Unit D.4: Induction单元 D.4：电磁感应 HL only

The data-booklet formula $\Phi = BA\cos\theta$ defines $\theta$ as the angle between $\vec{B}$ and the area normal $\hat{n}$. If a question instead gives $\alpha$, the angle between $\vec{B}$ and the plane of the loop, then $\theta = 90^{\circ} - \alpha$ and

数据手册公式 $\Phi = BA\cos\theta$ 中 $\theta$ 定义为 $\vec{B}$ 与面法线 $\hat{n}$ 的夹角。若题目给的是 $\vec{B}$ 与回路平面的夹角 $\alpha$，则 $\theta = 90^{\circ} - \alpha$，于是

A field lying in the plane ($\alpha = 0$) gives zero flux; a field perpendicular to the plane ($\alpha = 90^{\circ}$) gives maximum flux. Reading this wrong is the single most common D.4 error: it flips $\sin$ and $\cos$ and silently corrupts every later step.

场在平面内（$\alpha = 0$）磁通量为零；场垂直于平面（$\alpha = 90^{\circ}$）磁通量最大。读错这一点是 D.4 最常见的错误：它会把 $\sin$ 与 $\cos$ 互换，并悄悄污染后续每一步。

Place the rod on rails separated by $L$, sweeping out area as it moves. In time $\Delta t$ the rod travels $\Delta x = v\,\Delta t$, so the circuit area changes by $\Delta A = L\,\Delta x = Lv\,\Delta t$. With $B$ perpendicular to the plane, the flux change is

把棒放在间距为 $L$ 的导轨上，运动时扫出面积。在时间 $\Delta t$ 内棒移动 $\Delta x = v\,\Delta t$，故回路面积变化 $\Delta A = L\,\Delta x = Lv\,\Delta t$。当 $B$ 垂直于平面时，磁通量变化为

By Faraday's law (single turn), $|\varepsilon| = |\Delta\Phi / \Delta t| = BLv$. So $\varepsilon = BvL$ is not a separate law: it is Faraday's law applied to a circuit whose area is changing at a steady rate. This is worth stating in a Paper 2 derivation; it earns the "rate of change of flux" reasoning mark.

由法拉第定律（单匝），$|\varepsilon| = |\Delta\Phi / \Delta t| = BLv$。所以 $\varepsilon = BvL$ 不是独立定律：它是法拉第定律应用于面积匀速变化的回路。Paper 2 推导中值得写出这一点，可拿到"磁通量变化率"的推理分。

For a smoothly varying flux, the average-rate form $\Delta\Phi / \Delta t$ becomes the instantaneous derivative:

对平滑变化的磁通量，平均率形式 $\Delta\Phi / \Delta t$ 变为瞬时导数：

On a flux-linkage-vs-time graph, the emf at any instant is the (negative of the) tangent gradient. A linear ramp of flux gives a constant emf; a flat region (constant flux) gives zero emf; a sinusoidal flux gives a cosinusoidal emf, $90^{\circ}$ ahead in phase. The IB does not require calculus manipulation here, but the "emf is the slope" reading is exactly what data-response graph questions test.

在磁链-时间图上，任一时刻的电动势为切线斜率（取负）。磁通量线性上升给出恒定电动势；平直段（磁通量不变）给出零电动势；正弦磁通量给出余弦电动势，相位超前 $90^{\circ}$。IB 不要求在此做微积分，但"电动势即斜率"的读法正是数据题图像所考。

Suppose, for contradiction, that the induced current flowed the "wrong" way, reinforcing the flux that created it. Then pushing the magnet a little toward the coil would increase the flux, drive a current that pulls the magnet in further, increasing the flux still more. The magnet would accelerate without any external input, and the resistor would dissipate ever-growing heat. Energy would appear from nothing.

反证：假设感应电流朝"错误"方向流动，强化产生它的磁通量。那么把磁铁稍微推向线圈会增大磁通量，驱动的电流又把磁铁进一步吸入，磁通量更增。磁铁将在无外部输入下加速，电阻耗散越来越多的热。能量将凭空出现。

Because that is impossible, the induced current must oppose the change. The minus sign in Faraday's law is therefore not an arbitrary convention; it is the bookkeeping symbol for energy conservation. Every Lenz's-law direction argument can be cross-checked by asking: "does my answer require energy input to change the flux?" If yes, it is consistent.

既然那不可能，感应电流必然反抗变化。故法拉第定律中的负号不是随意约定，而是能量守恒的记账符号。每个楞次定律判向都可这样复核："我的答案是否要求输入能量才能改变磁通量？"若是，则一致。

With $\Phi = BA\cos(\omega t)$, the flux is maximal at $t = 0$ (coil face-on). At that instant the coil is momentarily flat against the field and the flux is, for a brief moment, barely changing, so $\varepsilon = 0$. A quarter-period later the coil plane is parallel to $\vec{B}$: the flux is zero but it is sweeping through zero at its fastest rate, so $\varepsilon$ is maximal.

取 $\Phi = BA\cos(\omega t)$，磁通量在 $t = 0$（线圈正对）最大。此刻线圈瞬间正对场，磁通量在极短时间内几乎不变，故 $\varepsilon = 0$。再过四分之一周期，线圈平面平行于 $\vec{B}$：磁通量为零但正以最快速率穿过零，故 $\varepsilon$ 最大。

Formally, $\varepsilon = -N\,d\Phi/dt = -NBA \cdot (-\omega \sin\omega t) = NBA\omega\sin(\omega t)$. The derivative of a cosine is a (negative) sine, which is the cosine shifted by $90^{\circ}$. This phase relationship is what graph-sketching questions in Paper 2 reward: plot $\Phi$ and $\varepsilon$ on the same axes, with $\varepsilon$ peaking exactly where $\Phi$ crosses zero.

形式上，$\varepsilon = -N\,d\Phi/dt = -NBA \cdot (-\omega \sin\omega t) = NBA\omega\sin(\omega t)$。余弦的导数是（负）正弦，即余弦平移 $90^{\circ}$。这一相位关系正是 Paper 2 图像题所考：在同一坐标上画 $\Phi$ 与 $\varepsilon$，$\varepsilon$ 的峰值恰在 $\Phi$ 过零处。

Power lost as heat in a transmission cable of resistance $R$ carrying current $I$ is $P_{\text{loss}} = I^{2}R$. To deliver a fixed power $P = VI$, raising the transmission voltage $V$ lowers the current $I = P/V$, and because the loss goes as $I^{2}$, halving the current quarters the loss.

电阻为 $R$、电流为 $I$ 的输电线以热损失的功率为 $P_{\text{loss}} = I^{2}R$。要输送固定功率 $P = VI$，提高输电电压 $V$ 会降低电流 $I = P/V$；由于损耗按 $I^{2}$ 变化，电流减半则损耗降为四分之一。

This is exactly why transformers are indispensable: a step-up transformer at the power station raises the voltage to hundreds of kilovolts for long-distance transmission at low current, and step-down transformers near the consumer return it to a safe $230\ \mathrm{V}$. The whole high-voltage grid is built on Faraday's law, and it works only with AC, which is the historical reason mains electricity is alternating.

这正是变压器不可或缺的原因：发电站的升压变压器把电压升到数百千伏，以低电流远距离输送；用户附近的降压变压器再把它降回安全的 $230\ \mathrm{V}$。整个高压电网建立在法拉第定律之上，且只对交流有效——这正是市电采用交流的历史原因。