mathrm{m}$ wire carrying $4.0\ \mathrm{A}$ sits perpendicular to a $0.20\ \mathrm{T}$ field. The force on it is:载流 $4.0\ \mathrm{A}$ 的 $0.50\ \mathrm{m}$ 导线垂直置于 $0.20\ \mathrm{T}$ 磁场中。其受力为:
D3.1 · Q1
$0.10\ \mathrm{N}$
$0.20\ \mathrm{N}$
$0.40\ \mathrm{N}$
$1.6\ \mathrm{N}$
Perpendicular means $\theta = 90^\circ$, $\sin\theta = 1$. $F = BIL = (0.20)(4.0)(0.50) = 0.40\ \mathrm{N}$.垂直即 $\theta = 90^\circ$、$\sin\theta = 1$。$F = BIL = (0.20)(4.0)(0.50) = 0.40\ \mathrm{N}$。
Use $F = BIL\sin\theta$ with $\sin 90^\circ = 1$. Multiply all three quantities: $0.20 \times 4.0 \times 0.50$.用 $F = BIL\sin\theta$,$\sin 90^\circ = 1$。三量相乘:$0.20 \times 4.0 \times 0.50$。
A current-carrying wire is placed exactly parallel to a uniform magnetic field. The force on the wire is:一根载流导线恰好平行放置于匀强磁场中。导线受力为:
D3.1 · Q2
Maximum, equal to $BIL$最大,等于 $BIL$
Half of $BIL$$BIL$ 的一半
Directed along the wire沿导线方向
Zero为零
When the wire is parallel to $\vec{B}$, $\theta = 0^\circ$ and $\sin 0^\circ = 0$, so $F = BIL\sin\theta = 0$. The motor effect needs a component of current perpendicular to the field.导线平行于 $\vec{B}$ 时 $\theta = 0^\circ$、$\sin 0^\circ = 0$,故 $F = BIL\sin\theta = 0$。电动机效应需要电流有垂直于磁场的分量。
$F$ scales with $\sin\theta$. Parallel means $\theta = 0$, so $\sin\theta = 0$ and the force vanishes.$F$ 与 $\sin\theta$ 成正比。平行即 $\theta = 0$,$\sin\theta = 0$,力消失。

Magnetic Force on a Single Moving Charge单个运动电荷的磁力(洛伦兹力) D.3 SL+HL

The Lorentz (magnetic) force. A charge $q$ moving at speed $v$ at angle $\theta$ to a field $B$ feels $$ F = q v B \sin\theta. $$
  • 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.
Direction. Use Fleming's left-hand rule with "current" pointing along the direction of conventional current. For a positive charge, current is along $\vec{v}$. For a negative charge (e.g. an electron), conventional current points opposite to $\vec{v}$, so the force reverses.
洛伦兹力(磁力)。电荷 $q$ 以速率 $v$、与磁场 $B$ 成 $\theta$ 角运动时受力 $$ F = q v B \sin\theta. $$
  • 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。
方向。用弗莱明左手定则,"电流"沿常规电流方向。对电荷,电流沿 $\vec{v}$。对电荷(如电子),常规电流方向与 $\vec{v}$ 相反,故受力方向反转。
Worked Example D3.2 (positive vs negative charge)D3.2 例题(正电荷与负电荷)

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)。

An electron and a proton enter the same magnetic field with the same velocity (perpendicular to $\vec{B}$). Compared with the proton, the magnetic force on the electron has:电子与质子以相同速度(垂直于 $\vec{B}$)进入同一磁场。与质子相比,电子受到的磁力:
D3.2 · Q1
Same magnitude, opposite direction大小相同,方向相反
Same magnitude, same direction大小相同,方向相同
Larger magnitude, opposite direction大小更大,方向相反
Zero为零
The magnitudes of their charges are equal, so $F = qvB$ gives the same size. But the electron's negative charge reverses the conventional-current direction, flipping the force.两者电荷大小相等,故 $F = qvB$ 给出相同大小。但电子带负电使常规电流方向反向,从而力方向相反。
$F = qvB$ depends on the magnitude of charge (equal here) so sizes match. The negative sign of the electron reverses direction only.$F = qvB$ 取决于电荷大小(此处相等)故大小一致。电子的负号只反转方向。
A charged particle moves exactly along the direction of a uniform magnetic field. The magnetic force on it is:带电粒子恰好沿匀强磁场方向运动。它受到的磁力为:
D3.2 · Q2
$qvB$
$qvB$, perpendicular to $\vec{v}$$qvB$,垂直于 $\vec{v}$
Maximum, along $\vec{v}$最大,沿 $\vec{v}$
Zero为零
$\vec{v}\parallel\vec{B}$ means $\theta = 0^\circ$, so $\sin\theta = 0$ and $F = qvB\sin\theta = 0$. The particle continues in a straight line.$\vec{v}\parallel\vec{B}$ 即 $\theta = 0^\circ$,$\sin\theta = 0$,故 $F = qvB\sin\theta = 0$。粒子继续沿直线运动。
The force carries a $\sin\theta$ factor. Motion along the field gives $\theta = 0$, so the force is zero.力含 $\sin\theta$ 因子。沿磁场运动时 $\theta = 0$,故力为零。

Circular Motion of a Charged Particle带电粒子的圆周运动 D.3 SL+HL

Why a circle? When $\vec{v} \perp \vec{B}$, the magnetic force $qvB$ stays perpendicular to $\vec{v}$ and constant in magnitude (speed is constant). A constant force perpendicular to velocity is exactly the centripetal condition, so the path is a circle. Radius (gyroradius). Set magnetic force equal to centripetal force: $$ q v B = \frac{m v^{2}}{r} \;\Rightarrow\; r = \frac{m v}{q B}. $$ Period. From $T = 2\pi r / v$: $$ T = \frac{2\pi m}{q B}, \qquad f = \frac{qB}{2\pi m}. $$
  • $r$ 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.
为何是圆?当 $\vec{v} \perp \vec{B}$ 时,磁力 $qvB$ 始终垂直于 $\vec{v}$ 且大小恒定(速率不变)。垂直于速度的恒力正是向心力条件,故轨迹为圆。 回旋半径。令磁力等于向心力: $$ q v B = \frac{m v^{2}}{r} \;\Rightarrow\; r = \frac{m v}{q B}. $$ 周期。由 $T = 2\pi r / v$: $$ T = \frac{2\pi m}{q B}, \qquad f = \frac{qB}{2\pi m}. $$
  • $r$ 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。
Worked Example D3.3 (orbit radius)D3.3 例题(轨道半径)

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。

A charged particle in a uniform $\vec{B}$ has its speed doubled (still entering perpendicular to $\vec{B}$). Its orbit radius and period become:匀强磁场 $\vec{B}$ 中一带电粒子速率加倍(仍垂直入射)。其轨道半径与周期变为:
D3.3 · Q1
Radius doubles, period doubles半径加倍,周期加倍
Radius doubles, period unchanged半径加倍,周期不变
Radius unchanged, period halves半径不变,周期减半
Both unchanged两者都不变
$r = mv/(qB)$ is proportional to $v$, so doubling $v$ doubles $r$. But $T = 2\pi m/(qB)$ contains no $v$, so the period is unchanged.$r = mv/(qB)$ 正比于 $v$,故 $v$ 加倍则 $r$ 加倍。但 $T = 2\pi m/(qB)$ 不含 $v$,故周期不变。
Radius scales with $v$; period does not. Look at which formula contains $v$: only $r = mv/(qB)$.半径随 $v$ 变化;周期不随。看哪个公式含 $v$:只有 $r = mv/(qB)$。
Two particles of equal speed enter the same $\vec{B}$ perpendicular. Particle X has twice the mass and twice the charge of particle Y. The ratio $r_X / r_Y$ is:两个等速粒子垂直进入同一磁场 $\vec{B}$。X 的质量与电荷都是 Y 的两倍。比值 $r_X / r_Y$ 为:
D3.3 · Q2
$4$
$2$
$1$
$\tfrac{1}{2}$
$r = mv/(qB)$, so $r \propto m/q$. Doubling both $m$ and $q$ leaves $m/q$ unchanged, so $r_X/r_Y = 1$.$r = mv/(qB)$,故 $r \propto m/q$。$m$ 与 $q$ 同时加倍则 $m/q$ 不变,故 $r_X/r_Y = 1$。
Radius depends on the ratio $m/q$, not $m$ or $q$ alone. If both double, the ratio is unchanged.半径取决于比值 $m/q$,而非单独的 $m$ 或 $q$。两者都加倍时比值不变。

Charged Particle in a Uniform Electric Field匀强电场中的带电粒子 D.3 SL+HL

Constant force, constant acceleration. A charge $q$ in a uniform field $E$ feels a constant force $$ F = qE, \qquad a = \frac{qE}{m}. $$ Between parallel plates separated by $d$ at potential difference $V$, the field is $E = V/d$ (data booklet 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).
恒力、恒加速度。电荷 $q$ 在匀强电场 $E$ 中受恒力 $$ F = qE, \qquad a = \frac{qE}{m}. $$ 在相距 $d$、电势差为 $V$ 的平行板之间,场强 $E = V/d$(数据手册 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)。
Worked Example D3.4 (energy gain through a p.d.)D3.4 例题(经电势差获得能量)

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。

A charge $q$ accelerated from rest through a p.d. $V$ gains kinetic energy:电荷 $q$ 从静止经电势差 $V$ 加速,获得的动能为:
D3.4 · Q1
$\dfrac{1}{2}qV$
$qV$
$qV^{2}$
$\dfrac{q}{V}$
Work done by the field on the charge is $W = qV$ (data booklet), and from rest all of it becomes kinetic energy. So $E_k = qV$.电场对电荷做的功 $W = qV$(数据手册),从静止出发全部转为动能。故 $E_k = qV$。
The work-energy relation here is simply $W = qV$, with all the work converting to kinetic energy from rest.这里的功能关系就是 $W = qV$,从静止出发全部功转为动能。
An electron beam enters a uniform electric field perpendicular to the field lines. The path the beam follows inside the field is:电子束垂直于电场线进入匀强电场。束在场内的轨迹是:
D3.4 · Q2
Parabolic抛物线
Circular圆形
A straight line along the field沿场方向的直线
Helical螺旋形
Constant velocity across the field plus constant acceleration along it gives projectile-like motion: the path is a parabola. (A magnetic field would give a circle; an electric field gives a parabola.)沿垂直方向匀速、沿场方向匀加速,构成类抛体运动:轨迹为抛物线。(磁场给出圆,电场给出抛物线。)
A constant force perpendicular to the initial velocity (like gravity on a projectile) produces a parabola, not a circle.垂直于初速度的恒力(如抛体所受重力)产生抛物线,而非圆。

Forces Between Two Parallel Current-Carrying Wires两条平行载流导线间的力 D.3 SL · HL深化

Attract or repel? Each wire sits in the magnetic field created by the other and feels a motor-effect force.
  • Currents in the same direction → wires attract.
  • Currents in opposite directions → wires repel.
Magnitude (HL). HL The field of one long straight wire at distance $r$ is $B = \dfrac{\mu_0 I_1}{2\pi r}$. The force per unit length on the second wire (current $I_2$) is $$ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}. $$ Here $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ is the permeability of free space (data booklet). The force grows with both currents and falls off as $1/r$.
吸引还是排斥?每根导线都处在另一根产生的磁场中,受到电动机效应力(安培力)。
  • 电流同向 → 两线相吸。
  • 电流反向 → 两线相斥。
大小(HL)。HL 一根长直导线在距离 $r$ 处的磁场为 $B = \dfrac{\mu_0 I_1}{2\pi r}$。第二根导线(电流 $I_2$)单位长度受力为 $$ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}. $$ 其中 $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ 是真空磁导率(数据手册)。力随两电流增大,随距离按 $1/r$ 减小。
Worked Example D3.5 (direction + magnitude per metre)D3.5 例题(方向 + 每米受力)

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——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。

Two long parallel wires carry currents in opposite directions. The wires:两条长平行导线载有方向相反的电流。两线:
D3.5 · Q1
Repel each other相互排斥
Attract each other相互吸引
Exert no force on each other彼此无作用力
Rotate to become perpendicular转动至相互垂直
Opposite-direction (antiparallel) currents repel; same-direction currents attract. Apply Fleming's left-hand rule in each wire's mutual field to confirm.反向(反平行)电流相斥;同向电流相吸。在彼此磁场中用弗莱明左手定则即可确认。
Remember the pair rule: same direction → attract, opposite direction → repel. Opposite currents here means repulsion.记住配对规则:同向相吸,反向相斥。此处反向即排斥。
HL Two parallel wires feel a force per length $F/L$. If the separation $r$ is doubled and both currents are unchanged, the new force per length is:HL 两平行导线单位长度受力 $F/L$。若间距 $r$ 加倍、两电流不变,新的单位长度受力为:
D3.5 · Q2
$4\times$ as large变为 $4$ 倍
$2\times$ as large变为 $2$ 倍
Half as large变为一半
Unchanged不变
$F/L = \mu_0 I_1 I_2 / (2\pi r) \propto 1/r$. Doubling $r$ halves the force per length.$F/L = \mu_0 I_1 I_2 / (2\pi r) \propto 1/r$。$r$ 加倍则单位长度受力减半。
The force per length goes as $1/r$. Doubling $r$ therefore halves it (not a square law).单位长度受力按 $1/r$ 变化。$r$ 加倍故减半(不是平方关系)。

Combined Fields: the Velocity Selector复合场:速度选择器 D.3 SL · HL深化

Crossed (perpendicular) $\vec{E}$ and $\vec{B}$. Arrange a uniform electric field and a uniform magnetic field perpendicular to each other, and send a charge through perpendicular to both. The two forces oppose:
  • Electric force $F_E = qE$ (independent of speed).
  • Magnetic force $F_B = qvB$ (grows with speed).
The velocity selector. Only particles for which the forces balance ($qE = qvB$) pass straight through undeflected: $$ qE = qvB \;\Rightarrow\; v = \frac{E}{B}. $$ The selected speed depends only on the field ratio, not on the charge or mass — any particle, fast or slow, light or heavy, is passed only if its speed equals $E/B$. Applications. The mass spectrometer (a velocity selector feeding charges into a magnetic field where $r = mv/(qB)$ sorts by mass) and the cathode-ray tube / CRT.
正交(垂直)的 $\vec{E}$ 与 $\vec{B}$。布置相互垂直的匀强电场与匀强磁场,让电荷垂直于二者射入。两个力相互对抗:
  • 电场力 $F_E = qE$(与速率无关)。
  • 磁场力 $F_B = qvB$(随速率增大)。
速度选择器。只有使两力平衡($qE = qvB$)的粒子才不偏转、直线通过: $$ qE = qvB \;\Rightarrow\; v = \frac{E}{B}. $$ 所选速率仅取决于场强之比,电荷或质量无关——任何粒子,无论快慢轻重,只有速率等于 $E/B$ 才能通过。 应用。质谱仪(速度选择器把电荷送入磁场,由 $r = mv/(qB)$ 按质量分离)以及阴极射线管 / CRT。
Worked Example D3.6 (selected speed)D3.6 例题(被选速率)

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。

In a velocity selector with perpendicular $\vec{E}$ and $\vec{B}$, a particle passes straight through. Its speed must equal:在 $\vec{E}$ 与 $\vec{B}$ 垂直的速度选择器中,一粒子直线通过。其速率必等于:
D3.6 · Q1
$EB$
$\dfrac{E}{B}$
$\dfrac{B}{E}$
$\dfrac{qE}{B}$
Undeflected means $qE = qvB$. The charge cancels, leaving $v = E/B$ — independent of charge and mass.不偏转即 $qE = qvB$。电荷消去,得 $v = E/B$——与电荷、质量无关。
Balance the forces: $qE = qvB$. Cancel $q$ and solve for $v$ to get $v = E/B$.令两力平衡:$qE = qvB$。消去 $q$ 解得 $v = E/B$。
In a mass spectrometer, ions leave a velocity selector (all at the same speed) and enter a uniform magnetic field. Heavier ions land:质谱仪中,离子离开速度选择器(速率相同)后进入匀强磁场。较重的离子落点:
D3.6 · Q2
At the same point as light ions与轻离子相同
Closer to the entry point (smaller radius)更靠近入口(半径更小)
They are not deflected at all完全不偏转
Farther out (larger radius)更靠外(半径更大)
With $v$, $q$ and $B$ fixed, $r = mv/(qB) \propto m$. Heavier ions have a larger radius and land farther out, which is how the device separates masses.在 $v$、$q$、$B$ 固定时,$r = mv/(qB) \propto m$。较重离子半径更大、落点更外,仪器据此分离质量。
All ions share the same selected $v$, so $r = mv/(qB)$ is proportional to mass. Heavier means larger radius.所有离子共享同一被选 $v$,故 $r = mv/(qB)$ 正比于质量。越重半径越大。

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Hand rules and charge sign (every paper)手定则与电荷符号(每张试卷)
  • 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。
Don't mix the $\sin\theta$ factor不要漏掉 $\sin\theta$ 因子
  • 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。
Circular motion (Paper 2 standard)圆周运动(Paper 2 常考)
  • 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 vs magnetic fields (compare-and-contrast)电场与磁场(对比题)
  • 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 $v = E/B$。

Flashcards闪卡

0 / 14 flipped已翻 0 / 14
Force on a current-carrying wire?载流导线受力?
$$F = B I L \sin\theta$$
Force on a moving charge?运动电荷受力?
$$F = q v B \sin\theta$$
Fleming's left-hand rule assignments?弗莱明左手定则对应?
Thumb = force, First finger = field $\vec{B}$, seCond finger = Current.拇指=力,食指=磁场 $\vec{B}$,中指=电流。
Gyroradius in a magnetic field?磁场中回旋半径?
$$r = \frac{m v}{q B}$$
Period of circular orbit?圆轨道周期?
$$T = \frac{2\pi m}{q B}$$Independent of speed.与速率无关。
Does a magnetic field do work?磁场做功吗?
No. Force $\perp \vec{v}$: speed constant, only direction turns.不做功。力垂直于 $\vec{v}$:速率恒定,只改变方向。
Force on a negative charge vs positive (same $\vec{v}$, $\vec{B}$)?负电荷与正电荷受力(同 $\vec{v}$、$\vec{B}$)?
Same magnitude, opposite direction.大小相同,方向相反。
Force on a charge in a uniform $\vec{E}$?匀强电场 $\vec{E}$ 中电荷受力?
$$F = qE, \quad a = \frac{qE}{m}$$
Work / energy gain through a p.d. $V$?经电势差 $V$ 的功 / 能量增益?
$$W = qV = \tfrac12 m v^{2}$$
Path of a charge in a uniform $\vec{E}$ (perp. entry)?电荷在匀强 $\vec{E}$ 中(垂直入射)的轨迹?
Parabolic deflection (projectile-like).抛物线偏转(类抛体)。
Parallel wires: same direction?平行导线:电流同向?
Attract.相吸。
Force per length between parallel wires? HL平行导线单位长度受力?HL
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$$
Velocity selector: selected speed?速度选择器:被选速率?
$$v = \frac{E}{B}$$From $qE = qvB$.由 $qE = qvB$ 得。
Mass spectrometer: how is mass found?质谱仪:如何求质量?
Fixed $v$ via selector, then $r = mv/(qB) \propto m$.选择器定 $v$,再由 $r = mv/(qB) \propto m$。

Unit D.3 Practice Quiz单元 D.3 练习测验

A wire of length $0.20\ \mathrm{m}$ carrying $6.0\ \mathrm{A}$ lies perpendicular to a $0.50\ \mathrm{T}$ field. The force on the wire is:载流 $6.0\ \mathrm{A}$ 的 $0.20\ \mathrm{m}$ 导线垂直于 $0.50\ \mathrm{T}$ 磁场。导线受力为:
Q1
$0.30\ \mathrm{N}$
$1.2\ \mathrm{N}$
$0.60\ \mathrm{N}$
$6.0\ \mathrm{N}$
$F = BIL\sin 90^\circ = (0.50)(6.0)(0.20)(1) = 0.60\ \mathrm{N}$.$F = BIL\sin 90^\circ = (0.50)(6.0)(0.20)(1) = 0.60\ \mathrm{N}$。
Perpendicular gives $\sin\theta = 1$, so $F = BIL = 0.50\times6.0\times0.20$.垂直时 $\sin\theta = 1$,故 $F = BIL = 0.50\times6.0\times0.20$。
A proton enters a uniform magnetic field perpendicular to $\vec{B}$ and follows a circle of radius $r$. If $B$ is doubled (speed unchanged), the new radius is:质子垂直于 $\vec{B}$ 进入匀强磁场,做半径为 $r$ 的圆周。若 $B$ 加倍(速率不变),新半径为:
Q2
$r/2$
$r$
$2r$
$4r$
$r = mv/(qB) \propto 1/B$. Doubling $B$ halves the radius, giving $r/2$.$r = mv/(qB) \propto 1/B$。$B$ 加倍则半径减半,得 $r/2$。
Radius is inversely proportional to $B$. Stronger field, tighter circle: doubling $B$ gives $r/2$.半径与 $B$ 成反比。场越强圆越紧:$B$ 加倍给出 $r/2$。
An alpha particle (charge $+2e$) is accelerated from rest through $1000\ \mathrm{V}$. The kinetic energy it gains is ($e = 1.6\times10^{-19}\ \mathrm{C}$):α 粒子(电荷 $+2e$)从静止经 $1000\ \mathrm{V}$ 加速。获得的动能为($e = 1.6\times10^{-19}\ \mathrm{C}$):
Q3
$1.6\times10^{-19}\ \mathrm{J}$
$1.6\times10^{-16}\ \mathrm{J}$
$8.0\times10^{-17}\ \mathrm{J}$
$3.2\times10^{-16}\ \mathrm{J}$
$E_k = qV = (2\times1.6\times10^{-19})(1000) = 3.2\times10^{-16}\ \mathrm{J}$. The charge is $2e$, not $e$.$E_k = qV = (2\times1.6\times10^{-19})(1000) = 3.2\times10^{-16}\ \mathrm{J}$。电荷是 $2e$,不是 $e$。
Use $W = qV$ with $q = 2e$ for an alpha particle: $(3.2\times10^{-19})(1000)$.对 α 粒子用 $W = qV$ 且 $q = 2e$:$(3.2\times10^{-19})(1000)$。
In a velocity selector, $E = 2.0\times10^{4}\ \mathrm{V\,m^{-1}}$ and $B = 0.50\ \mathrm{T}$ are perpendicular. The speed of ions passing undeflected is:速度选择器中 $E = 2.0\times10^{4}\ \mathrm{V\,m^{-1}}$ 与 $B = 0.50\ \mathrm{T}$ 垂直。不偏转通过的离子速率为:
Q4
$1.0\times10^{4}\ \mathrm{m\,s^{-1}}$
$4.0\times10^{4}\ \mathrm{m\,s^{-1}}$
$2.5\times10^{4}\ \mathrm{m\,s^{-1}}$
$1.0\times10^{5}\ \mathrm{m\,s^{-1}}$
$v = E/B = (2.0\times10^{4}) / 0.50 = 4.0\times10^{4}\ \mathrm{m\,s^{-1}}$, independent of charge and mass.$v = E/B = (2.0\times10^{4}) / 0.50 = 4.0\times10^{4}\ \mathrm{m\,s^{-1}}$,与电荷、质量无关。
Balance forces: $qE = qvB \Rightarrow v = E/B$. Divide $E$ by $B$.令力平衡:$qE = qvB \Rightarrow v = E/B$。用 $E$ 除以 $B$。
HL Two long parallel wires carry $8.0\ \mathrm{A}$ and $4.0\ \mathrm{A}$ in the same direction, separated by $0.040\ \mathrm{m}$. The force per unit length, and its sense, are ($\mu_0/2\pi = 2\times10^{-7}$):HL 两条长平行导线同向载流 $8.0\ \mathrm{A}$ 与 $4.0\ \mathrm{A}$,相距 $0.040\ \mathrm{m}$。单位长度受力及其指向为($\mu_0/2\pi = 2\times10^{-7}$):
Q5
$8.0\times10^{-4}\ \mathrm{N\,m^{-1}}$, repulsive$8.0\times10^{-4}\ \mathrm{N\,m^{-1}}$,排斥
$1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$, attractive$1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$,吸引
$1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$, attractive$1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$,吸引
$3.2\times10^{-4}\ \mathrm{N\,m^{-1}}$, repulsive$3.2\times10^{-4}\ \mathrm{N\,m^{-1}}$,排斥
$F/L = \mu_0 I_1 I_2 /(2\pi r) = (2\times10^{-7})(8.0)(4.0)/0.040 = 1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$. Same-direction currents attract.$F/L = \mu_0 I_1 I_2 /(2\pi r) = (2\times10^{-7})(8.0)(4.0)/0.040 = 1.6\times10^{-4}\ \mathrm{N\,m^{-1}}$。同向电流相吸。
Compute $(2\times10^{-7})(8.0)(4.0)/0.040 = 1.6\times10^{-4}$, and same-direction currents attract.算 $(2\times10^{-7})(8.0)(4.0)/0.040 = 1.6\times10^{-4}$,且同向电流相吸。

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