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Magnetism and Electromagnetic Induction · Solutions磁场与电磁感应 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 19 marksAP 选择题 + 安/卑省考短答 · 共 19 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Magnetic field lines磁场线 · HS-PS2-5 [3 marks][3 分]

Outside a bar magnet, field lines are drawn so that they...?在条形磁铁外部，磁场线的画法满足什么？

Answer:答案：  (B)  exit the north pole, enter the south pole从北极出发，进入南极

(a) Apply the field-line convention套用磁场线约定 M1·A1·A1

By convention, in the space outside a magnet the field lines emerge from (exit) the north pole and enter the south pole. Inside the magnet they run from south back to north, forming complete closed loops. Option (B).按约定，在磁铁外部空间，磁场线从北极发出（出发），进入南极。在磁铁内部，它们从南极回到北极，构成完整的闭合回路。选 (B)。
Why the distractors fail.干扰项分析。
(A): reverses the convention; this is the direction inside the magnet, not outside.颠倒了约定；这是磁铁内部的方向，而非外部。
(C): field lines never start or stop in empty space; they always form closed loops ($\nabla\cdot\vec B = 0$).磁场线绝不会起止于空旷处；它们总是构成闭合回路（$\nabla\cdot\vec B = 0$）。
(D): field lines never cross; if they did, the field would have two directions at one point.磁场线绝不相交；若相交，则某点磁场会有两个方向。
Field lines exit N, enter S, and always close on themselves.磁场线出 N 入 S，并始终自我闭合。 Three rules govern every magnetic field diagram: (1) outside a magnet, lines run N to S; (2) lines form continuous closed loops because there are no magnetic monopoles ($\nabla\cdot\vec B = 0$, Gauss's law for magnetism); (3) the density of lines indicates field strength, and lines never cross. Contrast this with electric field lines, which begin on positive charges and end on negative charges. The "no monopole" rule is exactly why every cut piece of a magnet still has both a north and a south pole.每张磁场图都遵循三条规则：(1) 在磁铁外部，磁场线由 N 指向 S；(2) 磁场线构成连续的闭合回路，因为不存在磁单极（$\nabla\cdot\vec B = 0$，磁场的高斯定律）；(3) 磁场线的疏密表示场强，且永不相交。与之相对，电场线起始于正电荷、终止于负电荷。"无磁单极"规则正是为何磁铁被切成任意小块后仍同时具有北极和南极的原因。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Lorentz force洛伦兹力 · HS-PS2-5 [3 marks][3 分]

Proton at $2.0\times10^6$ m/s perpendicular to a $0.30$ T field. Magnitude of magnetic force?质子以 $2.0\times10^6$ m/s 垂直于 $0.30$ T 磁场运动。磁力大小？

Answer:答案：  (A)  $9.6\times10^{-14}\ \text{N}$

(a) Apply the Lorentz-force formula $F = qvB\sin\theta$套用洛伦兹力公式 $F = qvB\sin\theta$ M1·A1·A1

Perpendicular motion means $\theta = 90°$, so $\sin\theta = 1$:垂直运动意味着 $\theta = 90°$，故 $\sin\theta = 1$： $$ F \;=\; qvB \;=\; (1.6\times10^{-19})(2.0\times10^6)(0.30) \;=\; 9.6\times10^{-14}\ \text{N}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $4.8\times10^{-20}\ \text{N}$: drops the factor of $v = 2.0\times10^6$ (computes $qB$ only, off by the speed).漏掉速率 $v = 2.0\times10^6$（只算 $qB$）。
(C) $6.0\times10^{5}\ \text{N}$: computes $vB$ but omits the charge $q$.算出 $vB$ 但漏掉电荷量 $q$。
(D) $0\ \text{N}$: would be correct only if $\vec v \parallel \vec B$; here they are perpendicular.仅当 $\vec v \parallel \vec B$ 时成立；此处两者垂直。
$F = qvB\sin\theta$ peaks at $\theta = 90°$ and vanishes at $\theta = 0°$.$F = qvB\sin\theta$ 在 $\theta = 90°$ 时最大，在 $\theta = 0°$ 时为零。 The magnetic force on a moving charge depends on the angle between velocity and field through $\sin\theta$. Maximum when motion is perpendicular to $\vec B$; zero when parallel. Carry units carefully: coulombs times metres-per-second times tesla yields newtons. A reliable habit is to substitute the perpendicular case ($\sin 90° = 1$) explicitly so you never silently drop the factor. This same expression, summed over all the drifting charges in a wire, becomes $F = BIL$ in the next question.运动电荷所受磁力通过 $\sin\theta$ 取决于速度与磁场的夹角。运动垂直于 $\vec B$ 时最大；平行时为零。注意单位：库仑乘米每秒乘特斯拉得到牛顿。一个可靠的习惯是显式代入垂直情形（$\sin 90° = 1$），这样就不会悄悄漏掉该因子。把这一表达式对导线中所有漂移电荷求和，便得到下一题的 $F = BIL$。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Force on a wire导线受力 · SPH3U F2 [4 marks][4 分]

Wire $0.50$ m, $I = 6.0$ A north, in a $0.40$ T field pointing up. (a) Force magnitude. (b) Direction. (c) Reverse the current.导线 $0.50$ m，$I = 6.0$ A 向北，置于 $0.40$ T 向上的磁场中。(a) 力的大小。(b) 方向。(c) 电流反向。

Answer:答案：  (a) $F = 1.2\ \text{N}$  ·  (b) east向东  ·  (c) same size, reversed to west大小不变，反向为向西

(a) Force magnitude using $F = BIL\sin\theta$用 $F = BIL\sin\theta$ 求力的大小 M1·A1

Current (north) is perpendicular to field (up), so $\theta = 90°$:电流（向北）垂直于磁场（向上），故 $\theta = 90°$： $$ F \;=\; BIL \;=\; (0.40)(6.0)(0.50) \;=\; 1.2\ \text{N}. $$

(b) Direction by the right-hand rule用右手定则确定方向 A1

Point the right-hand fingers north (current), curl them upward (field); the thumb points east. The force on the wire is directed east.右手手指向北（电流），弯向上方（磁场）；拇指指向东。导线受力方向向东。

(c) Effect of reversing the current电流反向的影响 A1

Reversing $I$ reverses the force direction (now west) while its magnitude stays $1.2$ N, since $F = BIL$ does not depend on the current's sign.电流反向使力的方向反转（变为向西），大小仍为 $1.2$ N，因为 $F = BIL$ 与电流的正负无关。
$F = BIL$ is the Lorentz force summed over a whole wire; the right-hand rule fixes its direction.$F = BIL$ 是洛伦兹力在整根导线上的总和；右手定则确定其方向。 A current is just charge in motion, so the force on each drifting charge ($qvB$) adds up across the wire to give $F = BIL$ when current and field are perpendicular. The right-hand rule is the same gesture as for a single positive charge: fingers along the conventional (positive) current, curl toward $\vec B$, thumb gives the force. This east-pointing force is exactly the mechanism that turns a motor coil. Reversing either the current or the field flips the force, which is why a motor needs a commutator to keep the torque turning the same way.电流不过是运动中的电荷，因此当电流与磁场垂直时，每个漂移电荷所受的力（$qvB$）在整根导线上累加，得到 $F = BIL$。右手定则的手势与单个正电荷相同：手指沿惯例（正）电流方向，弯向 $\vec B$，拇指给出力的方向。这个向东的力正是使电动机线圈转动的机制。电流或磁场任一反向都会翻转力的方向，这正是电动机需要换向器来保持力矩沿同一方向旋转的原因。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Field of a straight wire直导线的磁场 · HS-PS2-5 [4 marks][4 分]

Long straight wire, $I = 10$ A. Field at $r = 5.0$ cm?长直导线，$I = 10$ A。$r = 5.0$ cm 处的磁场？

Answer:答案：  (A)  $4.0\times10^{-5}\ \text{T}$

(a) Apply $B = \mu_0 I /(2\pi r)$ after converting units换算单位后套用 $B = \mu_0 I /(2\pi r)$ M1·A1·A1

Convert $r = 5.0\ \text{cm} = 0.050$ m. The $\pi$ cancels:换算 $r = 5.0\ \text{cm} = 0.050$ m。$\pi$ 消去： $$ B \;=\; \frac{\mu_0 I}{2\pi r} \;=\; \frac{(4\pi\times10^{-7})(10)}{2\pi(0.050)} \;=\; \frac{2\times10^{-7}\times 10}{0.050} \;=\; 4.0\times10^{-5}\ \text{T}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $1.3\times10^{-4}\ \text{T}$: forgets to convert $r$, using $5.0$ cm as $5.0$ m... actually it drops the $2\pi$ and divides by $r$ alone.漏掉 $2\pi$，只除以 $r$。
(C) $4.0\times10^{-3}\ \text{T}$: leaves $r$ in centimetres ($0.050 \to 5.0$), shifting the answer by $10^2$.未将 $r$ 换算为米（把 $0.050$ 当作 $5.0$），结果偏差 $10^2$。
(D) $2.0\times10^{-6}\ \text{T}$: uses $r = 1$ m or otherwise mishandles the factor of $2$.用 $r = 1$ m 或错误处理因子 $2$。
The field of a long straight wire falls off as $1/r$ and circles the wire (right-hand rule).长直导线的磁场随 $1/r$ 衰减，并环绕导线（右手定则）。 $B = \mu_0 I/(2\pi r)$ has two examiner-favourite traps: (1) leaving the distance in centimetres, and (2) forgetting that the $\pi$ in $\mu_0 = 4\pi\times10^{-7}$ cancels the $\pi$ in $2\pi r$, leaving the clean factor $2\times10^{-7}$. Result: a $10$ A wire produces about $4\times10^{-5}$ T at $5$ cm, comparable to Earth's field, so a nearby compass would deflect. The direction wraps around the wire: grip the wire with your right hand, thumb along the current, and your fingers curl in the direction of $\vec B$.$B = \mu_0 I/(2\pi r)$ 有两个阅卷常见陷阱：(1) 距离仍以厘米代入；(2) 忘记 $\mu_0 = 4\pi\times10^{-7}$ 中的 $\pi$ 与 $2\pi r$ 中的 $\pi$ 相消，留下干净的因子 $2\times10^{-7}$。结果：$10$ A 导线在 $5$ cm 处产生约 $4\times10^{-5}$ T，与地球磁场相当，故附近的指南针会偏转。方向环绕导线：右手握住导线，拇指沿电流，四指弯曲方向即 $\vec B$ 的方向。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §1-2 Right-hand rule右手定则 · Physics 12 [5 marks][5 分]

Electron moving east at $3.0\times10^5$ m/s; field points up. (a) Force direction. (b) Why no work + path. (c) Force if it moved up (parallel).电子以 $3.0\times10^5$ m/s 向东运动；磁场向上。(a) 力的方向。(b) 为何不做功 + 路径。(c) 若改为向上（平行）运动时的力。

Answer:答案：  (a) north向北  ·  (b) $\vec F \perp \vec v$, so KE constant; circular path$\vec F \perp \vec v$，动能不变；圆形路径  ·  (c) $F = 0$

(a) Direction by the right-hand rule, then flip for the negative charge用右手定则，再对负电荷反向 M1·A1

For a positive charge moving east with the field pointing up: fingers east, curl upward, thumb points south. The electron is negative, so its force is reversed: it points north.对一个向东运动、磁场向上的正电荷：手指向东，弯向上方，拇指指向南。电子带负电，故受力方向相反：指向北。

(b) No work; the resulting path不做功；运动路径 A1·A1

The magnetic force is always perpendicular to the velocity ($\vec F \perp \vec v$). Work $= Fd\cos 90° = 0$, so the speed and kinetic energy never change. With a constant-magnitude force perpendicular to a constant-speed velocity, the electron follows a circular path (in this geometry, a circle in the horizontal plane).磁力始终垂直于速度（$\vec F \perp \vec v$）。功 $= Fd\cos 90° = 0$，故速率和动能始终不变。在大小恒定且垂直于恒定速率速度的力作用下，电子做圆周运动（在此几何中为水平面内的圆）。

(c) Force when moving parallel to the field沿磁场方向运动时的力 A1

If the electron moves straight up (parallel to $\vec B$), then $\theta = 0°$ and $F = qvB\sin 0° = 0$. No magnetic force acts.若电子竖直向上（平行于 $\vec B$）运动，则 $\theta = 0°$，$F = qvB\sin 0° = 0$。不受磁力。
The magnetic force does no work, so it curves the path without changing the speed.磁力不做功，因此只弯曲路径而不改变速率。 Two ideas combine here. First, direction: use the right-hand rule for a positive charge and then reverse the result for an electron, since it carries negative charge. Second, energetics: because $\vec F \perp \vec v$ at every instant, the force can never add or remove kinetic energy, so the speed is constant and the path is a circle (radius $r = mv/qB$). This is why mass spectrometers, cyclotrons, and the aurora all rely on magnetic fields to steer charged particles without speeding them up. When velocity is parallel to the field, $\sin\theta = 0$ and the particle sails straight through, undeflected.这里结合了两个要点。第一，方向：先用右手定则求正电荷的方向，再对电子（带负电）反向。第二，能量：由于每一瞬间 $\vec F \perp \vec v$，该力无法增减动能，故速率恒定、路径为圆（半径 $r = mv/qB$）。这正是质谱仪、回旋加速器和极光都依靠磁场在不加速带电粒子的前提下使其偏转的原因。当速度平行于磁场时，$\sin\theta = 0$，粒子直线穿过、不发生偏转。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Charge in a field (circular motion)磁场中电荷（圆周运动） · HS-PS2-5 [8 marks][8 分]

Proton ($m_p = 1.67\times10^{-27}$ kg, $q = 1.6\times10^{-19}$ C), $v = 5.0\times10^5$ m/s perpendicular to $B = 0.20$ T. (a) Force. (b) Radius. (c) Period. (d) Effect of doubling $B$.质子（$m_p = 1.67\times10^{-27}$ kg，$q = 1.6\times10^{-19}$ C），$v = 5.0\times10^5$ m/s 垂直于 $B = 0.20$ T。(a) 力。(b) 半径。(c) 周期。(d) $B$ 加倍的影响。

Answer:答案：  (a) $F = 1.6\times10^{-14}\ \text{N}$  ·  (b) $r = 2.6\times10^{-2}\ \text{m}$  ·  (c) $T = 3.3\times10^{-7}\ \text{s}$  ·  (d) $r$ halves, $T$ halves$r$ 减半，$T$ 减半

(a) Magnetic force $F = qvB$磁力 $F = qvB$ M1·A1

$$ F \;=\; qvB \;=\; (1.6\times10^{-19})(5.0\times10^5)(0.20) \;=\; 1.6\times10^{-14}\ \text{N.} $$

(b) Radius using $r = mv/(qB)$用 $r = mv/(qB)$ 求半径 M1·A1

$$ r \;=\; \frac{mv}{qB} \;=\; \frac{(1.67\times10^{-27})(5.0\times10^5)}{(1.6\times10^{-19})(0.20)} \;=\; \frac{8.35\times10^{-22}}{3.2\times10^{-20}} \;=\; 2.6\times10^{-2}\ \text{m.} $$

(c) Period using $T = 2\pi m/(qB)$用 $T = 2\pi m/(qB)$ 求周期 M1·A1

$$ T \;=\; \frac{2\pi m}{qB} \;=\; \frac{2\pi (1.67\times10^{-27})}{(1.6\times10^{-19})(0.20)} \;=\; \frac{1.049\times10^{-26}}{3.2\times10^{-20}} \;=\; 3.3\times10^{-7}\ \text{s.} $$

(d) Doubling the field strength磁场强度加倍 A1·A1

Both $r = mv/(qB)$ and $T = 2\pi m/(qB)$ are inversely proportional to $B$. Doubling $B$ therefore halves the radius and halves the period. (The speed $v$ is unchanged because the magnetic force does no work.)$r = mv/(qB)$ 与 $T = 2\pi m/(qB)$ 都与 $B$ 成反比。因此 $B$ 加倍会使半径减半、周期减半。（速率 $v$ 不变，因为磁力不做功。）
The orbital period $T = 2\pi m/(qB)$ is independent of speed; this is the cyclotron principle.回旋周期 $T = 2\pi m/(qB)$ 与速率无关；这就是回旋加速器原理。 Setting the magnetic force equal to the centripetal requirement, $qvB = mv^2/r$, gives $r = mv/(qB)$; dividing the circumference $2\pi r$ by $v$ gives $T = 2\pi m/(qB)$, in which the speed cancels. A faster proton simply travels a bigger circle in the same time. This speed-independence is exactly what lets a cyclotron accelerate particles with a fixed-frequency driving voltage. Doubling $B$ tightens the orbit and quickens it in equal measure. A reliable check on part (b): $2.6$ cm is a sensible laboratory-scale radius for a slow proton in a moderate field.令磁力等于向心力 $qvB = mv^2/r$，得 $r = mv/(qB)$；用周长 $2\pi r$ 除以 $v$ 得 $T = 2\pi m/(qB)$，其中速率相消。更快的质子只是在相同时间内走更大的圆。这种与速率无关的特性正是回旋加速器能用固定频率的驱动电压加速粒子的原因。$B$ 加倍会等比例地收紧并加快轨道。(b) 的可靠核验：对于中等磁场中的慢速质子，$2.6$ cm 是合理的实验室尺度半径。
Q7MEDIUM中Honors荣誉级 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Faraday's law法拉第定律 · SPH4U D2 [8 marks][8 分]

Coil $N = 200$, $A = 0.015\ \text{m}^2$, perpendicular to a field falling from $0.80$ T to $0.20$ T in $0.50$ s; $R = 10\ \Omega$. (a) Flux change. (b) Induced EMF. (c) Induced current. (d) How to double the EMF.线圈 $N = 200$，$A = 0.015\ \text{m}^2$，垂直于磁场，磁场在 $0.50$ s 内从 $0.80$ T 降至 $0.20$ T；$R = 10\ \Omega$。(a) 磁通量变化。(b) 感应电动势。(c) 感应电流。(d) 如何使电动势加倍。

Answer:答案：  (a) $\Delta\Phi = -9.0\times10^{-3}\ \text{Wb}$  ·  (b) $|\varepsilon| = 3.6\ \text{V}$  ·  (c) $I = 0.36\ \text{A}$  ·  (d) halve the time (e.g. drop $B$ in $0.25$ s)把时间减半（如 $0.25$ s 内完成变化）

(a) Change in flux through one turn, $\Delta\Phi = \Delta B \cdot A$一匝的磁通量变化 $\Delta\Phi = \Delta B \cdot A$ M1·A1

The field is perpendicular to the coil ($\theta = 0$), so $\Phi = BA$:磁场垂直于线圈（$\theta = 0$），故 $\Phi = BA$： $$ \Delta\Phi \;=\; (\Delta B)\,A \;=\; (0.20 - 0.80)(0.015) \;=\; -9.0\times10^{-3}\ \text{Wb.} $$

(b) Induced EMF, $|\varepsilon| = N|\Delta\Phi/\Delta t|$感应电动势 $|\varepsilon| = N|\Delta\Phi/\Delta t|$ M1·A1·A1

$$ |\varepsilon| \;=\; N\left|\frac{\Delta\Phi}{\Delta t}\right| \;=\; 200 \times \frac{9.0\times10^{-3}}{0.50} \;=\; 200 \times 0.018 \;=\; 3.6\ \text{V.} $$

(c) Induced current, $I = |\varepsilon|/R$感应电流 $I = |\varepsilon|/R$ M1·A1

$$ I \;=\; \frac{|\varepsilon|}{R} \;=\; \frac{3.6}{10} \;=\; 0.36\ \text{A.} $$

(d) One change that doubles the EMF使电动势加倍的一种改动 A1

Halve the time over which the field changes (e.g. complete the drop in $0.25$ s). Equivalently, double $N$, double the area $A$, or double the size of the field change $\Delta B$; each doubles $|\varepsilon|$.把磁场变化所用的时间减半（如在 $0.25$ s 内完成下降）。等效地，将 $N$ 加倍、面积 $A$ 加倍或磁场变化量 $\Delta B$ 加倍；每一种都会使 $|\varepsilon|$ 加倍。
Faraday's law sets the EMF magnitude; the rate of flux change is what matters, not the field value.法拉第定律决定电动势大小；关键是磁通量的变化率，而非磁场的数值。 $|\varepsilon| = N|\Delta\Phi/\Delta t|$ shows the EMF depends on four levers: the number of turns $N$, the area $A$, the size of the field change $\Delta B$, and the time $\Delta t$. A huge but steady field induces nothing; only a changing flux drives an EMF. The negative sign in $\varepsilon = -N\,d\Phi/dt$ (Lenz's law) gives direction, not magnitude, so we drop it when computing the size. Once the EMF is known, the coil behaves as a battery of EMF $\varepsilon$ in a circuit of resistance $R$, so Ohm's law gives the current. Under US NGSS this analysis is the honors extension; HS-PS2-5 itself is assessed only qualitatively.$|\varepsilon| = N|\Delta\Phi/\Delta t|$ 表明电动势取决于四个杠杆：匝数 $N$、面积 $A$、磁场变化量 $\Delta B$ 与时间 $\Delta t$。巨大但恒定的磁场什么也感应不出；只有变化的磁通量才驱动电动势。$\varepsilon = -N\,d\Phi/dt$（楞次定律）中的负号给出方向而非大小，故求大小时舍去。求出电动势后，线圈在电阻为 $R$ 的电路中相当于一个电动势为 $\varepsilon$ 的电池，由欧姆定律得电流。在 US NGSS 中，这一分析属荣誉级拓展；HS-PS2-5 本身仅作定性考查。
Q8HARD难Honors荣誉级 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Lenz's law楞次定律 · Physics 12 [8 marks][8 分]

North pole of a magnet pushed toward a coil; flux through the coil increases. (a) Lenz's law + induced-current direction. (b) Pole presented + attract/repel. (c) Energy-conservation argument. (d) Aluminium plate swung through a magnet gap.磁铁北极推向线圈；穿过线圈的磁通量增大。(a) 楞次定律 + 感应电流方向。(b) 呈现的磁极 + 吸引/排斥。(c) 能量守恒论证。(d) 铝板摆过磁铁间隙。

Answer:答案：  (a) counterclockwise (seen from the magnet)逆时针（从磁铁侧看）  ·  (b) north pole; repelled北极；被排斥  ·  (c) work in $=$ electrical energy out输入的功 $=$ 输出的电能  ·  (d) eddy-current (magnetic) braking涡流（磁）制动

(a) State Lenz's law and find the current direction陈述楞次定律并求电流方向 M1·A1

Lenz's law: the induced current flows so as to oppose the change in flux that produces it. Here the flux is increasing (north pole approaching), so the induced current must create a field that opposes the incoming field, i.e. pointing back toward the magnet. By the right-hand rule, the current is counterclockwise as seen from the magnet's side.楞次定律：感应电流的流向使其阻碍引起它的磁通量变化。此处磁通量增大（北极靠近），故感应电流必须产生一个与来场相反的磁场，即指回磁铁方向。由右手定则，电流为从磁铁侧看的逆时针方向。

(b) Pole presented and resulting interaction呈现的磁极及相互作用 A1·A1

The counterclockwise current makes the coil's near face a north pole. North faces north, so the coil repels the approaching magnet.逆时针电流使线圈靠近磁铁的一面成为北极。北极对北极，故线圈排斥靠近的磁铁。

(c) Why energy conservation requires this为何能量守恒要求如此 M1·A1

Because the coil repels the magnet, you must do positive work to keep pushing it in. That mechanical work is exactly what supplies the electrical energy dissipated in the coil. If the induced current instead attracted the magnet, the magnet would accelerate on its own, generating ever more current and energy from nothing , a perpetual-motion machine. Opposition is the only outcome consistent with conservation of energy.由于线圈排斥磁铁，你必须做正功才能持续把它推入。这份机械功正好提供了在线圈中耗散的电能。若感应电流反而吸引磁铁，磁铁就会自行加速，凭空产生越来越多的电流和能量 , 即永动机。阻碍是唯一与能量守恒相符的结果。

(d) Name and explain the plate effect命名并解释铝板效应 A1·A1

This is eddy-current (magnetic) braking. As the aluminium plate moves through the field, the changing flux through it induces circulating eddy currents. By Lenz's law these currents oppose the plate's motion, producing a retarding force that slows it , even though aluminium is not ferromagnetic.这是涡流（磁）制动。当铝板穿过磁场时，穿过它的磁通量变化感应出环流涡流。由楞次定律，这些电流阻碍铝板的运动，产生使其减速的阻力 , 即使铝并非铁磁性材料。
Lenz's law is the directional half of induction, and it is energy conservation in disguise.楞次定律是电磁感应中决定方向的那一半，本质上就是伪装的能量守恒。 Faraday's law gives the size of an induced EMF; Lenz's law (the minus sign) gives its direction: the induced effect always opposes the change that caused it. The mechanical consequence is a force that resists relative motion, so an external agent must do work, and that work becomes the electrical (and ultimately thermal) energy in the circuit. Eddy-current braking is the same physics in a solid conductor instead of a wire loop: contactless, wear-free, and used in roller coasters, train brakes, and MRI patient tables. If Lenz's law had the opposite sign, energy would be created from nothing, which nature forbids.法拉第定律给出感应电动势的大小；楞次定律（负号）给出方向：感应效应总是阻碍引起它的变化。其力学后果是一个阻碍相对运动的力，因此外界必须做功，而这份功转化为电路中的电能（最终为热能）。涡流制动是同一物理在固体导体（而非导线回路）中的体现：无接触、无磨损，用于过山车、列车制动和 MRI 检查台。若楞次定律取相反的符号，能量将凭空产生，而这是自然界所禁止的。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 + §7 AC generator交流发电机 · HS-PS2-5 (quantitative)（定量） [8 marks][8 分]

AC generator: $N = 100$ turns, $A = 0.020\ \text{m}^2$, $B = 0.25$ T, $f = 60$ Hz; $\varepsilon_0 = NBA\omega$, $\omega = 2\pi f$. (a) $\omega$. (b) Peak EMF. (c) Why EMF zero at flux-max, max at flux-zero. (d) How to double the peak EMF.交流发电机：$N = 100$ 匝，$A = 0.020\ \text{m}^2$，$B = 0.25$ T，$f = 60$ Hz；$\varepsilon_0 = NBA\omega$，$\omega = 2\pi f$。(a) $\omega$。(b) 峰值电动势。(c) 为何磁通量最大时电动势为零、磁通量为零时电动势最大。(d) 如何使峰值电动势加倍。

Answer:答案：  (a) $\omega \approx 377\ \text{rad/s}$  ·  (b) $\varepsilon_0 \approx 188\ \text{V}$  ·  (c) EMF tracks the rate of flux change, not the flux电动势随磁通量变化率，而非磁通量本身  ·  (d) double $f$ (or $N$, $B$, or $A$)将 $f$ 加倍（或 $N$、$B$、$A$）

(a) Angular frequency $\omega = 2\pi f$角频率 $\omega = 2\pi f$ M1·A1

$$ \omega \;=\; 2\pi f \;=\; 2\pi (60) \;=\; 120\pi \;\approx\; 377\ \text{rad/s.} $$

(b) Peak EMF $\varepsilon_0 = NBA\omega$峰值电动势 $\varepsilon_0 = NBA\omega$ M1·A1·A1

$$ \varepsilon_0 \;=\; NBA\omega \;=\; (100)(0.25)(0.020)(377) \;=\; (0.50)(377) \;\approx\; 188\ \text{V.} $$

(c) Why the EMF and flux are a quarter-cycle out of phase为何电动势与磁通量相差四分之一周期 A1·A1

Faraday's law makes the EMF proportional to the rate of change of flux, $\varepsilon = -N\,d\Phi/dt$, not to the flux itself. When the coil plane is perpendicular to $\vec B$, the flux $\Phi = BA$ is at its maximum, but it is momentarily not changing ($d\Phi/dt = 0$), so the EMF is zero. A quarter-turn later the plane is parallel to $\vec B$, the flux is zero but sweeping through its fastest rate of change, so the EMF is maximum. The flux varies as $\cos(\omega t)$ and the EMF as $\sin(\omega t)$.法拉第定律使电动势正比于磁通量的变化率，$\varepsilon = -N\,d\Phi/dt$，而非磁通量本身。当线圈平面垂直于 $\vec B$ 时，磁通量 $\Phi = BA$ 最大，但此刻瞬时不变（$d\Phi/dt = 0$），故电动势为零。再转四分之一圈，平面平行于 $\vec B$，磁通量为零但正以最快速率变化，故电动势最大。磁通量按 $\cos(\omega t)$ 变化，电动势按 $\sin(\omega t)$ 变化。

(d) One change that doubles the peak EMF使峰值电动势加倍的一种改动 A1

Since $\varepsilon_0 = NBA\omega$, doubling the rotation frequency $f$ (hence $\omega$) doubles the peak EMF. Doubling $N$, $B$, or $A$ would do the same.由于 $\varepsilon_0 = NBA\omega$，将旋转频率 $f$（从而 $\omega$）加倍即可使峰值电动势加倍。将 $N$、$B$ 或 $A$ 加倍也能达到同样效果。
A generator is Faraday's law turned into a machine: rotating a coil makes the flux vary sinusoidally, inducing an AC EMF.发电机就是把法拉第定律变成机器：旋转线圈使磁通量按正弦变化，感应出交流电动势。 Spinning a coil in a steady field continuously changes the angle $\theta$ in $\Phi = BA\cos\theta$, so the flux oscillates and Faraday's law produces an alternating EMF with peak value $NBA\omega$. The output is largest not when the flux is largest, but when the flux is changing fastest, which is the conceptual heart of the question. North American grids run at $f = 60$ Hz ($\omega \approx 377$ rad/s); Europe and China use $50$ Hz. The four levers $N$, $B$, $A$, $\omega$ are exactly the design knobs an engineer turns to set a generator's voltage. Run the same machine backward, feeding current in, and it becomes a motor.在恒定磁场中旋转线圈会持续改变 $\Phi = BA\cos\theta$ 中的角度 $\theta$，故磁通量振荡，法拉第定律产生峰值为 $NBA\omega$ 的交变电动势。输出最大并非在磁通量最大时，而是在磁通量变化最快时，这正是本题的概念核心。北美电网频率为 $f = 60$ Hz（$\omega \approx 377$ rad/s）；欧洲与中国用 $50$ Hz。四个杠杆 $N$、$B$、$A$、$\omega$ 正是工程师设定发电机电压时调节的旋钮。把同一台机器反向运行、输入电流，它便成为电动机。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 30 marks阿省毕业考 + 通用题型 · 共 30 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 Solenoid / electromagnet螺线管 / 电磁铁 · 30-B3.8k [9 marks][9 分]

Solenoid: $N = 500$ turns over $L = 0.50$ m, $I = 4.0$ A. (a) Turns per metre. (b) Interior field. (c) Factor change when $I\times3$, $n\times2$. (d) Electromagnet advantage.螺线管：$N = 500$ 匝，$L = 0.50$ m，$I = 4.0$ A。(a) 每米匝数。(b) 内部磁场。(c) $I$ 增至三倍、$n$ 增至两倍时的变化倍数。(d) 电磁铁优点。

Answer:答案：  (a) $n = 1000\ \text{turns/m}$  ·  (b) $B \approx 5.0\times10^{-3}\ \text{T}$  ·  (c) $\times 6$  ·  (d) field switches on/off and is adjustable磁场可开关、可调节

(a) Turns per metre, $n = N/L$每米匝数 $n = N/L$ M1·A1

$$ n \;=\; \frac{N}{L} \;=\; \frac{500}{0.50} \;=\; 1000\ \text{turns/m.} $$

(b) Interior field, $B = \mu_0 n I$内部磁场 $B = \mu_0 n I$ M1·A1·A1

$$ B \;=\; \mu_0 n I \;=\; (4\pi\times10^{-7})(1000)(4.0) \;=\; 16\pi\times10^{-4} \;\approx\; 5.0\times10^{-3}\ \text{T.} $$

(c) Factor change when current is tripled and turns/m doubled电流增至三倍、每米匝数增至两倍时的倍数 M1·A1

Since $B = \mu_0 n I$ is linear in both $n$ and $I$, the new field is $3 \times 2 = 6$ times the original. ($B$ becomes about $3.0\times10^{-2}$ T.)由于 $B = \mu_0 n I$ 对 $n$ 与 $I$ 均为线性，新磁场是原来的 $3 \times 2 = 6$ 倍。（$B$ 约变为 $3.0\times10^{-2}$ T。）

(d) Advantage of an electromagnet电磁铁的优点 A1·A1

An electromagnet can be switched on and off (current on $=$ field on, current off $=$ field off) and its strength can be adjusted by changing the current. A permanent magnet is always on at fixed strength. This is why scrapyard cranes use electromagnets: drop the load by cutting the current.电磁铁可以开关（通电有场、断电无场），且可通过改变电流调节强度。永磁体始终开启且强度固定。这就是废料场起重机使用电磁铁的原因：切断电流即可放下负载。
Inside a long solenoid the field is nearly uniform and set by $B = \mu_0 n I$ , turns density times current.在长螺线管内部，磁场近乎均匀，由 $B = \mu_0 n I$ 决定 , 匝密度乘电流。 A solenoid stacks the circular fields of many loops so that they add inside and cancel outside, producing a uniform interior field that depends only on the turns per metre $n = N/L$ and the current $I$, not on the wire's diameter or the solenoid's radius. Because $B$ is linear in both $n$ and $I$, scaling factors simply multiply , here $3\times2 = 6$. The controllable, switchable nature of this field is what makes electromagnets indispensable in motors, relays, MRI machines, and lifting cranes, where a permanent magnet's fixed field would be useless.螺线管把许多线圈的圆形磁场叠加起来，使它们在内部相加、在外部相消，产生只取决于每米匝数 $n = N/L$ 与电流 $I$ 的均匀内部磁场，而与导线直径或螺线管半径无关。由于 $B$ 对 $n$ 与 $I$ 均为线性，缩放因子直接相乘 , 此处为 $3\times2 = 6$。这种可控、可开关的磁场正是电磁铁在电动机、继电器、MRI 机器和起重机中不可或缺的原因，而永磁体的固定磁场在这些场合毫无用处。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §7 Transformer变压器 · 30-B3.9k [9 marks][9 分]

Transformer: $N_p = 300$, $V_p = 120$ V, $N_s = 1500$, $I_p = 8.0$ A, ideal. (a) Secondary voltage + type. (b) Secondary current. (c) Power check. (d) Why AC, not DC.变压器：$N_p = 300$，$V_p = 120$ V，$N_s = 1500$，$I_p = 8.0$ A，理想。(a) 次级电压 + 类型。(b) 次级电流。(c) 功率核验。(d) 为何用交流而非直流。

Answer:答案：  (a) $V_s = 600\ \text{V}$ (step-up)（升压）  ·  (b) $I_s = 1.6\ \text{A}$  ·  (c) $P_p = P_s = 960\ \text{W}$  ·  (d) only AC gives changing flux只有交流才有变化磁通量

(a) Secondary voltage, $V_s = V_p (N_s/N_p)$次级电压 $V_s = V_p (N_s/N_p)$ M1·A1·A1

$$ V_s \;=\; V_p \frac{N_s}{N_p} \;=\; 120 \times \frac{1500}{300} \;=\; 120 \times 5 \;=\; 600\ \text{V.} $$ Since $N_s > N_p$, this is a step-up transformer.由于 $N_s > N_p$，这是一台升压变压器。

(b) Secondary current, $I_s = I_p (N_p/N_s)$次级电流 $I_s = I_p (N_p/N_s)$ M1·A1

$$ I_s \;=\; I_p \frac{N_p}{N_s} \;=\; 8.0 \times \frac{300}{1500} \;=\; 8.0 \times 0.20 \;=\; 1.6\ \text{A.} $$

(c) Power conservation check功率守恒核验 M1·A1

$$ P_p \;=\; V_p I_p \;=\; 120 \times 8.0 \;=\; 960\ \text{W}; \qquad P_s \;=\; V_s I_s \;=\; 600 \times 1.6 \;=\; 960\ \text{W.} $$ Equal, as required for an ideal transformer.相等，符合理想变压器的要求。

(d) Why transformers need AC变压器为何需要交流 A1·A1

A transformer works only when the flux through the secondary is changing, because Faraday's law induces an EMF from $d\Phi/dt$. An alternating current produces a continuously changing flux. A steady direct current makes a constant flux ($d\Phi/dt = 0$), so no EMF is induced in the secondary and no power is transferred.变压器只有在穿过次级的磁通量变化时才工作，因为法拉第定律由 $d\Phi/dt$ 感应电动势。交流电产生持续变化的磁通量。恒定的直流电产生恒定磁通量（$d\Phi/dt = 0$），故次级不产生电动势，也不传输功率。
An ideal transformer trades voltage for current at constant power: $V_s/V_p = N_s/N_p$ and $V_p I_p = V_s I_s$.理想变压器在功率不变的前提下以电压换电流：$V_s/V_p = N_s/N_p$ 且 $V_p I_p = V_s I_s$。 Stepping the voltage up by a factor of 5 steps the current down by the same factor, keeping $P = VI$ fixed. This is the linchpin of the power grid: generators step up to hundreds of kilovolts for long-distance transmission (tiny current means tiny $I^2R$ losses), then step down near homes. The deep reason a transformer cannot run on DC is the same Faraday's law behind every question in this set: no change in flux, no induced EMF. Always verify with the power check; a mismatch between $V_p I_p$ and $V_s I_s$ flags an arithmetic slip.电压升高 5 倍，电流就降低同样的倍数，保持 $P = VI$ 不变。这是电网的关键：发电机升压至数十万伏进行远距离输电（电流极小意味着 $I^2R$ 损耗极小），再在住宅附近降压。变压器无法用直流运行的根本原因，与本套题每道题背后的法拉第定律相同：磁通量不变，则无感应电动势。务必用功率核验：$V_p I_p$ 与 $V_s I_s$ 不相等即提示计算有误。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 + §7 Motor force & grid transmission电动机受力与电网输电 · HS-PS2-5 [12 marks][12 分]

Motor coil side $L = 0.20$ m, $I = 4.0$ A, perpendicular to $B = 0.50$ T. Same plant sends $P = 120$ kW down a line of $R = 0.20\ \Omega$. (a) Force on the side. (b) How the torque arises. (c) Line current and loss at $240$ V then $4800$ V. (d) Loss factor + transformer role.电动机线圈边 $L = 0.20$ m，$I = 4.0$ A，垂直于 $B = 0.50$ T。同一发电厂沿 $R = 0.20\ \Omega$ 的输电线输送 $P = 120$ kW。(a) 该边受力。(b) 力矩如何产生。(c) $240$ V 与 $4800$ V 时的线路电流与损耗。(d) 损耗倍数 + 变压器作用。

Answer:答案：  (a) $F = 0.40\ \text{N}$  ·  (c) $500\ \text{A}, 50\ \text{kW}$ vs对比 $25\ \text{A}, 125\ \text{W}$  ·  (d) losses drop by $\times 400$损耗降为 $1/400$

(a) Force on the coil side, $F = BIL$线圈边受力 $F = BIL$ M1·A1

$$ F \;=\; BIL \;=\; (0.50)(4.0)(0.20) \;=\; 0.40\ \text{N.} $$

(b) How the torque arises力矩如何产生 A1·A1

Current flows in opposite directions along the two opposite sides of the coil, so by the right-hand rule the magnetic forces on them point in opposite directions. This pair of equal and opposite forces, acting on opposite sides of the rotation axis, forms a couple that produces a turning effect (torque), spinning the coil. A commutator reverses the current every half-turn so the torque keeps driving rotation the same way.电流沿线圈两条对边的方向相反，故由右手定则，作用于它们的磁力方向相反。这对大小相等、方向相反、作用在转轴两侧的力构成力偶，产生转动效应（力矩），使线圈旋转。换向器每半圈反转一次电流，使力矩始终沿同一方向驱动旋转。

(c) Line current and power loss at each voltage各电压下的线路电流与损耗功率 M1·A1·A1·A1·A1

At $240$ V:在 $240$ V 时： $$ I \;=\; \frac{P}{V} \;=\; \frac{120{,}000}{240} \;=\; 500\ \text{A}, \qquad P_{\text{loss}} \;=\; I^2 R \;=\; (500)^2(0.20) \;=\; 50{,}000\ \text{W} \;=\; 50\ \text{kW.} $$ At $4800$ V:在 $4800$ V 时： $$ I \;=\; \frac{120{,}000}{4800} \;=\; 25\ \text{A}, \qquad P_{\text{loss}} \;=\; (25)^2(0.20) \;=\; 125\ \text{W.} $$

(d) Loss-reduction factor and the transformer's role损耗下降倍数与变压器的作用 M1·A1·A1

Losses drop from $50{,}000$ W to $125$ W, a factor of $50{,}000/125 = 400$ , the square of the $20\times$ voltage increase. A transformer makes high-voltage transmission practical: a step-up transformer raises the generator voltage for the long line (cutting $I$ and so $I^2R$), and a step-down transformer lowers it again to safe levels near the consumer, all while conserving power.损耗从 $50{,}000$ W 降至 $125$ W，倍数为 $50{,}000/125 = 400$ , 即电压提高 $20$ 倍的平方。变压器使高压输电切实可行：升压变压器为长输电线升高发电机电压（减小 $I$ 从而减小 $I^2R$），降压变压器在靠近用户处再将其降至安全水平，整个过程功率守恒。
Two faces of magnetism in one grid: $F = BIL$ spins the motor, and high-voltage transmission ($P_{\text{loss}} = I^2R$) delivers the power that drives it.电网中磁学的两副面孔：$F = BIL$ 驱动电动机，高压输电（$P_{\text{loss}} = I^2R$）输送驱动它的电力。 The same force law that turns a motor coil is what makes generators, loudspeakers, and meters work. On the supply side, because line loss scales as $I^2$, transmitting at high voltage (hence low current) is the single most effective way to cut losses: a $20\times$ voltage step gives a $400\times$ loss reduction. Note the $50$ kW loss at $240$ V is nearly half the transmitted power, which is exactly why low-voltage long-distance transmission is infeasible and why the whole grid runs on transformers and AC. This problem ties the unit together: force on currents, induction, and the transformer all serve the electrical grid.驱动电动机线圈的同一个力定律，也使发电机、扬声器和电表得以工作。在供电侧，由于线路损耗按 $I^2$ 变化，高压（从而低电流）输电是削减损耗最有效的单一手段：电压提高 $20$ 倍，损耗降低 $400$ 倍。注意 $240$ V 时 $50$ kW 的损耗几乎是输送功率的一半，这正是低压远距离输电不可行、整个电网依赖变压器和交流电的原因。本题把整个单元串联起来：电流受力、电磁感应与变压器都服务于电网。