High School Physics

Magnetism and Electromagnetic Induction磁场与电磁感应

Magnetism and electromagnetic induction form the bridge between electricity and light. This guide builds from permanent magnets and the concept of the magnetic field through the Lorentz force on moving charges ($F = qvB\sin\theta$), the force on current-carrying wires, the Biot-Savart picture of fields produced by currents, and then the centrepiece: Faraday's law ($\varepsilon = -N\tfrac{d\Phi}{dt}$) and Lenz's law, which together explain how changing flux drives current. The final section shows how these laws power generators, motors, and transformers — the machines of the electrical grid. Worked examples use real numbers throughout.磁场与电磁感应是连接电学与光学的桥梁。本指南从永磁体（permanent magnet，永磁体）与磁场（磁场）概念出发，经洛伦兹力（洛伦兹力，$F = qvB\sin\theta$）、载流导线所受磁力、电流产生磁场的毕奥-萨伐尔图像，到核心内容：法拉第定律（法拉第定律，$\varepsilon = -N\tfrac{d\Phi}{dt}$）与楞次定律（楞次定律）——二者共同解释变化磁通量如何驱动电流。最后一节展示这些定律如何驱动发电机（发电机）、电动机（电动机）和变压器（变压器）——电力网络的核心机器。全部例题均用真实数字演算。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Honors blocks on Faraday & Lenz法拉第与楞次定律为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Magnetism and electromagnetic induction appear in all four curricula we map to, but at different grade levels and depths. US NGSS HS-PS2-5 is deliberately kept qualitative and experimental — it asks students to design investigations showing the current-field relationship, not to compute flux. The Faraday's law and Lenz's law content in §5 and §6 therefore carries the Honors chip for the US NGSS track. Ontario SPH3U (Strand F) and SPH4U (Strand D) address magnetic fields and induction across both Grade 11 and Grade 12; Alberta Physics 30 Unit B GO3 is quantitative and calculative. BC Physics 12 covers magnetic field, electromagnetic induction, and applications of induction explicitly as Content. The table below tells you which sections are core for you; each row cites the curriculum document it was checked against.磁场与电磁感应出现在我们对照的四套大纲中，但年级与深度各不相同。US NGSS HS-PS2-5 刻意保持定性与实验性——要求学生设计展示电流-场关系的调查，而非计算磁通量。因此 §5 和 §6 中的法拉第定律与楞次定律内容在 US NGSS 轨道上标 Honors。安大略 SPH3U（F 单元）与 SPH4U（D 单元）在 11 年级与 12 年级分别涉及磁场与电磁感应；阿尔伯塔 Physics 30 B 单元 GO3 是定量与计算性的。BC Physics 12 将磁场、电磁感应及其应用明确列为内容。下表告诉你哪些节属于你的核心；每行都注明所依据的课纲文件。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1 through §4 (magnets, Lorentz force on charges and wires, magnetic fields from currents) — the qualitative-experimental core under HS-PS2-5§1 至 §4（磁体、洛伦兹力、载流导线受力、电流产生磁场）—— HS-PS2-5 下的定性实验核心 §5 and §6 (Faraday's law and Lenz's law, flux calculations): valuable but above the NGSS assessment floor, which is limited to qualitative experimental investigation§5 和 §6（法拉第定律与楞次定律、磁通量计算）：很有价值，但高于 NGSS 定性实验调查的评估下限 ngss_hs_ps_extract.md — HS-PS2-5 PE + Assessment Boundary— HS-PS2-5 表现期望及评估边界
🇨🇦 ON Grade 11 & 12 — SPH3U / SPH4U安大略 11/12 年级 — SPH3U / SPH4U §1 through §7. SPH3U Strand F (Electricity and Magnetism) covers magnetic fields and induction at Grade 11; SPH4U Strand D extends this to gravitational, electric, and magnetic fields at Grade 12§1 至 §7。SPH3U F 单元（电学与磁学）在 11 年级涵盖磁场与电磁感应；SPH4U D 单元在 12 年级将其延伸至重力场、电场与磁场 Nothing — the full unit is in scope across both courses无 — 全单元在两门课程范围内 science_11-12_physics_extract.md — SPH3U Strand F Overall Expectations F1–F3; SPH4U Strand D D1–D3— SPH3U F 单元总体期望 F1–F3；SPH4U D 单元 D1–D3
🇨🇦 BC Grade 12 — Physics 12BC 12 年级 — Physics 12 §1 through §7. Physics 12 Content explicitly lists "magnetic field and magnetic force," "electromagnetic induction," and "applications of electromagnetic induction" — covering all seven sections§1 至 §7。Physics 12 内容明确列出"磁场与磁力"、"电磁感应"和"电磁感应的应用"——覆盖全部 7 节 Nothing — BC Physics 12 is quantitative throughout; Faraday's law and transformer ratios are assessed无 — BC Physics 12 全程定量；法拉第定律与变压器比例均被评估 physics_11-12_extract.md — Physics 12 Content: magnetic field/force, electromagnetic induction, applications; right-hand rules— Physics 12 内容：磁场/磁力、电磁感应、应用；右手定则
🇨🇦 AB Grade 12 — Physics 30阿尔伯塔 12 年级 — Physics 30 §1 through §7. Physics 30 Unit B GO3 (30–B3.4k through 30–B3.9k) is explicitly quantitative: force on charges and wires ($F = qvB$ and $F = BIL$), field from currents, and moving a conductor in a field§1 至 §7。Physics 30 B 单元 GO3（30–B3.4k 至 30–B3.9k）明确要求定量：电荷与导线所受力（$F = qvB$ 与 $F = BIL$）、电流产生的场以及在场中移动导体 Nothing — AB Physics 30 treats induction quantitatively; Diploma-style problems involve numerical flux and EMF calculations无 — AB Physics 30 定量处理电磁感应；文凭考风格题包含数值化磁通量与电动势计算 physics_20-30_extract.md — Physics 30 Unit B GO3, knowledge outcomes 30–B3.4k through 30–B3.9k— Physics 30 B 单元 GO3，知识 outcome 30–B3.4k 至 30–B3.9k
🇺🇸 IB Physics HL feeder trackIB Physics HL 衔接轨道 All seven sections plus every going-deeper derivation. IB Physics HL Topic D (Fields) assumes fluent Lorentz force, Faraday's law, and transformer analysis from the start全部 7 节，并完成每个"深入"推导。IB Physics HL D 主题（场）一开始就默认你熟练洛伦兹力、法拉第定律与变压器分析 Nothing — this unit is the conceptual and quantitative foundation that IB Physics HL Topic D and the electromagnetic wave guides build on无 — 本单元是 IB Physics HL D 主题与电磁波各指南所依赖的概念与定量基础 ngss_hs_ps_extract.md — HS-PS2-5; IB feeder reads well beyond the NGSS qualitative floor— HS-PS2-5；IB 衔接读到远超 NGSS 定性下限之上

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four things: the Lorentz force $F = qvB\sin\theta$ (right-hand rule gives direction); the magnetic force on a wire $F = BIL\sin\theta$; Faraday's law $\varepsilon = -N\tfrac{d\Phi}{dt}$ (changing flux drives EMF); and Lenz's law (the induced current opposes the change that caused it). Read every cram-cheat box. Skip the going-deeper derivations.背熟四件事：洛伦兹力 $F = qvB\sin\theta$（右手定则给出方向）；导线受力 $F = BIL\sin\theta$；法拉第定律 $\varepsilon = -N\tfrac{d\Phi}{dt}$（变化磁通量驱动电动势）；以及楞次定律（感应电流方向阻碍引起它的变化）。读每个速记框，跳过深入推导。

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If you are going for the top mark如果你目标顶分

Master the right-hand rule in all three contexts (field around a wire, force on a charge, force on a wire). Practise computing magnetic flux $\Phi = BA\cos\theta$ for different orientations. For transformers, link the voltage and current ratios to the turns ratio: $\tfrac{V_s}{V_p} = \tfrac{N_s}{N_p}$ and $\tfrac{I_s}{I_p} = \tfrac{N_p}{N_s}$. AB Physics 30 and BC Physics 12 both expect quantitative EMF and force calculations; the IB feeder expects you to derive these results from first principles.掌握三种情境下的右手定则（导线周围的场、电荷受力、导线受力）。练习计算不同朝向下的磁通量 $\Phi = BA\cos\theta$。对于变压器，将电压与电流比联系到匝数比：$\tfrac{V_s}{V_p} = \tfrac{N_s}{N_p}$ 与 $\tfrac{I_s}{I_p} = \tfrac{N_p}{N_s}$。AB Physics 30 与 BC Physics 12 均要求定量的 EMF 与力计算；IB 衔接轨道要求你从第一性原理推导这些结果。

Honors flag.荣誉级标记。 Sections 5 and 6 (Faraday's law and Lenz's law with flux calculations) carry the Honors chip for the US NGSS track, where HS-PS2-5 is "limited to designing and conducting investigations" and does not require quantitative flux calculations. These sections are core, not honors, in Ontario SPH3U/SPH4U (Strand F/D), BC Physics 12 (Content lists "Faraday's law; Lenz's law" explicitly), and AB Physics 30 (GO3 knowledge outcomes 30–B3.9k). If your row above sends you to §5 and §6, treat them as required content, not enrichment.§5 和 §6（含磁通量计算的法拉第定律与楞次定律）在 US NGSS 轨道上标 Honors，因为 HS-PS2-5"限于设计和实施调查"，不要求定量磁通量计算。但在安大略 SPH3U/SPH4U（F/D 单元）、BC Physics 12（内容明确列出"法拉第定律；楞次定律"）、AB Physics 30（GO3 知识 outcome 30–B3.9k）中，它们是核心而非荣誉内容。如果你的行指向 §5 和 §6，就把它们视为必学，不是拓展。
Section 1 · Magnets and magnetic fields第 1 节 · 磁体与磁场

Magnets and Magnetic Fields磁体与磁场

The magnetic field $\vec B$ — the whole unit rests on this concept.磁场 $\vec B$ — 整个单元都建立在这一概念之上。
  • Permanent magnets永磁体 — have north and south poles. Like poles repel; unlike poles attract. Field lines exit the north pole, loop through space, and enter the south pole. You cannot isolate a single magnetic pole (no magnetic monopole).— 有北极（N）与南极（S）。同极相斥，异极相吸。磁场线从北极出发，经空间绕弧，进入南极。磁单极不存在（无法分离单个磁极）。
  • Magnetic field $\vec B$磁场 $\vec B$ — a vector field; SI unit is the tesla (T). Field lines are denser where $|\vec B|$ is stronger. Field lines never cross.— 矢量场；SI 单位是特斯拉（T）。磁场线越密，$|\vec B|$ 越强。磁场线互不相交。
  • Earth's magnetic field地球磁场 — roughly $5\times10^{-5}$ T at the surface; points roughly from geographic south to north (the geographic north pole is near a magnetic south pole, which is why compass needles point north).— 地球表面约 $5\times10^{-5}$ T；方向大致从地理南极指向北极（地理北极附近是磁南极，故指南针指北）。
BC Physics 12 Content names "magnetic field" as the first electromagnetism bullet: "vector field; induced by moving charges; interacts with polarity (north/south); attractive or repulsive; permanent magnets, straight wires, and solenoids." Alberta 30–B3.2k requires comparison of gravitational, electric, and magnetic fields in terms of their sources and directions.BC Physics 12 内容把"磁场"列为电磁学的第一条："矢量场；由运动电荷感应；与极性（N/S）相互作用；吸引或排斥；永磁体、直导线与螺线管。" 阿尔伯塔 30–B3.2k 要求比较重力场、电场与磁场的源与方向。
Worked Example 1 · Sketch the field of a bar magnet例题 1 · 画条形磁铁的磁场

A bar magnet lies horizontal with its north pole on the right. Describe the direction and relative strength of the magnetic field (a) midway between the poles along the axis, (b) directly above the centre, and (c) far from the magnet.一根条形磁铁水平放置，北极在右侧。描述磁场在 (a) 两极之间轴线中点、(b) 中心正上方、(c) 远离磁铁处的方向与相对强弱。

(a) On the axis between poles.(a) 两极之间轴线上。 Field lines travel from north to south through the external space, so on the axis between the poles the field points from right (N) to left (S) — i.e. leftward. This is where field lines are densest, so $|\vec B|$ is largest here.磁场线在外部空间从北极指向南极，故两极之间轴线上磁场方向从右（N）指向左（S）——即向左。这里磁场线最密，$|\vec B|$ 最大。

(b) Above the centre.(b) 中心正上方。 Field lines arch upward and curve outward from both poles; directly above the centre they point roughly leftward (from the N side to the S side) but with a component curving away from the axis. $|\vec B|$ is smaller here than on the axis.磁场线从两极向外弧形延伸；在中心正上方，它们大致指向左方（从 N 侧到 S 侧），但带有离轴弯曲分量。$|\vec B|$ 比轴线上小。

(c) Far from the magnet.(c) 远离磁铁处。 The field weakens rapidly with distance (like a dipole, $|\vec B| \propto 1/r^3$). The field lines are sparse and nearly parallel to the axis, pointing from north to south through the outside. At large distances $|\vec B| \approx 0$.磁场随距离迅速减弱（如偶极子，$|\vec B| \propto 1/r^3$）。磁场线稀疏，几乎平行于轴线，在外部从北极指向南极。在很远处 $|\vec B| \approx 0$。

Magnetic field lines exit which pole of a bar magnet and what do they do at that pole?条形磁铁的磁场线从哪个极出发？它们在该极如何？
§1 · Q1
South pole; they converge inward南极；向内汇聚
North pole; they diverge outward北极；向外发散
Both poles equally; they form closed loops inside only两极等量；仅在内部形成闭合回路
North pole; they converge inward北极；向内汇聚
By convention, field lines exit (diverge from) the north pole and enter (converge into) the south pole in the external space. Inside the magnet they complete the loop from south to north.按约定，磁场线在外部空间从北极出发（发散），进入南极（汇聚）。在磁铁内部，它们从南极到北极完成回路。
Magnetic field lines exit the north pole and enter the south pole (in the space outside the magnet). They form complete closed loops that pass through the magnet from S to N internally.磁场线从北极出发，进入南极（在磁铁外部空间）。它们形成完整的闭合回路，在磁铁内部从南极到北极穿行。
What happens when two north poles of bar magnets are brought close together?两根条形磁铁的北极相互靠近会发生什么？
§1 · Q2
They attract each other它们相互吸引
No force acts between them它们之间没有力
They repel each other它们相互排斥
They attract, then repel at very close range先吸引，近距离时再排斥
Like poles repel; unlike poles attract. Two north poles facing each other produce a repulsive force.同极相斥；异极相吸。两个北极相对产生排斥力。
Like poles always repel; unlike poles always attract. This is analogous to electric charges but magnetic poles cannot be isolated.同极总是相斥；异极总是相吸。这与电荷类似，但磁极不能单独存在。
Going deeper — why magnetic monopoles do not exist (and what that implies)深入 — 为何磁单极不存在（及其意义）

Every magnet cut in half produces two smaller magnets, each with a north and a south pole. This is captured by one of Maxwell's equations: $\nabla \cdot \vec B = 0$ (Gauss's law for magnetism), which states that magnetic field lines always form closed loops — they never start or end on a "magnetic charge." In contrast, electric field lines start on positive charges and end on negative charges ($\nabla \cdot \vec E = \rho/\varepsilon_0$). At HS and AP level this means: unlike gravity and electricity, you cannot have an isolated magnetic pole, so all magnetic fields are produced either by moving charges (currents) or by the intrinsic spin of electrons in materials. AB Physics 30 codes 30–B3.1k and 30–B3.2k ask you to compare magnetic fields with gravitational and electric fields in terms of their sources — this is the key distinction.每块磁铁切成两半都会产生两块更小的磁铁，各自都有北极和南极。这由麦克斯韦方程之一描述：$\nabla \cdot \vec B = 0$（磁场的高斯定律），表明磁场线总是形成闭合回路——从不起始或终止于"磁荷"。相比之下，电场线从正电荷出发，终止于负电荷（$\nabla \cdot \vec E = \rho/\varepsilon_0$）。在高中与 AP 层级，这意味着：与重力和电力不同，孤立磁极不存在，所有磁场要么由运动电荷（电流）产生，要么由材料中电子的固有自旋产生。AB Physics 30 代码 30–B3.1k 和 30–B3.2k 要求你从源与方向角度比较磁场与重力场、电场——这是关键区别。

Section 2 · Magnetic force on moving charges第 2 节 · 运动电荷所受磁力（洛伦兹力）

Magnetic Force on Moving Charges: the Lorentz Force运动电荷所受磁力：洛伦兹力

The Lorentz force — the master formula for magnetism.洛伦兹力 — 磁学的核心公式。 $$ F = qvB\sin\theta $$
  • Variables.变量。 $q$ = charge (C), $v$ = speed (m/s), $B$ = magnetic field strength (T), $\theta$ = angle between $\vec v$ and $\vec B$. Force is maximum when $\vec v \perp \vec B$ ($\sin 90° = 1$) and zero when $\vec v \parallel \vec B$ ($\sin 0° = 0$).$q$ = 电荷量（C），$v$ = 速率（m/s），$B$ = 磁场强度（T），$\theta$ = $\vec v$ 与 $\vec B$ 之间的夹角。当 $\vec v \perp \vec B$ 时力最大（$\sin 90° = 1$），当 $\vec v \parallel \vec B$ 时力为零（$\sin 0° = 0$）。
  • Direction — right-hand rule (RHR).方向 — 右手定则（RHR）。 Point the fingers of your right hand along $\vec v$, curl them toward $\vec B$; your thumb points in the direction of $\vec F$ for a positive charge. Reverse for a negative charge.右手手指沿 $\vec v$ 方向伸出，弯向 $\vec B$；拇指所指即正电荷所受 $\vec F$ 的方向。负电荷方向相反。
  • The magnetic force does no work.磁力不做功。 $\vec F$ is always perpendicular to $\vec v$, so the speed (and kinetic energy) of the charge never changes. The force only changes the direction of motion.$\vec F$ 始终垂直于 $\vec v$，故电荷的速率（和动能）不变。磁力只改变运动方向。
Alberta 30–B3.5k: "explain, qualitatively and quantitatively, how a uniform magnetic field affects a moving electric charge, using the relationships among charge, motion, field direction and strength, when motion and field directions are mutually perpendicular."阿尔伯塔 30–B3.5k："在运动与场方向互相垂直时，利用电荷量、运动、场方向与场强之间的关系，定性与定量地解释匀强磁场对运动电荷的影响。"
Worked Example 2 · Force on a proton in a magnetic field例题 2 · 磁场中质子所受的力

A proton ($q = 1.6\times10^{-19}$ C) moves east at $3.0\times10^6$ m/s into a uniform magnetic field of $0.50$ T directed vertically upward. Find the magnitude and direction of the magnetic force on the proton.一个质子（$q = 1.6\times10^{-19}$ C）以 $3.0\times10^6$ m/s 向东运动，进入竖直向上的 $0.50$ T 匀强磁场。求质子所受磁力的大小与方向。

Magnitude.大小。 $\theta = 90°$ because $\vec v$ (east) $\perp$ $\vec B$ (up).$\theta = 90°$，因为 $\vec v$（向东）$\perp$ $\vec B$（向上）。

$$ F = qvB\sin\theta = (1.6\times10^{-19})(3.0\times10^6)(0.50)\sin 90° = 2.4\times10^{-13}\ \mathrm{N}. $$

Direction via right-hand rule.用右手定则确定方向。 Point fingers east (along $\vec v$), curl them upward (toward $\vec B$): the thumb points south. The magnetic force on the proton is directed south.手指向东（沿 $\vec v$），弯向上方（朝 $\vec B$）：拇指指向南。质子所受磁力方向向南。

Sanity check.合理性核验。 The force is perpendicular to both $\vec v$ and $\vec B$, as expected. It will curve the proton southward without changing its speed.力垂直于 $\vec v$ 与 $\vec B$ 两者，符合预期。它将使质子向南偏转而不改变速率。

An electron moves at $2.0\times10^5$ m/s perpendicular to a $0.40$ T magnetic field. What is the magnitude of the magnetic force on it? ($e = 1.6\times10^{-19}$ C)一个电子以 $2.0\times10^5$ m/s 垂直于 $0.40$ T 磁场运动。其所受磁力大小是多少？（$e = 1.6\times10^{-19}$ C）
§2 · Q1
$1.3\times10^{-14}$ N
$8.0\times10^{-25}$ N
$3.2\times10^{-19}$ N
$1.28\times10^{-14}$ N
$F = qvB\sin 90° = (1.6\times10^{-19})(2.0\times10^5)(0.40)(1) = 1.28\times10^{-14}$ N.$F = qvB\sin 90° = (1.6\times10^{-19})(2.0\times10^5)(0.40)(1) = 1.28\times10^{-14}$ N。
Use $F = qvB\sin\theta$ with $\theta = 90°$ (perpendicular), so $\sin\theta = 1$. Multiply $q$, $v$, and $B$.用 $F = qvB\sin\theta$，$\theta = 90°$（垂直），故 $\sin\theta = 1$。将 $q$、$v$、$B$ 相乘。
A positive charge moves parallel to (in the same direction as) a magnetic field. What magnetic force does it experience?一个正电荷沿磁场方向平行运动。它所受磁力是多少？
§2 · Q2
Zero force零力
Maximum force, in the direction of motion最大力，方向与运动相同
Maximum force, perpendicular to motion最大力，垂直于运动
A force opposite to the direction of motion与运动方向相反的力
When $\vec v \parallel \vec B$, the angle $\theta = 0°$ and $\sin 0° = 0$, so $F = qvB\times 0 = 0$. A charge moving along the field experiences no magnetic force.当 $\vec v \parallel \vec B$ 时，夹角 $\theta = 0°$，$\sin 0° = 0$，故 $F = qvB\times 0 = 0$。沿磁场方向运动的电荷不受磁力。
$F = qvB\sin\theta$. Parallel means $\theta = 0°$, so $\sin\theta = 0$ and $F = 0$. The force is zero whenever the velocity and field are parallel.$F = qvB\sin\theta$。平行意味着 $\theta = 0°$，故 $\sin\theta = 0$，$F = 0$。速度与磁场平行时，磁力始终为零。
Going deeper — circular motion of a charge in a uniform magnetic field深入 — 匀强磁场中电荷的圆周运动

Because $\vec F \perp \vec v$ at every instant, the magnetic force acts as a centripetal force, bending the path into a circle. Setting $F = qvB$ equal to the centripetal requirement $mv^2/r$:由于每一瞬间 $\vec F \perp \vec v$，磁力充当向心力，将路径弯成圆形。令 $F = qvB$ 等于向心力 $mv^2/r$：

$$ qvB = \frac{mv^2}{r} \;\Longrightarrow\; r = \frac{mv}{qB}. $$

This radius is called the cyclotron radius (or Larmor radius). It grows with momentum $mv$ and shrinks with field strength $B$ or charge $q$. The period of one orbit is $T = 2\pi r/v = 2\pi m/(qB)$ — independent of speed. This is the principle behind mass spectrometers (different masses curve with different radii in the same field), cyclotron accelerators, and the aurora (charged particles from the sun spiral along Earth's field lines into the polar regions). AB Physics 30 GO3 and BC Physics 12 both expect you to solve force and radius problems using these relationships.这个半径称为回旋半径（或拉莫尔半径）。它随动量 $mv$ 增大，随场强 $B$ 或电荷量 $q$ 减小。一圈的周期为 $T = 2\pi r/v = 2\pi m/(qB)$——与速率无关。这是质谱仪（不同质量的粒子在同一磁场中以不同半径偏转）、回旋加速器和极光（来自太阳的带电粒子沿地球磁场线螺旋进入极地区域）的工作原理。AB Physics 30 GO3 与 BC Physics 12 均要求你用这些关系求解力与半径问题。

Section 3 · Magnetic force on current-carrying wires第 3 节 · 载流导线所受磁力

Magnetic Force on Current-Carrying Wires载流导线所受磁力

A current is a flow of charge — so a wire in a field feels a force.电流是电荷的定向流动 — 因此磁场中的导线受力。 $$ F = BIL\sin\theta $$
  • Variables.变量。 $B$ = field strength (T), $I$ = current (A), $L$ = length of wire in the field (m), $\theta$ = angle between the current direction and $\vec B$. Maximum force when current $\perp$ field; zero when parallel.$B$ = 场强（T），$I$ = 电流（A），$L$ = 在场中的导线长度（m），$\theta$ = 电流方向与 $\vec B$ 的夹角。电流垂直于磁场时力最大；平行时力为零。
  • Direction — right-hand rule.方向 — 右手定则。 Point fingers along the current direction, curl toward $\vec B$; thumb gives the direction of the force on the wire. (Conventional current direction, not electron flow.)手指沿电流方向伸出，弯向 $\vec B$；拇指即导线受力方向。（按惯例电流方向，非电子流方向。）
  • Origin.来源。 This is just $F = qvB\sin\theta$ applied to all the drifting charges in the wire at once; $I = nqv_d A$ and $L = $ wire length gives the same formula with $IL$ replacing $qv$.这不过是 $F = qvB\sin\theta$ 同时作用于导线中所有漂移电荷；$I = nqv_d A$，$L$ 为导线长度，用 $IL$ 代替 $qv$ 便得到同一公式。
Alberta 30–B3.8k: "explain, quantitatively, the effect of an external magnetic field on a current-carrying conductor." BC Physics 12 Content: "magnetic force: acting on a moving charge or current carrying wire within a magnetic field; right-hand rules."阿尔伯塔 30–B3.8k："定量解释外部磁场对载流导体的影响。" BC Physics 12 内容："磁力：作用于磁场中运动电荷或载流导线；右手定则。"
Worked Example 3 · Force on a wire in a motor coil例题 3 · 电动机线圈中导线受力

A horizontal wire of length $0.30$ m carries a current of $5.0$ A directed north. The wire sits in a uniform magnetic field of $0.80$ T directed vertically upward. Find the magnitude and direction of the force on the wire.一根长 $0.30$ m 的水平导线通有 $5.0$ A 向北的电流，置于竖直向上的 $0.80$ T 匀强磁场中。求导线所受力的大小与方向。

Magnitude.大小。 Current (north) $\perp$ field (up), so $\theta = 90°$.电流（向北）$\perp$ 磁场（向上），故 $\theta = 90°$。

$$ F = BIL\sin\theta = (0.80)(5.0)(0.30)\sin 90° = 1.2\ \mathrm{N}. $$

Direction via right-hand rule.用右手定则确定方向。 Point fingers north (current), curl upward (field): thumb points east. The force on the wire is directed east.手指向北（电流），弯向上方（磁场）：拇指指向东。导线受力方向向东。

Physical interpretation.物理解读。 This east-directed force on the wire is exactly the mechanism that makes an electric motor spin. In a motor, the geometry is arranged so that this force creates a torque that continuously rotates the coil.这个向东的力正是电动机旋转的机制。在电动机中，几何结构的安排使这个力产生持续旋转线圈的力矩。

A $0.50$ m wire carries $3.0$ A of current perpendicular to a $0.60$ T magnetic field. What is the magnetic force on the wire?一根 $0.50$ m 的导线通有 $3.0$ A 电流，垂直于 $0.60$ T 磁场。导线所受磁力是多少？
§3 · Q1
$0.30$ N
$0.90$ N
$9.0$ N
$0.10$ N
$F = BIL\sin 90° = (0.60)(3.0)(0.50)(1) = 0.90$ N.$F = BIL\sin 90° = (0.60)(3.0)(0.50)(1) = 0.90$ N。
$F = BIL\sin\theta$ with $\theta = 90°$. Multiply $B \times I \times L$: $0.60 \times 3.0 \times 0.50$.$F = BIL\sin\theta$，$\theta = 90°$。将 $B \times I \times L$ 相乘：$0.60 \times 3.0 \times 0.50$。
If the current in the wire above is doubled and the field is halved, what happens to the force?若上题中电流加倍、磁场减半，力如何变化？
§3 · Q2
Quadrupled变为原来的四倍
Halved减半
Doubled变为原来的两倍
Unchanged不变
$F = BIL\sin\theta$. Doubling $I$ multiplies $F$ by 2; halving $B$ divides $F$ by 2. Net effect: $2 \times \tfrac{1}{2} = 1$ — the force is unchanged.$F = BIL\sin\theta$。电流加倍使 $F$ 乘以 2；磁场减半使 $F$ 除以 2。净效果：$2 \times \tfrac{1}{2} = 1$ — 力不变。
$F \propto B \times I$. Doubling $I$ and halving $B$ leaves the product $BI$ unchanged, so $F$ is unchanged.$F \propto B \times I$。电流加倍、磁场减半，乘积 $BI$ 不变，故 $F$ 不变。
Section 4 · Magnetic fields from currents第 4 节 · 电流产生磁场

Magnetic Fields from Currents: Oersted's Discovery电流产生磁场：奥斯特的发现

Moving charges produce magnetic fields — currents are moving charges.运动电荷产生磁场 — 电流就是运动电荷。
  • Long straight wire.长直导线。 The field circles the wire in concentric rings. Direction: right-hand rule — wrap your right hand around the wire with thumb pointing in the current direction; fingers curl in the direction of $\vec B$. Strength: $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0 = 4\pi\times10^{-7}$ T·m/A.磁场以同心圆环绕导线。方向：右手定则——右手拇指沿电流方向，四指弯曲方向即 $\vec B$ 方向。场强：$B = \frac{\mu_0 I}{2\pi r}$，其中 $\mu_0 = 4\pi\times10^{-7}$ T·m/A。
  • Solenoid (coil).螺线管（线圈）。 Inside a long solenoid the field is nearly uniform and parallel to the axis: $B = \mu_0 n I$, where $n$ = number of turns per metre. The field outside is nearly zero. A solenoid is an electromagnet: current on, field on; current off, field off.在长螺线管内部，磁场几乎均匀且平行于轴线：$B = \mu_0 n I$，$n$ 为每米匝数。外部磁场几乎为零。螺线管就是电磁铁：通电有场，断电无场。
  • Oersted's 1820 discovery.奥斯特 1820 年的发现。 A compass needle deflected when placed near a current-carrying wire, proving for the first time that electricity and magnetism are related. Alberta 30–B3.3k requires you to "describe how the discoveries of Oersted and Faraday form the foundation of the theory relating electricity to magnetism."置于载流导线旁的指南针偏转，首次证明电与磁相互关联。阿尔伯塔 30–B3.3k 要求你"描述奥斯特与法拉第的发现如何构成电磁关联理论的基础"。
Worked Example 4 · Field near a long wire例题 4 · 长导线附近的磁场

A long straight wire carries a current of $8.0$ A. Find the magnetic field strength at a perpendicular distance of $4.0$ cm from the wire. ($\mu_0 = 4\pi\times10^{-7}$ T·m/A)一根长直导线通有 $8.0$ A 电流。求距导线垂直距离 $4.0$ cm 处的磁场强度。（$\mu_0 = 4\pi\times10^{-7}$ T·m/A）

Convert distance to metres.将距离换算为米。 $r = 4.0\ \mathrm{cm} = 0.040\ \mathrm{m}$.$r = 4.0\ \mathrm{cm} = 0.040\ \mathrm{m}$。

$$ B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(8.0)}{2\pi(0.040)} = \frac{4\pi\times10^{-7}\times 8.0}{2\pi\times0.040}. $$

Simplify.化简。 The $\pi$ cancels:$\pi$ 消去：

$$ B = \frac{4\times10^{-7}\times 8.0}{2\times0.040} = \frac{3.2\times10^{-6}}{0.080} = 4.0\times10^{-5}\ \mathrm{T}. $$

Sanity check.合理性核验。 This is similar in magnitude to Earth's field ($5\times10^{-5}$ T) at $4$ cm from an 8 A wire — a compass placed that close would be noticeably deflected.这与地球磁场（$5\times10^{-5}$ T）量级相近——在 8 A 导线 4 cm 处，指南针会明显偏转。

What is the direction of the magnetic field directly above a long horizontal wire that carries current flowing eastward? (Use the right-hand rule.)对于一根通有向东电流的水平长导线，其正上方磁场方向如何？（用右手定则。）
§4 · Q1
East向东
East then curving upward向东然后向上弯曲
North向北
South向南
Right-hand rule: thumb east (current), fingers wrap around the wire. Directly above, the fingers point north. So $\vec B$ points north at that location.右手定则：拇指向东（电流），手指绕导线弯曲。在正上方，手指指向北。故该处 $\vec B$ 指向北。
Wrap your right hand around the wire, thumb pointing east (current direction). Directly above the wire your fingers point northward — that is the field direction there.右手绕导线，拇指向东（电流方向）。在导线正上方，手指指向北 — 这就是该处的磁场方向。
A solenoid has $500$ turns over a $0.25$ m length and carries $2.0$ A. What is the field inside? ($\mu_0 = 4\pi\times10^{-7}$ T·m/A; use $\pi \approx 3.14$)一根螺线管在 $0.25$ m 长度内有 $500$ 匝，通有 $2.0$ A 电流。内部磁场是多少？（$\mu_0 = 4\pi\times10^{-7}$ T·m/A；取 $\pi \approx 3.14$）
§4 · Q2
$5.0\times10^{-3}$ T
$2.5\times10^{-3}$ T
$1.0\times10^{-2}$ T
$8.0\times10^{-4}$ T
$n = 500/0.25 = 2000$ turns/m. $B = \mu_0 n I = (4\pi\times10^{-7})(2000)(2.0) = 16\pi\times10^{-4} \approx 5.0\times10^{-3}$ T.$n = 500/0.25 = 2000$ 匝/m。$B = \mu_0 n I = (4\pi\times10^{-7})(2000)(2.0) = 16\pi\times10^{-4} \approx 5.0\times10^{-3}$ T。
First find $n = N/L = 500/0.25$, then apply $B = \mu_0 n I$.先求 $n = N/L = 500/0.25$，再代入 $B = \mu_0 n I$。
Going deeper — the Biot-Savart law and why the field falls off as $1/r$深入 — 毕奥-萨伐尔定律与磁场为何随 $1/r$ 衰减

The Biot-Savart law gives the contribution $d\vec B$ to the magnetic field from a tiny current element $I\,d\vec\ell$ at a displacement $\vec r$ from the element:毕奥-萨伐尔定律给出距离电流元 $I\,d\vec\ell$ 为 $\vec r$ 处的磁场贡献 $d\vec B$：

$$ d\vec B = \frac{\mu_0}{4\pi} \frac{I\,d\vec\ell \times \hat r}{r^2}. $$

Integrating this around an infinite straight wire gives $B = \mu_0 I/(2\pi r)$ — the formula above — confirming the $1/r$ dependence. By contrast, a magnetic dipole falls off as $1/r^3$ (much faster) because the north and south poles partially cancel at large distances. For a solenoid, the contributions from all the loops add constructively inside but cancel outside, giving a nearly uniform interior field and zero exterior — exactly why MRI machines use solenoids.对无限长直导线积分得到 $B = \mu_0 I/(2\pi r)$——即上述公式——验证了 $1/r$ 关系。相比之下，磁偶极子以 $1/r^3$ 衰减（快得多），因为远处北极与南极的贡献部分抵消。对于螺线管，所有线圈的贡献在内部相加，在外部相消，产生近似均匀的内部场和零外部场——这正是 MRI 机器使用螺线管的原因。

Section 5 · Electromagnetic induction and Faraday's law第 5 节 · 电磁感应与法拉第定律

Electromagnetic Induction and Faraday's Law Honors — US NGSS (qualitative assessed)电磁感应与法拉第定律 荣誉 — US NGSS（仅评估定性）

Curriculum note.课纲提示。 Faraday's law with quantitative flux calculations is core in Ontario SPH3U/SPH4U (Strand F/D), BC Physics 12 (Content: "Faraday's law; Lenz's law; current induced by a change in magnetic flux"), and AB Physics 30 (30–B3.9k). It carries the Honors chip for the US NGSS track only, where HS-PS2-5 is limited to qualitative experimental investigation.含定量磁通量计算的法拉第定律在安大略 SPH3U/SPH4U（F/D 单元）、BC Physics 12（内容："法拉第定律；楞次定律；磁通量变化感应电流"）与 AB Physics 30（30–B3.9k）中均为核心内容。仅在 US NGSS 轨道上标 Honors，因为 HS-PS2-5 限于定性实验调查。
A changing magnetic flux induces an EMF — this is Faraday's law.变化的磁通量感应电动势 — 这就是法拉第定律。 $$ \varepsilon = -N\frac{d\Phi}{dt} \qquad \text{where} \quad \Phi = BA\cos\theta $$
  • Magnetic flux $\Phi$.磁通量 $\Phi$。 $\Phi = BA\cos\theta$ (SI unit: weber, Wb $=$ T·m²). $B$ = field strength, $A$ = area of loop, $\theta$ = angle between $\vec B$ and the normal to the loop. Flux is maximum when $\vec B \perp$ loop plane ($\theta = 0$); zero when $\vec B \parallel$ loop plane ($\theta = 90°$).$\Phi = BA\cos\theta$（SI 单位：韦伯，Wb $=$ T·m²）。$B$ = 场强，$A$ = 线圈面积，$\theta$ = $\vec B$ 与线圈法线的夹角。$\vec B$ 垂直于线圈平面时（$\theta = 0$）磁通量最大；$\vec B$ 平行于线圈平面时（$\theta = 90°$）为零。
  • EMF $\varepsilon$.电动势 $\varepsilon$。 The magnitude $|\varepsilon| = N|\Delta\Phi/\Delta t|$ (for uniform rate of change). $N$ = number of turns. The negative sign (Lenz's law, §6) tells you the direction. A larger $N$ or faster flux change gives a larger EMF.大小 $|\varepsilon| = N|\Delta\Phi/\Delta t|$（均匀变化时）。$N$ = 匝数。负号（楞次定律，§6）告诉你方向。$N$ 越大或磁通量变化越快，电动势越大。
  • Three ways to change flux.改变磁通量的三种方式。 Change $B$, change $A$ (stretch or compress the loop), or change $\theta$ (rotate the loop). Any of these induces an EMF.改变 $B$、改变 $A$（拉伸或压缩线圈），或改变 $\theta$（旋转线圈）。任何一种都能感应电动势。
BC Physics 12 Elaboration: "Faraday's law; Lenz's law; current induced by a change in magnetic flux; moving a bar, wire, coil, single charge within a changing magnetic field (strength, polarity, or area)." Alberta 30–B3.3k: "describe how the discoveries of Oersted and Faraday form the foundation of the theory relating electricity to magnetism."BC Physics 12 细化说明："法拉第定律；楞次定律；磁通量变化感应电流；在变化磁场（场强、极性或面积变化）中移动导体棒、导线、线圈、单个电荷"。阿尔伯塔 30–B3.3k："描述奥斯特与法拉第的发现如何构成电磁关联理论的基础"。
Worked Example 5 · EMF from a shrinking loop例题 5 · 收缩线圈中的电动势

A circular loop of wire ($N = 1$) sits perpendicular to a uniform $0.40$ T magnetic field. Its area decreases at a steady rate from $0.10\ \mathrm{m^2}$ to $0.04\ \mathrm{m^2}$ in $0.30$ s. Calculate the magnitude of the induced EMF.一个圆形导线线圈（$N = 1$）垂直置于 $0.40$ T 匀强磁场中。其面积在 $0.30$ s 内从 $0.10\ \mathrm{m^2}$ 均匀减小到 $0.04\ \mathrm{m^2}$。计算感应电动势的大小。

Rate of change of flux.磁通量的变化率。 Since $\vec B \perp$ loop plane, $\theta = 0$ and $\Phi = BA$.因为 $\vec B$ 垂直于线圈平面，$\theta = 0$，$\Phi = BA$。

$$ \frac{\Delta\Phi}{\Delta t} = B\frac{\Delta A}{\Delta t} = (0.40)\frac{0.04 - 0.10}{0.30} = (0.40)\frac{-0.06}{0.30} = -0.080\ \mathrm{Wb/s}. $$

Induced EMF magnitude.感应电动势大小。

$$ |\varepsilon| = N\left|\frac{\Delta\Phi}{\Delta t}\right| = (1)(0.080) = 0.080\ \mathrm{V}. $$

Interpretation.解读。 As the area shrinks, the flux through the loop decreases. By Lenz's law (§6) the induced current will try to maintain the original flux — so the induced current creates a magnetic field in the same direction as $\vec B$.面积收缩时，穿过线圈的磁通量减小。由楞次定律（§6），感应电流将试图维持原来的磁通量——故感应电流产生与 $\vec B$ 同向的磁场。

A $50$-turn coil has an area of $0.020\ \mathrm{m^2}$ perpendicular to a magnetic field that increases from $0.10$ T to $0.50$ T in $0.40$ s. What is the magnitude of the induced EMF?一个 $50$ 匝线圈面积为 $0.020\ \mathrm{m^2}$，垂直于磁场，磁场在 $0.40$ s 内从 $0.10$ T 增大到 $0.50$ T。感应电动势大小是多少？
§5 · Q1
$0.040$ V
$0.50$ V
$1.0$ V
$2.0$ V
$\Delta\Phi = (0.50-0.10)(0.020) = 0.008$ Wb. $|\varepsilon| = N|\Delta\Phi/\Delta t| = 50 \times 0.008/0.40 = 50 \times 0.020 = 1.0$ V.$\Delta\Phi = (0.50-0.10)(0.020) = 0.008$ Wb。$|\varepsilon| = N|\Delta\Phi/\Delta t| = 50 \times 0.008/0.40 = 50 \times 0.020 = 1.0$ V。
Compute $\Delta\Phi = \Delta B \times A$ (since field is perpendicular to loop), then $|\varepsilon| = N \Delta\Phi / \Delta t$.先求 $\Delta\Phi = \Delta B \times A$（磁场垂直于线圈），再算 $|\varepsilon| = N \Delta\Phi / \Delta t$。
Which of the following does NOT induce an EMF in a conducting loop?下列哪种情况不会在导线线圈中感应出电动势？
§5 · Q2
A constant magnetic field with the loop stationary匀强磁场中线圈静止
A decreasing magnetic field passing through the loop穿过线圈的磁场减弱
The loop moving into a region of stronger field线圈移入更强磁场区域
The loop rotating in a uniform magnetic field线圈在匀强磁场中转动
A constant field with a stationary loop means $\Delta\Phi = 0$ (no change in $B$, $A$, or $\theta$), so $\varepsilon = 0$. All other options change the flux through the loop.匀强磁场中线圈静止意味着 $\Delta\Phi = 0$（$B$、$A$、$\theta$ 均不变），故 $\varepsilon = 0$。其他选项均改变穿过线圈的磁通量。
EMF requires a change in flux ($\Delta\Phi \ne 0$). Only a constant field with a stationary loop produces no flux change and hence no EMF.电动势需要磁通量变化（$\Delta\Phi \ne 0$）。只有匀强磁场中的静止线圈不产生磁通量变化，因而无电动势。
Section 6 · Lenz's law第 6 节 · 楞次定律

Lenz's Law Honors — US NGSS (qualitative assessed)楞次定律 荣誉 — US NGSS（仅评估定性）

Lenz's law gives the direction of the induced current.楞次定律给出感应电流的方向。

Statement: The induced current flows in the direction that opposes the change in magnetic flux that caused it.表述：感应电流的流向使其所产生的磁场阻碍引起它的磁通量变化。

  • Flux increasing?磁通量增加？ The induced current creates a field opposing the original field (to reduce the net increase). By the right-hand rule, this determines the current direction in the loop.感应电流产生与原磁场相反的场（减缓净增加）。由右手定则，这决定了线圈中的电流方向。
  • Flux decreasing?磁通量减少？ The induced current creates a field in the same direction as the original field (to resist the decrease).感应电流产生与原磁场同向的场（阻止减少）。
  • Energy conservation.能量守恒。 Lenz's law is a consequence of energy conservation: the induced current always opposes the motion that creates it, so you must do work to maintain the flux change. BC Physics 12 elaborations note "Lenz's law" as a named content item. Alberta STS outcome 30–B3.1sts relates Lenz's law to conservation of energy.楞次定律是能量守恒的结果：感应电流总是阻碍产生它的运动，因此必须做功来维持磁通量变化。BC Physics 12 细化说明将"楞次定律"列为具名内容项。阿尔伯塔 STS outcome 30–B3.1sts 将楞次定律与能量守恒相联系。
Worked Example 6 · Direction of induced current as a magnet approaches a coil例题 6 · 磁铁靠近线圈时感应电流的方向

The north pole of a bar magnet is pushed toward the face of a circular coil (the field points toward the coil and is increasing). Using Lenz's law, determine the direction of the induced current as seen from the magnet's side.条形磁铁的北极推向圆形线圈正面（磁场指向线圈且在增强）。用楞次定律，判断从磁铁侧看到的感应电流方向。

Identify the flux change.判断磁通量变化。 The north pole approaching increases the flux through the coil (field points into the page as seen from the magnet).北极靠近使穿过线圈的磁通量增加（从磁铁侧看，磁场向右进入线圈）。

Apply Lenz's law.应用楞次定律。 The induced current must create a field that opposes the increasing flux — i.e. the induced field must point away from the magnet (out of the page as seen from the magnet's side, toward the magnet).感应电流必须产生阻碍磁通量增加的场——即感应磁场必须指向远离磁铁方向（从磁铁侧看向外，朝向磁铁）。

Right-hand rule for current direction.用右手定则确定电流方向。 For the induced field to point toward the magnet (out of the left face of the coil), the induced current flows counterclockwise as seen from the magnet's side. This counterclockwise current presents a north pole to the approaching magnet, repelling it — consistent with energy conservation (you must push the magnet to do work).要使感应磁场指向磁铁（从线圈左侧面向外），感应电流从磁铁侧看为逆时针。这个逆时针电流向靠近的磁铁呈现北极，产生排斥力——与能量守恒一致（你必须推动磁铁做功）。

A magnet is pulled away from a coil (so flux through the coil decreases). The induced current will produce a magnetic field that:一块磁铁从线圈旁拉开（穿过线圈的磁通量减少）。感应电流产生的磁场将：
§6 · Q1
Opposes the original field (to further reduce the flux)与原磁场相反（进一步减少磁通量）
Is in the same direction as the original field (to maintain the flux)与原磁场同向（维持磁通量）
Is perpendicular to the original field垂直于原磁场
Has zero magnitude because the flux is decreasing大小为零，因为磁通量在减少
Lenz's law: the induced current opposes the change. When flux decreases, the induced current tries to maintain the flux by producing a field in the same direction as the original field.楞次定律：感应电流阻碍变化。磁通量减少时，感应电流产生与原磁场同向的场，试图维持磁通量。
Lenz's law says the induced current opposes the change. Decreasing flux means the induced field tries to replace the lost flux — same direction as the original.楞次定律：感应电流阻碍变化。磁通量减少意味着感应磁场试图补充失去的磁通量——与原磁场同向。
Which physical law is Lenz's law a consequence of?楞次定律是哪个物理定律的推论？
§6 · Q2
Newton's second law牛顿第二定律
Coulomb's law库仑定律
Ohm's law欧姆定律
Conservation of energy能量守恒定律
Lenz's law is a consequence of energy conservation. If the induced current aided the change instead of opposing it, the current would accelerate the magnet and create more flux, producing more current in a runaway loop — violating energy conservation.楞次定律是能量守恒的推论。如果感应电流助长变化而非阻碍它，电流将加速磁铁、产生更多磁通量、进而产生更多电流，形成失控循环——违反能量守恒。
Lenz's law is rooted in energy conservation: the opposing current ensures you must do work (input energy) to maintain the flux change, preventing a perpetual-motion scenario.楞次定律植根于能量守恒：阻碍方向的电流确保你必须做功（输入能量）才能维持磁通量变化，防止永动机情形。
Going deeper — eddy currents and their applications深入 — 涡流及其应用

When a changing magnetic field penetrates a solid conductor (not just a wire loop), it induces circulating currents throughout the conductor's volume — eddy currents. By Lenz's law, these currents always oppose the cause. Key applications at HS and AP/IB level:当变化的磁场穿过固体导体（不只是导线线圈）时，它在导体体内感应出环流电流——涡流。由楞次定律，这些电流总是阻碍其成因。高中与 AP/IB 级的关键应用：

  • Magnetic braking.磁制动。 A conducting plate moving through a magnetic field experiences eddy currents whose Lenz-law force opposes the motion — a smooth, contactless brake. Used in roller coasters, MRI patient tables, and train brakes.在磁场中运动的导体板产生涡流，涡流的楞次力阻碍运动——平滑无接触制动。用于过山车、MRI 检查台和列车制动。
  • Induction heating.感应加热。 Eddy currents dissipate energy as heat ($P = I^2R$). Induction cooktops and industrial furnaces exploit this: the cooking pot itself becomes the resistor.涡流以热量耗散能量（$P = I^2R$）。电磁炉和工业炉利用这一点：锅具本身成为电阻器。
Section 7 · Generators, motors and transformers第 7 节 · 发电机、电动机与变压器

Generators, Motors and Transformers发电机、电动机与变压器

Three machines that convert between mechanical and electrical energy.三种在机械能与电能之间转换的机器。
  • Generator.发电机。 A coil rotated in a magnetic field. Rotating the coil changes the angle $\theta$ between $\vec B$ and the normal, so $\Phi = BA\cos\theta$ changes continuously, inducing a sinusoidal EMF: $\varepsilon = NBA\omega\sin(\omega t)$. Mechanical energy (from turbines, wind, water) $\to$ electrical energy. This is Faraday's law in action.在磁场中旋转的线圈。旋转改变 $\vec B$ 与法线的夹角 $\theta$，使 $\Phi = BA\cos\theta$ 持续变化，感应出正弦电动势：$\varepsilon = NBA\omega\sin(\omega t)$。机械能（来自涡轮、风、水）$\to$ 电能。这正是法拉第定律的实际应用。
  • Motor.电动机。 A current-carrying coil in a magnetic field. The force $F = BIL\sin\theta$ on the wire creates a torque that rotates the coil. Electrical energy $\to$ mechanical energy. A motor is a generator run in reverse.磁场中通有电流的线圈。导线所受力 $F = BIL\sin\theta$ 产生使线圈旋转的力矩。电能 $\to$ 机械能。电动机是反向运行的发电机。
  • Transformer.变压器。 Two coils sharing an iron core. An alternating current in the primary coil creates a changing flux that induces an EMF in the secondary. The voltage ratio equals the turns ratio:共用铁芯的两个线圈。初级线圈中的交变电流产生变化磁通量，在次级线圈中感应出电动势。电压比等于匝数比： $$ \frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad \text{and} \qquad \frac{I_s}{I_p} = \frac{N_p}{N_s} \quad (\text{ideal transformer}) $$ Step-up transformer: $N_s > N_p$, voltage increases, current decreases. Step-down: $N_s < N_p$. Power is conserved: $V_p I_p = V_s I_s$ for an ideal transformer.升压变压器：$N_s > N_p$，电压升高、电流减小。降压：$N_s < N_p$。功率守恒：理想变压器 $V_p I_p = V_s I_s$。
BC Physics 12 Elaboration: "applications of electromagnetic induction: back EMF, direct current (DC) motors, generators, transformers." Alberta 30–B3.7k–30–B3.9k covers the same scope quantitatively. Ontario SPH3U F3 explicitly names "technologies that use these properties and principles to produce and transmit electrical energy."BC Physics 12 细化说明："电磁感应的应用：反电动势、直流电动机、发电机、变压器"。阿尔伯塔 30–B3.7k–30–B3.9k 以定量方式涵盖同一内容。安大略 SPH3U F3 明确提及"利用这些性质和原理产生和传输电能的技术"。
Worked Example 7 · Transformer step-up calculation例题 7 · 升压变压器计算

A power plant generates $240$ V AC. A step-up transformer has $500$ turns on the primary and $10{,}000$ turns on the secondary. Find (a) the secondary voltage and (b) the secondary current if the primary current is $60$ A (assume an ideal transformer).一座发电厂产生 $240$ V 交流电。一台升压变压器初级有 $500$ 匝，次级有 $10{,}000$ 匝。求 (a) 次级电压和 (b) 若初级电流为 $60$ A 时的次级电流（假设理想变压器）。

(a) Secondary voltage.(a) 次级电压。

$$ V_s = V_p \frac{N_s}{N_p} = 240 \times \frac{10{,}000}{500} = 240 \times 20 = 4800\ \mathrm{V}. $$

(b) Secondary current (conservation of power).(b) 次级电流（功率守恒）。

$$ I_s = I_p \frac{N_p}{N_s} = 60 \times \frac{500}{10{,}000} = 60 \times 0.05 = 3.0\ \mathrm{A}. $$

Sanity check with power.用功率核验。 $P_p = V_p I_p = 240 \times 60 = 14{,}400\ \mathrm{W}$. $P_s = V_s I_s = 4800 \times 3.0 = 14{,}400\ \mathrm{W}$. Equal, as required for an ideal transformer. ✓$P_p = V_p I_p = 240 \times 60 = 14{,}400\ \mathrm{W}$。$P_s = V_s I_s = 4800 \times 3.0 = 14{,}400\ \mathrm{W}$。与理想变压器要求相等。✓

Physical meaning.物理意义。 High-voltage transmission ($4800$ V) reduces the current to $3.0$ A, so resistive losses in the power line ($P = I^2 R$) drop by a factor of $400$ compared to transmitting at $240$ V with $60$ A. This is why power grids operate at hundreds of kilovolts.高压输电（$4800$ V）将电流降至 $3.0$ A，因此输电线上的电阻损耗（$P = I^2 R$）相比 $240$ V、$60$ A 输电降低了 $400$ 倍。这就是为何电网在数十万伏特下运行。

A transformer has $200$ turns on the primary and $50$ turns on the secondary. If the primary voltage is $120$ V, what is the secondary voltage?一台变压器初级有 $200$ 匝，次级有 $50$ 匝。若初级电压为 $120$ V，次级电压是多少？
§7 · Q1
$480$ V
$30$ V
$120$ V
$24$ V
$V_s = V_p (N_s/N_p) = 120 \times (50/200) = 120 \times 0.25 = 30$ V. This is a step-down transformer ($N_s < N_p$).$V_s = V_p (N_s/N_p) = 120 \times (50/200) = 120 \times 0.25 = 30$ V。这是一台降压变压器（$N_s < N_p$）。
Use $V_s/V_p = N_s/N_p$. Here $N_s/N_p = 50/200 = 1/4$, so $V_s = 120/4$.用 $V_s/V_p = N_s/N_p$。这里 $N_s/N_p = 50/200 = 1/4$，故 $V_s = 120/4$。
What is the key physical reason that power grids transmit electricity at very high voltages?电力网络以极高电压输电的关键物理原因是什么？
§7 · Q2
Higher voltage reduces the current, so resistive losses ($P = I^2R$) in the wires are much smaller更高的电压减小电流，从而导线中的电阻损耗（$P = I^2R$）大幅降低
Higher voltage increases the current, delivering more power更高的电压增大电流，传输更多功率
High voltage prevents Lenz's law from opposing the transmission高电压防止楞次定律阻碍输电
Wire resistance decreases at high voltage due to superconductivity因超导性，导线电阻在高电压下降低
For a given power $P = VI$: higher $V$ means lower $I$. Line losses are $P_{\text{loss}} = I^2 R$, which scale as $I^2$ — a 20-fold increase in voltage reduces losses by a factor of $400$.对于给定功率 $P = VI$：更高的 $V$ 意味着更小的 $I$。线路损耗为 $P_{\text{loss}} = I^2 R$，随 $I^2$ 变化——电压提高 20 倍，损耗降低 $400$ 倍。
Power delivered equals $VI$: for fixed power, doubling $V$ halves $I$. Line losses are $I^2 R$, so halving $I$ reduces losses by a factor of 4. Higher voltage dramatically cuts transmission losses.传输功率为 $VI$：功率固定时，$V$ 加倍则 $I$ 减半。线路损耗为 $I^2 R$，$I$ 减半则损耗降为 $1/4$。高电压大幅削减输电损耗。
Going deeper — back-EMF in motors and why motors draw more current at startup深入 — 电动机中的反电动势与启动时为何电流更大

A running motor is also a generator: its rotating coil produces an EMF that opposes the supply voltage. This opposing EMF is called back-EMF ($\varepsilon_b$). The net voltage driving current through the coil resistance $R$ is $V_{\text{supply}} - \varepsilon_b$, so the motor current is运行中的电动机同时也是一台发电机：其旋转线圈产生与供电电压相反的电动势，称为反电动势（$\varepsilon_b$）。驱动电流流过线圈电阻 $R$ 的净电压为 $V_{\text{supply}} - \varepsilon_b$，故电动机电流为

$$ I = \frac{V_{\text{supply}} - \varepsilon_b}{R}. $$

At startup the rotor is stationary ($\varepsilon_b = 0$), so the full supply voltage drives current through only the coil resistance — the current is enormous, which is why motors trip circuit breakers if they fail to start. As the rotor spins faster, $\varepsilon_b$ grows, reducing the current. At full speed, $\varepsilon_b \approx V_{\text{supply}}$ and the current drops to just what is needed to overcome friction and carry the load. This is BC Physics 12's "back EMF" elaboration and Alberta's GO3 STS coverage of Lenz's law in motors.启动时转子静止（$\varepsilon_b = 0$），完整的供电电压只驱动电流流过线圈电阻——电流极大，这就是为何电动机无法启动时会跳闸。随着转子转速增加，$\varepsilon_b$ 增大，电流减小。满速时 $\varepsilon_b \approx V_{\text{supply}}$，电流降至仅克服摩擦和承载负载所需的量。这正是 BC Physics 12 的"反电动势"细化说明和阿尔伯塔 GO3 STS 对电动机中楞次定律的覆盖。

Exam Tactics考试策略

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

Right-hand rule discipline右手定则纪律
  • Three different contexts, three different gestures.三种情境，三种手势。 For the field around a straight wire: wrap your hand, thumb along current, fingers give field. For force on a charge or wire: fingers along $\vec v$ (or current), curl toward $\vec B$, thumb gives force direction (for positive charge or conventional current; reverse for negative).直导线周围的场：握拳，拇指沿电流，手指给出场方向。电荷或导线受力：手指沿 $\vec v$（或电流），弯向 $\vec B$，拇指给出力方向（正电荷/惯例电流；负电荷反向）。
  • Draw the geometry first.先画几何图。 Label $\vec v$, $\vec B$, and the angle $\theta$ before applying any formula. A sketch prevents the most common error: confusing which angle to use in $\sin\theta$.套用任何公式前，先标出 $\vec v$、$\vec B$ 及夹角 $\theta$。草图能防止最常见的错误：混淆 $\sin\theta$ 中哪个角。
  • Carry units.带单位计算。 Force in newtons (N), field in tesla (T), flux in weber (Wb), EMF in volts (V). A unit mismatch is the quickest way to lose marks.力的单位是牛顿（N），磁场单位是特斯拉（T），磁通量单位是韦伯（Wb），电动势单位是伏特（V）。单位不匹配是最快失分的方式。
Magnetic force pitfalls (§1–§4)磁力常见陷阱（§1–§4）
  • Magnetic force does no work.磁力不做功。 $\vec F \perp \vec v$ always, so the speed and kinetic energy of a charged particle in a pure magnetic field never change. The force only curves the path.$\vec F$ 始终垂直于 $\vec v$，故纯磁场中带电粒子的速率和动能永不改变。磁力只使路径弯曲。
  • Parallel means zero force.平行即零力。 $F = qvB\sin\theta = 0$ when $\theta = 0°$ (velocity parallel to field). A particle moving along a field line feels no magnetic force.当 $\theta = 0°$（速度平行于磁场）时，$F = qvB\sin\theta = 0$。沿磁场线运动的粒子不受磁力。
  • Conventional current direction matters.惯例电流方向很重要。 In $F = BIL\sin\theta$ the current $I$ uses conventional (positive-charge) flow direction. The force on the wire is the same regardless of whether positive or negative charges carry it.在 $F = BIL\sin\theta$ 中，电流 $I$ 使用惯例（正电荷）流动方向。无论正电荷还是负电荷携带电流，导线受力方向不变。
Faraday and Lenz pitfalls (§5–§6) Honors — US NGSS法拉第与楞次陷阱（§5–§6）荣誉 — US NGSS
  • Constant field, stationary loop: EMF $= 0$.恒定磁场，静止线圈：电动势 $= 0$。 EMF requires $\Delta\Phi \ne 0$. No change means no induction, even if $B$ is enormous.电动势需要 $\Delta\Phi \ne 0$。无变化则无感应，即使 $B$ 极大亦然。
  • Lenz says direction, Faraday says magnitude.楞次定律给方向，法拉第定律给大小。 Use $|\varepsilon| = N|\Delta\Phi/\Delta t|$ to compute the EMF, then use Lenz's law (opposing the flux change) to determine the current direction.用 $|\varepsilon| = N|\Delta\Phi/\Delta t|$ 计算电动势大小，再用楞次定律（阻碍磁通量变化）确定电流方向。
  • The angle in $\Phi = BA\cos\theta$.$\Phi = BA\cos\theta$ 中的角度。 $\theta$ is the angle between $\vec B$ and the normal to the loop. Flux is maximum when $\vec B \perp$ loop plane ($\theta = 0$), zero when $\vec B \parallel$ loop plane ($\theta = 90°$).$\theta$ 是 $\vec B$ 与线圈法线的夹角。$\vec B$ 垂直于线圈平面时（$\theta = 0$）磁通量最大，平行时（$\theta = 90°$）为零。
Answer hygiene作答规范
  • Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位；仅在最终答案处按题目要求精度四舍五入。
  • State the formula, substitute, solve in steps.写公式，代数值，逐步求解。 Examiners award method marks. $F = qvB\sin\theta = \ldots$ earns marks even if the arithmetic slips.评分者给方法分。写出 $F = qvB\sin\theta = \ldots$ 即使计算有误也可得分。
  • For transformers: check power conservation.变压器题：验证功率守恒。 After computing $V_s$ and $I_s$, verify $V_p I_p = V_s I_s$ as a sanity check. A mismatch signals an arithmetic error.计算出 $V_s$ 与 $I_s$ 后，验证 $V_p I_p = V_s I_s$ 作合理性检验。不匹配表明计算有误。
Active Recall主动复习

Flashcards闪卡

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Lorentz force magnitude?洛伦兹力大小？
$$F = qvB\sin\theta$$ max when $\vec v \perp \vec B$; zero when parallel$\vec v \perp \vec B$ 时最大；平行时为零
Force on a current-carrying wire?载流导线所受磁力？
$$F = BIL\sin\theta$$
Does magnetic force do work?磁力做功吗？
No. $\vec F \perp \vec v$ always, so speed and KE are unchanged. Force only changes direction.不做。$\vec F \perp \vec v$ 始终成立，速率和动能不变。磁力只改变运动方向。
Field of a long straight wire?长直导线的磁场？
$$B = \frac{\mu_0 I}{2\pi r}$$ circles the wire; RHR gives direction环绕导线；右手定则给出方向
Field inside a solenoid?螺线管内部磁场？
$$B = \mu_0 n I$$ $n$ = turns per metre; nearly uniform inside, ~0 outside$n$ = 每米匝数；内部近均匀，外部近零
Magnetic flux?磁通量？
$$\Phi = BA\cos\theta$$ $\theta$ = angle between $\vec B$ and loop normal; unit: Wb$\theta$ = $\vec B$ 与线圈法线夹角；单位：Wb
Faraday's law?法拉第定律？
$$\varepsilon = -N\frac{d\Phi}{dt}$$ changing flux induces EMF; larger $N$ or faster change → larger EMF变化的磁通量感应电动势；$N$ 更大或变化更快 → 电动势更大
Lenz's law?楞次定律？
Induced current opposes the change in flux that caused it. Consequence of energy conservation.感应电流阻碍引起它的磁通量变化。是能量守恒的推论。
Three ways to induce an EMF?感应电动势的三种方式？
Change $B$; change area $A$; change angle $\theta$ (rotate the loop). Any change in $\Phi$ works.改变 $B$；改变面积 $A$；改变角度 $\theta$（旋转线圈）。任何 $\Phi$ 的变化均可。
Transformer voltage ratio?变压器电压比？
$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$
Transformer current ratio (ideal)?变压器电流比（理想）？
$$\frac{I_s}{I_p} = \frac{N_p}{N_s}$$ power conserved: $V_p I_p = V_s I_s$功率守恒：$V_p I_p = V_s I_s$
Generator vs motor?发电机与电动机？
Generator: mechanical → electrical (rotating coil in field, Faraday's law). Motor: electrical → mechanical (force on wire, reversed). Same device, opposite energy flow.发电机：机械能 → 电能（线圈在场中旋转，法拉第定律）。电动机：电能 → 机械能（导线受力，反向运行）。同一装置，能量流向相反。
Why transmit at high voltage?为何高压输电？
Higher $V$ → lower $I$ for same power. Line losses $= I^2R$, so low $I$ slashes losses dramatically.更高的 $V$ → 同等功率下更低的 $I$。线路损耗 $= I^2R$，低 $I$ 大幅削减损耗。
Cyclotron (Larmor) radius?回旋半径（拉莫尔半径）？
$$r = \frac{mv}{qB}$$ charge curves into a circle; larger $B$ → tighter radius电荷弯成圆形；$B$ 越大 → 半径越小
Mixed Practice综合练习

Practice Quiz综合测验

A charged particle moves at $4.0\times10^6$ m/s perpendicular to a $0.25$ T magnetic field and follows a circular path. What type of energy change does the magnetic force cause? 🇺🇸 NGSS HS-PS2-5一带电粒子以 $4.0\times10^6$ m/s 垂直于 $0.25$ T 磁场运动，做圆周运动。磁力引起哪种能量变化？🇺🇸 NGSS HS-PS2-5
Q1
Kinetic energy increases动能增加
No energy change; speed is constant无能量变化；速率不变
Kinetic energy decreases动能减少
Potential energy increases势能增加
The magnetic force is always perpendicular to the velocity, so it does no work and causes no energy change. The speed (and thus KE) stays constant; only the direction changes.磁力始终垂直于速度，故不做功，不引起能量变化。速率（因而动能）保持不变；只有方向改变。
$\vec F \perp \vec v$ means zero work: $W = \vec F \cdot \Delta\vec r = 0$. The magnetic force cannot change speed or energy.$\vec F \perp \vec v$ 意味着零功：$W = \vec F \cdot \Delta\vec r = 0$。磁力不能改变速率或能量。
A $0.20$ m wire carrying $4.0$ A is at $30^{\circ}$ to a $0.50$ T magnetic field. What is the force on the wire?一根 $0.20$ m 的导线通有 $4.0$ A 电流，与 $0.50$ T 磁场成 $30^{\circ}$。导线所受力是多少？
Q2
$0.40$ N
$0.069$ N
$0.35$ N
$0.20$ N
$F = BIL\sin\theta = (0.50)(4.0)(0.20)\sin 30^{\circ} = (0.50)(4.0)(0.20)(0.5) = 0.20$ N.$F = BIL\sin\theta = (0.50)(4.0)(0.20)\sin 30^{\circ} = (0.50)(4.0)(0.20)(0.5) = 0.20$ N。
Use $F = BIL\sin\theta$. Don't forget $\sin 30^{\circ} = 0.5$.用 $F = BIL\sin\theta$。不要忘记 $\sin 30^{\circ} = 0.5$。
A long wire carries $10$ A. What is the magnetic field at $5.0$ cm from the wire? ($\mu_0 = 4\pi\times10^{-7}$ T·m/A; use $\pi \approx 3.14$) 🇨🇦 AB 30–B3.4k一根长导线通有 $10$ A 电流。距导线 $5.0$ cm 处磁场是多少？（$\mu_0 = 4\pi\times10^{-7}$ T·m/A；取 $\pi \approx 3.14$）🇨🇦 AB 30–B3.4k
Q3
$4.0\times10^{-5}$ T
$2.0\times10^{-5}$ T
$8.0\times10^{-5}$ T
$1.0\times10^{-4}$ T
$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(10)}{2\pi(0.050)} = \frac{4\times10^{-6}}{0.10} = 4.0\times10^{-5}$ T. ($\pi$ cancels.)$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(10)}{2\pi(0.050)} = \frac{4\times10^{-6}}{0.10} = 4.0\times10^{-5}$ T。（$\pi$ 消去。）
$B = \mu_0 I/(2\pi r)$. Convert $r$ to metres first ($5.0$ cm $= 0.050$ m), then the $\pi$ cancels.$B = \mu_0 I/(2\pi r)$。先将 $r$ 换算为米（$5.0$ cm $= 0.050$ m），则 $\pi$ 消去。
A $100$-turn circular coil of area $0.050\ \mathrm{m^2}$ is perpendicular to a field that drops from $0.80$ T to $0.20$ T in $0.50$ s. What is the induced EMF? 🇨🇦 BC Physics 12 / AB 30–B3.9k一个 $100$ 匝、面积 $0.050\ \mathrm{m^2}$ 的圆形线圈垂直于磁场，磁场在 $0.50$ s 内从 $0.80$ T 降到 $0.20$ T。感应电动势是多少？🇨🇦 BC Physics 12 / AB 30–B3.9k
Q4
$0.60$ V
$3.0$ V
$6.0$ V
$0.030$ V
$\Delta\Phi = \Delta B \times A = (0.20-0.80)(0.050) = -0.030$ Wb. $|\varepsilon| = N|\Delta\Phi/\Delta t| = 100 \times 0.030/0.50 = 100 \times 0.060 = 6.0$ V.$\Delta\Phi = \Delta B \times A = (0.20-0.80)(0.050) = -0.030$ Wb。$|\varepsilon| = N|\Delta\Phi/\Delta t| = 100 \times 0.030/0.50 = 100 \times 0.060 = 6.0$ V。
$|\Delta B| = 0.60$ T, $A = 0.050$ m², so $|\Delta\Phi| = 0.030$ Wb. Then $|\varepsilon| = N|\Delta\Phi|/\Delta t = 100\times 0.030/0.50$.$|\Delta B| = 0.60$ T，$A = 0.050$ m²，故 $|\Delta\Phi| = 0.030$ Wb。再算 $|\varepsilon| = N|\Delta\Phi|/\Delta t = 100\times 0.030/0.50$。
A bar magnet's north pole is thrust into a coil, increasing the flux through it. By Lenz's law, the induced current in the coil: 🇨🇦 ON SPH3U F3 / BC Physics 12条形磁铁北极插入线圈，穿过线圈的磁通量增加。由楞次定律，线圈中感应电流：🇨🇦 ON SPH3U F3 / BC Physics 12
Q5
Creates a field opposing the magnet's field (to resist the flux increase)产生与磁铁磁场相反的场（阻碍磁通量增加）
Creates a field reinforcing the magnet's field (to aid the flux increase)产生与磁铁磁场同向的场（助推磁通量增加）
Is zero because the magnet is moving为零，因为磁铁在运动
Is perpendicular to the magnet's motion垂直于磁铁的运动方向
Lenz's law: the induced current opposes the change. Flux is increasing, so the induced current creates a field opposing the original field. This presents a north pole to the approaching magnet, repelling it.楞次定律：感应电流阻碍变化。磁通量增加，故感应电流产生与原磁场相反的场。这向靠近的磁铁呈现北极，产生排斥力。
Lenz: induced current opposes the change. Increasing flux → induced field opposes original field. Never aids the change (that would violate energy conservation).楞次：感应电流阻碍变化。磁通量增加 → 感应磁场与原磁场相反。绝不助推变化（那将违反能量守恒）。
A step-up transformer has $N_p = 200$ turns and $N_s = 1000$ turns. The primary voltage is $120$ V. What is the secondary voltage? 🇨🇦 ON SPH3U F3 / BC Physics 12一台升压变压器有 $N_p = 200$ 匝初级和 $N_s = 1000$ 匝次级。初级电压为 $120$ V。次级电压是多少？🇨🇦 ON SPH3U F3 / BC Physics 12
Q6
$24$ V
$600$ V
$120$ V
$1200$ V
$V_s = V_p(N_s/N_p) = 120\times(1000/200) = 120\times5 = 600$ V.$V_s = V_p(N_s/N_p) = 120\times(1000/200) = 120\times5 = 600$ V。
$V_s/V_p = N_s/N_p$. Here $N_s/N_p = 5$, so $V_s = 5\times120$.$V_s/V_p = N_s/N_p$。这里 $N_s/N_p = 5$，故 $V_s = 5\times120$。
For the transformer in Q6, if the primary current is $10$ A, what is the secondary current (ideal transformer)?对于 Q6 中的变压器，若初级电流为 $10$ A，次级电流是多少（理想变压器）？
Q7
$50$ A
$10$ A
$5$ A
$2.0$ A
$I_s = I_p(N_p/N_s) = 10\times(200/1000) = 10\times0.2 = 2.0$ A. Check: $P_p = 120\times10 = 1200$ W; $P_s = 600\times2 = 1200$ W. ✓$I_s = I_p(N_p/N_s) = 10\times(200/1000) = 10\times0.2 = 2.0$ A。核验：$P_p = 120\times10 = 1200$ W；$P_s = 600\times2 = 1200$ W。✓
Current ratio inverts the turns ratio: $I_s/I_p = N_p/N_s$. Higher voltage means lower current for the same power.电流比是匝数比的倒数：$I_s/I_p = N_p/N_s$。相同功率下，电压越高则电流越小。
A proton ($q = 1.6\times10^{-19}$ C, $m = 1.67\times10^{-27}$ kg) moves at $2.0\times10^6$ m/s perpendicular to a $0.30$ T field. What is its circular orbit radius? 🇨🇦 AB 30–B3.5k一个质子（$q = 1.6\times10^{-19}$ C，$m = 1.67\times10^{-27}$ kg）以 $2.0\times10^6$ m/s 垂直于 $0.30$ T 磁场运动。其圆轨道半径是多少？🇨🇦 AB 30–B3.5k
Q8
$7.0\times10^{-3}$ m
$3.5\times10^{-2}$ m
$6.96\times10^{-2}$ m
$1.4\times10^{-1}$ m
$r = mv/(qB) = (1.67\times10^{-27})(2.0\times10^6)/[(1.6\times10^{-19})(0.30)] = 3.34\times10^{-21}/(4.8\times10^{-20}) \approx 0.0696$ m $\approx 6.96\times10^{-2}$ m.$r = mv/(qB) = (1.67\times10^{-27})(2.0\times10^6)/[(1.6\times10^{-19})(0.30)] = 3.34\times10^{-21}/(4.8\times10^{-20}) \approx 0.0696$ m $\approx 6.96\times10^{-2}$ m。
From $qvB = mv^2/r$: $r = mv/(qB)$. Plug in all values carefully, keeping track of powers of 10.由 $qvB = mv^2/r$：$r = mv/(qB)$。仔细代入所有数值，追踪 10 的幂次。
An electric current in a wire produces a magnetic field around it. This was first demonstrated experimentally by: 🇨🇦 AB 30–B3.3k导线中的电流在其周围产生磁场，这最初由谁通过实验证明的？🇨🇦 AB 30–B3.3k
Q9
Oersted (1820) — a compass deflected near a current-carrying wire奥斯特（1820 年）——指南针在载流导线旁偏转
Faraday — he discovered that changing fields induce currents法拉第——他发现变化磁场感应电流
Maxwell — he unified electricity and magnetism mathematically麦克斯韦——他用数学统一了电学与磁学
Ampere — he quantified the force between current-carrying wires安培——他定量研究了载流导线之间的力
Hans Christian Oersted observed in 1820 that a compass needle deflected when placed near a wire carrying current, proving for the first time that electricity produces magnetism. Alberta 30–B3.3k explicitly names "the discoveries of Oersted and Faraday."汉斯·克里斯蒂安·奥斯特于 1820 年观察到指南针在通有电流的导线旁发生偏转，首次证明电学产生磁学。阿尔伯塔 30–B3.3k 明确点名"奥斯特与法拉第的发现"。
Oersted's 1820 compass experiment was the first demonstration that electric current produces a magnetic field. Faraday later showed the reverse: changing magnetic fields induce currents.奥斯特 1820 年的指南针实验是首次证明电流产生磁场。法拉第后来证明了反向：变化的磁场感应电流。
A generator coil rotates in a magnetic field. What is the energy conversion taking place? 🇨🇦 BC Physics 12 — applications of EM induction发电机线圈在磁场中旋转。发生什么能量转换？🇨🇦 BC Physics 12 — 电磁感应的应用
Q10
Electrical energy $\to$ mechanical energy电能 $\to$ 机械能
Mechanical energy $\to$ electrical energy机械能 $\to$ 电能
Thermal energy $\to$ electrical energy热能 $\to$ 电能
Magnetic energy $\to$ electrical energy directly磁能 $\to$ 直接转化为电能
A generator converts mechanical energy (rotation of the coil, driven by a turbine, wind, water, etc.) into electrical energy via Faraday's law. The reverse (electrical to mechanical) is a motor.发电机通过法拉第定律将机械能（线圈旋转，由涡轮、风、水等驱动）转化为电能。反向（电能转为机械能）是电动机。
A generator: mechanical input → electrical output (Faraday's law). A motor is the reverse. Thermal → electrical is a thermoelectric device, not a standard generator.发电机：机械能输入 → 电能输出（法拉第定律）。电动机是反向。热能 → 电能是热电装置，不是标准发电机。
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Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Magnetism and electromagnetic induction sit at the intersection of electricity and optics. The Lorentz force and Faraday's law together predict the existence of electromagnetic waves (Maxwell's equations), which are the foundation of optics (Unit 7) and the full electromagnetic spectrum. Within the HS Physics sequence, this unit builds on Electric Fields (Unit 8) and Current Electricity (Unit 9), and prepares for waves and light. The connections below point at the next steps in this repo.磁场与电磁感应处于电学与光学的交汇处。洛伦兹力与法拉第定律共同预言了电磁波（麦克斯韦方程）的存在，电磁波是光学（Unit 7）和整个电磁波谱的基础。在 HS Physics 课程序列中，本单元建立在电场（Unit 8）和电流（Unit 9）之上，为波与光做准备。下方链接指向本仓库的后续步骤。

Within High School Physics.在 HS Physics 内部。

Unit 8 (Electrostatics) introduced Coulomb's law and electric fields, which are the electric counterpart to the magnetic field treated here. Unit 9 (Current Electricity) introduced the current $I$ and resistance $R$ that appear in transformer and solenoid calculations. Unit 7 (Light and Optics) extends electromagnetic theory to visible light: Maxwell showed that light itself is an electromagnetic wave, with speed $c = 1/\sqrt{\mu_0\varepsilon_0}$, directly derived from the constants of electromagnetism this unit introduces ($\mu_0$) and electrostatics ($\varepsilon_0$).Unit 8（静电学）介绍了库仑定律和电场，是本单元磁场的电学对应。Unit 9（电流电学）介绍了出现在变压器和螺线管计算中的电流 $I$ 和电阻 $R$。Unit 7（光与光学）将电磁理论延伸至可见光：麦克斯韦证明光本身是电磁波，速度 $c = 1/\sqrt{\mu_0\varepsilon_0}$，直接从本单元引入的电磁学常数（$\mu_0$）和静电学常数（$\varepsilon_0$）推导得出。

Across the IB Physics feeders in this repo.本仓库中的 IB Physics 衔接单元。

HS Physics Unit 8 · Electrostatics (Coulomb's law, $\vec E$ field — the electric parallel to this unit's $\vec B$ field)HS Physics Unit 8 · 静电学（库仑定律、$\vec E$ 场 — 本单元 $\vec B$ 场的电学对应） HS Physics Unit 9 · Current Electricity (current, resistance, and power — quantities used in transformer and solenoid calculations)HS Physics Unit 9 · 电流电学（电流、电阻与功率 — 变压器和螺线管计算中用到的量）

If you are aiming for IB Physics HL, Topic D (Fields) picks up exactly this scope — magnetic fields, induction, Faraday, Lenz — and extends it with electromagnetic induction in rotating frames and the full Maxwell equations. For AP Physics C: Electricity and Magnetism, the Lorentz force, Faraday's law, and transformer analysis are all explicitly tested. The quantitative depth in this guide (especially the Honors sections) matches the AP Physics C and IB HL expectation level.备考 IB Physics HL：D 主题（场）正是接续此范围——磁场、感应、法拉第、楞次——并以旋转参考系中的电磁感应和完整麦克斯韦方程加以拓展。对于 AP Physics C：电磁学，洛伦兹力、法拉第定律和变压器分析均被明确考查。本指南中的定量深度（尤其是 Honors 各节）与 AP Physics C 和 IB HL 的期望水平相匹配。

DINGRUI SCHOLARS

High School Physics · Magnetism and Electromagnetic Induction · v1高中物理 · 磁场与电磁感应 · v1