📚 A-Level Physics Jun 18 Markscheme 2 Formula Derivation | A-Level 物理 Jun 18 评分方案 2 公式推导
In the June 2018 A-Level Physics Paper 2, one question challenged students to derive the familiar transformer voltage relation from first principles. The relevant markscheme provided a clear step-by-step justification that relied on Faraday’s law and the concept of magnetic flux linkage. This article reconstructs that derivation, explains each assumption, and offers practical examples to help you master this essential skill for the exam.
在2018年6月的A-Level物理试卷2中，有一道题目要求学生从基本原理推导出我们所熟悉的变压器电压关系。相应的评分方案给出了一套清晰的逐步论证，其依据是法拉第定律和磁通链的概念。本文将重现这一推导过程，解释每一个假设条件，并提供实际例题，帮助你掌握这项考试必备的技能。
1. Context of the Question | 题目背景
The Paper 2 question presented a simple ideal transformer: a primary coil of Np turns and a secondary coil of Ns turns wound on a closed iron core. An alternating current in the primary produced a changing magnetic flux in the core. Students were asked to show that the ratio of secondary voltage Vs to primary voltage Vp equals the ratio of turns Ns/Np.
试卷2的这道题展示了一个简单的理想变压器：一个匝数为Np的原线圈和一个匝数为Ns的副线圈绕在闭合铁芯上。原线圈中的交变电流在铁芯中产生变化的磁通量。要求学生证明副边电压Vs与原边电压Vp之比等于匝数比Ns/Np。
2. Faraday’s Law of Electromagnetic Induction | 法拉第电磁感应定律
At the heart of the derivation lies Faraday’s law, which states that the magnitude of the induced emf in any circuit is equal to the rate of change of magnetic flux linkage through that circuit. Mathematically, for a coil of N turns, this is expressed as:
推导的核心是法拉第定律，该定律指出：任何回路中感应电动势的大小等于穿过该回路的磁通链的变化率。对于匝数为N的线圈，其数学表达式为：
ε = −N (ΔΦ / Δt)
Here ΔΦ/Δt is the rate of change of magnetic flux through one turn, N is the number of turns, and the negative sign indicates the direction of the induced emf (Lenz’s law). In the transformer derivation, we often work with magnitudes and reintroduce phase separately.
其中ΔΦ/Δt是穿过单匝线圈的磁通量变化率，N是线圈匝数，负号表示感应电动势的方向（楞次定律）。在变压器推导中，我们通常先考虑大小，而将相位关系单独处理。
3. Understanding Magnetic Flux Linkage | 理解磁通链
Magnetic flux linkage is simply the product of the number of turns N and the magnetic flux Φ passing through each turn. If all turns experience the same flux Φ, the total flux linkage is NΦ. In the transformer core, a time-varying flux Φ(t) threads both coils, giving each a flux linkage that changes with time.
磁通链就是线圈匝数N与穿过每匝的磁通量Φ的乘积。如果所有匝都经历相同的磁通量Φ，则总磁通链为NΦ。在变压器铁芯中，随时间变化的磁通量Φ(t)同时穿过两个线圈，使每个线圈的磁通链都随时间变化。
4. Applying Faraday’s Law to Each Coil | 对每个线圈应用法拉第定律
For the primary coil, the induced emf (which opposes the applied voltage) has magnitude |εp| = Np (ΔΦ/Δt). In an ideal transformer with zero winding resistance, the terminal voltage Vp equals this induced emf in magnitude, so we can write:
对于原线圈，感应电动势（与外加电压方向相反）的大小为 |εp| = Np (ΔΦ/Δt)。在绕组电阻为零的理想变压器中，端电压Vp的大小就等于这个感应电动势，因此我们可以写出：
Vp = Np (ΔΦ / Δt)
Similarly, for the secondary coil, the induced emf appears as the terminal voltage Vs, giving:
类似地，对于副线圈，感应电动势表现为端电压Vs，因此有：
Vs = Ns (ΔΦ / Δt)
At this stage we are using magnitudes, so the negative sign is omitted for clarity. The markscheme accepts this approach provided the candidate states the assumption of no voltage drop across winding resistance.
此时我们使用大小，因此为清晰起见省略了负号。只要考生说明假设绕组电阻上无压降，评分方案即认可这种处理。
5. Equality of Flux Change in an Ideal Core | 理想铁芯中磁通变化量相等
A crucial feature of the ideal transformer is that the entire magnetic flux is confined within the iron core. There is no leakage flux; every field line that passes through one turn of the primary also passes through one turn of the secondary. Consequently, the same flux Φ threads each turn of both coils, and the rate of change ΔΦ/Δt is identical for the primary and the secondary.
理想变压器的一个关键特征是，所有磁通都局限于铁芯内部，没有漏磁；每条穿过原线圈一匝的磁力线同样穿过副线圈一匝。因此，同一个磁通Φ同时穿过两个线圈的每一匝，原边和副边的磁通变化率ΔΦ/Δt完全相同。
This equality allows us to equate the flux change terms in the two voltage equations. It is the single most important step in the derivation.
正是这个相等关系使我们能够将两个电压方程中的磁通变化项等同起来。这是整个推导中最重要的一步。
6. Deriving the Voltage Ratio | 推导电压比
From the two equations Vp = Np (ΔΦ/Δt) and Vs = Ns (ΔΦ/Δt), we can rearrange each to express the common factor:
由两个方程 Vp = Np (ΔΦ/Δt) 和 Vs = Ns (ΔΦ/Δt)，我们可以分别整理，表示出共同因子：
ΔΦ/Δt = Vp / Np = Vs / Ns
Cross-multiplying yields the standard transformer equation:
交叉相乘即得到标准的变压器公式：
Vs / Vp = Ns / Np
This is precisely the relationship that examiners expected in the Jun 18 markscheme. It shows that if the secondary has more turns than the primary, the voltage is stepped up, and vice versa.
这正是2018年6月评分方案中期望得到的关系式。它表明，如果副线圈匝数多于原线圈，电压就会升高；反之则降低。
7. The Negative Sign and Phase | 负号与相位
If we retain the negative sign from Faraday’s law, the equations become Vp = −Np (ΔΦ/Δt) and Vs = −Ns (ΔΦ/Δt). The ratio Vs/Vp remains Ns/Np in magnitude, but the negative signs indicate that Vs and Vp are 180° out of phase in an ideal transformer under resistive load. Many markschemes award credit for acknowledging this phase relationship, even if the derivation focuses on magnitudes.
如果保留法拉第定律的负号，方程就变为 Vp = −Np (ΔΦ/Δt) 和 Vs = −Ns (ΔΦ/Δt)。大小上 Vs/Vp 仍为 Ns/Np，但负号表明在理想变压器带纯电阻负载时，Vs 与 Vp 相位相差180°。即使推导侧重于大小，许多评分方案仍会为承认这一相位关系而给分。
8. Assumptions and Their Implications | 假设条件及其影响
The derivation rests on three main assumptions: (i) the core has infinite permeability, so all flux is confined and there is no leakage; (ii) the windings have zero electrical resistance; and (iii) the core does not saturate, and eddy current and hysteresis losses are negligible. In real transformers, these assumptions break down to some extent, causing the actual voltage ratio to deviate slightly from the ideal turns ratio.
推导建立在三个主要假设之上：(i) 铁芯磁导率无穷大，因此所有磁通都被约束在铁芯内，没有漏磁；(ii) 绕组电阻为零；(iii) 铁芯不饱和，且涡流和磁滞损耗可以忽略。在真实变压器中，这些假设在一定程度上不成立，导致实际电压比与理想匝数比略有偏差。
The Jun 18 markscheme specifically credited candidates who identified these assumptions and discussed their effect on the validity of the derived formula.
2018年6月的评分方案明确对能指出这些假设，并讨论其对推导公式有效性影响的考生给予加分。
9. Worked Example from a Typical Paper 2 Question | 典型试卷2例题
A transformer has 400 turns on the primary coil and 80 turns on the secondary. The primary is connected to a 230 V AC supply. Calculate the secondary voltage, assuming the transformer is ideal.
一台变压器原线圈400匝，副线圈80匝，原边接230 V交流电源。假设变压器为理想变压器，计算副边电压。
Using the formula Vs/Vp = Ns/Np:
应用公式 Vs/Vp = Ns/Np：
Vs = Vp × (Ns/Np) = 230 V × (80 / 400) = 46 V
This straightforward calculation mirrors the kind of numerical follow-up often seen in Paper 2 after a derivation part.
这个简单的计算再现了试卷2中在推导部分之后常出现的数值计算题型。
10. Common Mistakes and Marking Points | 常见错误与得分点
When answering such derivation questions, students frequently lose marks by omitting the justification that ΔΦ/Δt is the same for both coils. Some also mix up the turns ratio, writing Np/Ns instead of Ns/Np. Another common error is attempting to apply the formula to direct current: since DC produces no changing flux after the initial transient, the transformer equation does not apply in steady-state DC conditions.
在回答这类推导题时，学生常因遗漏“ΔΦ/Δt对两个线圈相同”的论证而失分。有些学生还会混淆匝数比，写成 Np/Ns 而非 Ns/Np。另一个常见错误是试图将该公式用于直流电：直流电在初始瞬态过后不会产生变化的磁通，因此变压器公式不适用于稳态直流情况。
Examiners’ reports from the Jun 18 series highlighted that clearly stating the equality of flux change rate and referencing Faraday’s law were essential for full credit.
2018年6月系列考试的考官报告强调，清晰地说明磁通变化率相等并引用法拉第定律是获得满分的必要条件。
11. Summary and Revision Tips | 总结与复习建议
The transformer voltage equation follows logically from Faraday’s law and the shared flux in an ideal core. To secure all marks in a derivation question, you should: state Faraday’s law, express the induced emf for each coil, assert that ΔΦ/Δt is the same, and then equate V/N terms to obtain the ratio. Practice writing the derivation in your own words, and always note the underlying assumptions.
变压器电压公式是从法拉第定律和理想铁芯中的共享磁通合乎逻辑地得出的。要在推导题中稳拿全部分数，你应该：写出法拉第定律，分别表示两个线圈的感应电动势，明确ΔΦ/Δt相等，然后将V/N项联立求得比值。练习用自己的话写出推导过程，并始终注明背后的假设。
When revising, refer back to the Jun 18 markscheme structure: the examiner rewards a clear logical sequence, not just the final formula. Use past papers to spot similar derivation tasks, and you will build confidence for Paper 2.
复习时，应回顾2018年6月评分方案的结构：考官奖励
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