Topic 5 电磁学 / Electricity & Magnetism
静电场 · 电路 · 磁场 · 电磁感应
IB SL / HLTopic 5电磁学是物理学的基础
电磁学是物理学的核心分支之一，研究电荷之间的相互作用及其效应。本章从静电场（Coulomb's Law, 电场强度）出发，学习电路（Ohm's Law, Kirchhoff's Laws, 内阻）的分析方法，进而探讨磁场对电荷和电流的作用力（Lorentz Force），最后学习电磁感应（Faraday's Law, Lenz's Law）——这是发电机和变压器的基础。电磁学构成了现代电气工程和电子技术的基础。
Electromagnetism is a core branch of physics studying interactions between electric charges. This chapter begins with electric fields (Coulomb's Law, field strength), then covers electric circuits (Ohm's Law, Kirchhoff's Laws, internal resistance), explores magnetic forces on charges and currents (Lorentz Force), and finally electromagnetic induction (Faraday's Law, Lenz's Law) — the basis for generators and transformers. Electromagnetism forms the foundation of modern electrical engineering and electronics.
5.1 静电场 / Electric Fields
IB 5.1核心基础库仑定律 + 电场强度
电荷与库仑定律 / Charge and Coulomb's Law
$$F = k\frac{q_1 q_2}{r^2} \qquad k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9\ \text{N·m}^2/\text{C}^2$$
库仑定律（Coulomb's Law）描述两个点电荷之间的静电力：力的大小与电荷量的乘积成正比，与距离的平方成反比。同号电荷相斥，异号电荷相吸。$k$ 为库仑常数，$\epsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N·m}^2$ 为真空介电常数。
Coulomb's Law describes the electrostatic force between two point charges: proportional to the product of charges and inversely proportional to the square of the distance. Like charges repel, opposite charges attract. $k$ is Coulomb's constant, $\epsilon_0$ is the permittivity of free space.
🎯 IB 考试要点：库仑定律与万有引力定律形式相似（都是 $1/r^2$ 形式），但静电力可以吸引或排斥，而引力只有吸引。计算时电荷量用绝对值，方向通过"同斥异吸"判断。
🎯 IB Exam Tip: Coulomb's Law and Newton's Law of Gravitation have the same $1/r^2$ form, but electric forces can be attractive or repulsive while gravity is only attractive. Use absolute values for magnitude; determine direction by like-repel/opposite-attract.
电场强度 / Electric Field Strength
$$E = \frac{F}{q} \qquad E = k\frac{Q}{r^2} \quad \text{（点电荷）}$$
电场强度（Electric Field Strength）$E$ 定义为放置在电场中某点的试探电荷所受的静电力 $F$ 与该电荷量 $q$ 的比值。电场是矢量，方向与正电荷在该点所受力的方向相同。点电荷产生的电场强度 $E = kQ/r^2$，方向为：正电荷向外辐射，负电荷向内汇聚。
Electric field strength $E$ is defined as the force per unit positive charge at a point: $E = F/q$. The electric field is a vector — direction is the direction of force on a positive test charge. For a point charge: $E = kQ/r^2$, radially outward for positive $Q$, inward for negative $Q$.
📐 电场叠加原理 / Superposition of Electric Fields
电场叠加原理（Principle of Superposition）：多个点电荷产生的合电场强度等于各电荷单独产生的电场强度的矢量之和。
设空间中有 $n$ 个点电荷 $Q_1, Q_2, \ldots, Q_n$，在某点 $P$ 处的合电场强度为：
$$\vec{E} = \vec{E}_1 + \vec{E}_2 + \ldots + \vec{E}_n = \sum_{i=1}^n k\frac{Q_i}{r_i^2}\hat{r}_i$$
其中 $\hat{r}_i$ 是从 $Q_i$ 指向 $P$ 点的单位矢量。
解题步骤：
① 确定每个电荷在 $P$ 点产生的电场的方向
② 计算每个电场的大小 $E_i = k|Q_i|/r_i^2$
③ 将各电场矢量分解到坐标轴上
④ 分别求和得到 $E_x$ 和 $E_y$
⑤ 合电场 $E = \sqrt{E_x^2 + E_y^2}$，方向 $\theta = \arctan(E_y/E_x)$
设空间中有 $n$ 个点电荷 $Q_1, Q_2, \ldots, Q_n$，在某点 $P$ 处的合电场强度为：
$$\vec{E} = \vec{E}_1 + \vec{E}_2 + \ldots + \vec{E}_n = \sum_{i=1}^n k\frac{Q_i}{r_i^2}\hat{r}_i$$
其中 $\hat{r}_i$ 是从 $Q_i$ 指向 $P$ 点的单位矢量。
解题步骤：
① 确定每个电荷在 $P$ 点产生的电场的方向
② 计算每个电场的大小 $E_i = k|Q_i|/r_i^2$
③ 将各电场矢量分解到坐标轴上
④ 分别求和得到 $E_x$ 和 $E_y$
⑤ 合电场 $E = \sqrt{E_x^2 + E_y^2}$，方向 $\theta = \arctan(E_y/E_x)$
Superposition Principle: The net electric field at a point is the vector sum of fields from each individual charge.
For $n$ point charges $Q_1, Q_2, \ldots, Q_n$ at point $P$:
$$\vec{E} = \vec{E}_1 + \vec{E}_2 + \ldots + \vec{E}_n = \sum_{i=1}^n k\frac{Q_i}{r_i^2}\hat{r}_i$$
Steps:
① Determine direction of each field at $P$
② Compute magnitude $E_i = k|Q_i|/r_i^2$
③ Decompose into components
④ Sum $E_x$ and $E_y$
⑤ $E = \sqrt{E_x^2 + E_y^2}$, direction $\theta = \arctan(E_y/E_x)$
For $n$ point charges $Q_1, Q_2, \ldots, Q_n$ at point $P$:
$$\vec{E} = \vec{E}_1 + \vec{E}_2 + \ldots + \vec{E}_n = \sum_{i=1}^n k\frac{Q_i}{r_i^2}\hat{r}_i$$
Steps:
① Determine direction of each field at $P$
② Compute magnitude $E_i = k|Q_i|/r_i^2$
③ Decompose into components
④ Sum $E_x$ and $E_y$
⑤ $E = \sqrt{E_x^2 + E_y^2}$, direction $\theta = \arctan(E_y/E_x)$
电场线 / Electric Field Lines
电场线（electric field lines / lines of force）用图形表示电场的方向和强弱：
• 电场线上每一点的切线方向与该点的电场方向相同
• 电场线的疏密表示电场强度的大小——线越密，场越强
• 电场线从正电荷出发，终止于负电荷
• 电场线不相交
• 匀强电场（uniform electric field）的电场线是平行等间距的直线
典型电场线图：孤立正电荷（向外辐射）、孤立负电荷（向内汇聚）、等量异号电荷（偶极子 dipole）、平行板电容器（匀强电场）。
• 电场线上每一点的切线方向与该点的电场方向相同
• 电场线的疏密表示电场强度的大小——线越密，场越强
• 电场线从正电荷出发，终止于负电荷
• 电场线不相交
• 匀强电场（uniform electric field）的电场线是平行等间距的直线
典型电场线图：孤立正电荷（向外辐射）、孤立负电荷（向内汇聚）、等量异号电荷（偶极子 dipole）、平行板电容器（匀强电场）。
Electric field lines visually represent the field:
• Tangent at any point gives field direction
• Density of lines indicates field strength
• Lines start at positive charges, end at negative charges
• Lines never cross
• A uniform field has parallel, equally spaced lines
Typical patterns: isolated positive charge (radially outward), isolated negative charge (radially inward), electric dipole, parallel plates (uniform field).
• Tangent at any point gives field direction
• Density of lines indicates field strength
• Lines start at positive charges, end at negative charges
• Lines never cross
• A uniform field has parallel, equally spaced lines
Typical patterns: isolated positive charge (radially outward), isolated negative charge (radially inward), electric dipole, parallel plates (uniform field).
电场中的电势能 / Electric Potential Energy
$$E_p = k\frac{q_1 q_2}{r} \qquad V = \frac{E_p}{q} \qquad V = k\frac{Q}{r} \quad \text{（点电荷电势）}$$
电势能（Electric Potential Energy）：两个点电荷 $q_1$ 和 $q_2$ 相距 $r$ 时的电势能 $E_p = kq_1q_2/r$。同号电荷 $E_p > 0$（排斥，势能为正），异号电荷 $E_p < 0$（吸引，势能为负）。
电势（Electric Potential）$V$ 是单位正电荷在某点的电势能：$V = E_p/q$。点电荷产生的电势 $V = kQ/r$（标量）。电势差（voltage / potential difference）$V_{AB} = V_A - V_B$。
电势（Electric Potential）$V$ 是单位正电荷在某点的电势能：$V = E_p/q$。点电荷产生的电势 $V = kQ/r$（标量）。电势差（voltage / potential difference）$V_{AB} = V_A - V_B$。
Electric Potential Energy for two point charges: $E_p = kq_1q_2/r$. Like charges: $E_p > 0$ (repulsive, positive). Unlike charges: $E_p < 0$ (attractive, negative).
Electric Potential $V$ is potential energy per unit charge: $V = E_p/q$. For a point charge: $V = kQ/r$ (scalar). Potential difference (voltage) $V_{AB} = V_A - V_B$.
Electric Potential $V$ is potential energy per unit charge: $V = E_p/q$. For a point charge: $V = kQ/r$ (scalar). Potential difference (voltage) $V_{AB} = V_A - V_B$.
电荷在电场中的运动 / Charge Motion in Electric Fields
$$F = qE \qquad a = \frac{qE}{m} \qquad W = qV = \frac12 mv^2$$
电荷在匀强电场中受到恒力 $F = qE$，做匀加速运动。如果电荷从静止通过电势差 $V$ 加速：$qV = \frac12 mv^2$，所以 $v = \sqrt{2qV/m}$。这是粒子加速器的基本原理。
电荷垂直射入匀强电场时做抛物线运动（类似抛体运动）：水平方向匀速，竖直方向匀加速。
电荷垂直射入匀强电场时做抛物线运动（类似抛体运动）：水平方向匀速，竖直方向匀加速。
A charge in a uniform field experiences a constant force $F = qE$, undergoing uniform acceleration. If accelerated from rest through potential difference $V$: $qV = \frac12 mv^2$, giving $v = \sqrt{2qV/m}$. This is the principle behind particle accelerators.
A charge entering a uniform field perpendicularly follows a parabolic path (similar to projectile motion).
A charge entering a uniform field perpendicularly follows a parabolic path (similar to projectile motion).
🎯 IB 考试要点：① 电场强度 $E$ 是矢量（有方向），电势 $V$ 是标量；② 匀强电场中 $E = V/d$（$d$ 为平行板间距）；③ 电场力做功与路径无关，只与始末位置的电势差有关；④ 点电荷的电势 $V = kQ/r$ 是标量求和（注意正负号），而电场是矢量求和。
🎯 IB Exam Tips: ① $E$ is a vector (direction matters), $V$ is a scalar; ② In a uniform field $E = V/d$ (parallel plates); ③ Work done by electric field is path-independent — depends only on potential difference; ④ $V = kQ/r$ is a scalar sum (include signs), while $E$ is a vector sum.
例 5.1 / Example 5.1
两个点电荷 $q_1 = +2.0\ \mu\text{C}$ 和 $q_2 = -3.0\ \mu\text{C}$ 相距 0.10 m。求它们之间的静电力大小和方向。
Two charges $q_1 = +2.0\ \mu\text{C}$ and $q_2 = -3.0\ \mu\text{C}$ are 0.10 m apart. Find the electrostatic force magnitude and direction.
解题过程 / Solution
$F = k|q_1 q_2|/r^2 = (8.99 \times 10^9) \times (2.0 \times 10^{-6}) \times (3.0 \times 10^{-6}) / (0.10)^2$
$F = 8.99 \times 10^9 \times 6.0 \times 10^{-12} / 0.01$
$F = 8.99 \times 10^9 \times 6.0 \times 10^{-10} = 5.39$ N
因为 $q_1$ 为正、$q_2$ 为负（异号），所以是吸引力。$q_1$ 受到 $q_2$ 的吸引力指向 $q_2$，$q_2$ 受到 $q_1$ 的吸引力指向 $q_1$。
$F = 8.99 \times 10^9 \times 6.0 \times 10^{-12} / 0.01$
$F = 8.99 \times 10^9 \times 6.0 \times 10^{-10} = 5.39$ N
因为 $q_1$ 为正、$q_2$ 为负（异号），所以是吸引力。$q_1$ 受到 $q_2$ 的吸引力指向 $q_2$，$q_2$ 受到 $q_1$ 的吸引力指向 $q_1$。
$F = k|q_1 q_2|/r^2 = 8.99 \times 10^9 \times 2.0 \times 10^{-6} \times 3.0 \times 10^{-6} / (0.10)^2 = 5.39$ N
Since charges are opposite, the force is attractive — each charge pulls the other toward it.
Since charges are opposite, the force is attractive — each charge pulls the other toward it.
例 5.2 / Example 5.2
两块平行板间距 0.020 m，电势差 500 V。求：(a) 板间电场强度；(b) 一个电子从负极板静止释放到达正极板时的速度。电子质量 $m_e = 9.11 \times 10^{-31}$ kg，电荷量 $e = 1.60 \times 10^{-19}$ C。
Parallel plates: $d = 0.020$ m, $V = 500$ V. Find: (a) $E$ between plates; (b) speed of an electron accelerated from rest across the gap.
解题过程 / Solution
(a) $E = V/d = 500 / 0.020 = 25000$ V/m = $2.5 \times 10^4$ N/C
(b) 能量守恒：$qV = \frac12 mv^2$
$v = \sqrt{2qV/m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 500 / 9.11 \times 10^{-31}}$
$v = \sqrt{1.60 \times 10^{-16} / 9.11 \times 10^{-31}} = \sqrt{1.756 \times 10^{14}} = 1.33 \times 10^7$ m/s
电子最终速度约为 $1.33 \times 10^7$ m/s（约光速的 4.4%）。
(b) 能量守恒：$qV = \frac12 mv^2$
$v = \sqrt{2qV/m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 500 / 9.11 \times 10^{-31}}$
$v = \sqrt{1.60 \times 10^{-16} / 9.11 \times 10^{-31}} = \sqrt{1.756 \times 10^{14}} = 1.33 \times 10^7$ m/s
电子最终速度约为 $1.33 \times 10^7$ m/s（约光速的 4.4%）。
(a) $E = V/d = 500 / 0.020 = 2.5 \times 10^4$ N/C
(b) Conservation of energy: $qV = \frac12 mv^2$
$v = \sqrt{2qV/m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 500 / 9.11 \times 10^{-31}} = 1.33 \times 10^7$ m/s
The electron reaches about 4.4% of the speed of light.
(b) Conservation of energy: $qV = \frac12 mv^2$
$v = \sqrt{2qV/m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 500 / 9.11 \times 10^{-31}} = 1.33 \times 10^7$ m/s
The electron reaches about 4.4% of the speed of light.
例 5.3 / Example 5.3
两个点电荷 $q_1 = +4.0\ \mu\text{C}$ 和 $q_2 = +1.0\ \mu\text{C}$ 相距 0.30 m。求连线上电场为零的位置。
Two charges $q_1 = +4.0\ \mu\text{C}$ and $q_2 = +1.0\ \mu\text{C}$, 0.30 m apart. Find the point on the line where $E = 0$.
解题过程 / Solution
设 $P$ 点距离 $q_1$ 为 $x$，则距离 $q_2$ 为 $0.30 - x$。在 $P$ 点合电场为零：
$E_1 = E_2$
$k(4.0 \times 10^{-6})/x^2 = k(1.0 \times 10^{-6})/(0.30 - x)^2$
$4/x^2 = 1/(0.30 - x)^2$
取平方根（正负号）：$2/x = 1/(0.30 - x)$
$2(0.30 - x) = x$
$0.60 - 2x = x$
$0.60 = 3x$
$x = 0.20$ m
所以电场为零的点在距离 $q_1$ 为 0.20 m（距离 $q_2$ 为 0.10 m）处。因为两个电荷都是正的，该点在它们之间的连线上。
$E_1 = E_2$
$k(4.0 \times 10^{-6})/x^2 = k(1.0 \times 10^{-6})/(0.30 - x)^2$
$4/x^2 = 1/(0.30 - x)^2$
取平方根（正负号）：$2/x = 1/(0.30 - x)$
$2(0.30 - x) = x$
$0.60 - 2x = x$
$0.60 = 3x$
$x = 0.20$ m
所以电场为零的点在距离 $q_1$ 为 0.20 m（距离 $q_2$ 为 0.10 m）处。因为两个电荷都是正的，该点在它们之间的连线上。
Let $P$ be $x$ from $q_1$, then $0.30 - x$ from $q_2$. At $P$, $E_1 = E_2$:
$kq_1/x^2 = kq_2/(0.30 - x)^2$
$4/x^2 = 1/(0.30 - x)^2$
$2/x = 1/(0.30 - x)$
$2(0.30 - x) = x$ → $x = 0.20$ m
Point is 0.20 m from $q_1$, 0.10 m from $q_2$.
$kq_1/x^2 = kq_2/(0.30 - x)^2$
$4/x^2 = 1/(0.30 - x)^2$
$2/x = 1/(0.30 - x)$
$2(0.30 - x) = x$ → $x = 0.20$ m
Point is 0.20 m from $q_1$, 0.10 m from $q_2$.
5.2 电路 / Electric Circuits
IB 5.2SL & HL 重点欧姆定律 + 基尔霍夫 + 内阻
电流与欧姆定律 / Current and Ohm's Law
$$I = \frac{\Delta q}{\Delta t} \qquad V = IR \qquad R = \frac{\rho L}{A}$$
电流（electric current）$I$ 是单位时间内通过导体横截面的电荷量，单位安培（A）。电流方向定义为正电荷移动的方向（与电子移动方向相反）。
欧姆定律（Ohm's Law）：导体两端的电压 $V$ 与通过导体的电流 $I$ 成正比，比例常数是电阻 $R$。$V = IR$。
电阻（resistance）$R$ 取决于导体的材料、长度和横截面积：$R = \rho L/A$，其中 $\rho$ 是电阻率（resistivity）。
欧姆定律（Ohm's Law）：导体两端的电压 $V$ 与通过导体的电流 $I$ 成正比，比例常数是电阻 $R$。$V = IR$。
电阻（resistance）$R$ 取决于导体的材料、长度和横截面积：$R = \rho L/A$，其中 $\rho$ 是电阻率（resistivity）。
Electric current $I$ is the rate of flow of charge: $I = \Delta q/\Delta t$, measured in amperes (A). Conventional current flows from positive to negative (opposite to electron flow).
Ohm's Law: Voltage across a conductor is proportional to the current through it: $V = IR$.
Resistance depends on material, length, and cross-sectional area: $R = \rho L/A$, where $\rho$ is resistivity.
Ohm's Law: Voltage across a conductor is proportional to the current through it: $V = IR$.
Resistance depends on material, length, and cross-sectional area: $R = \rho L/A$, where $\rho$ is resistivity.
电阻的串并联 / Resistors in Series and Parallel
$$R_s = R_1 + R_2 + R_3 + \ldots \qquad \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$$
串联（series）：总电阻等于各电阻之和。串联电路中电流处处相等，电压按电阻大小分配。
并联（parallel）：总电阻的倒数等于各电阻倒数之和。并联电路中各支路两端电压相等，电流按电阻大小反比分配。
并联（parallel）：总电阻的倒数等于各电阻倒数之和。并联电路中各支路两端电压相等，电流按电阻大小反比分配。
Series: total resistance = sum of individual resistances. Current is the same through all resistors; voltage divides proportionally to resistance.
Parallel: reciprocal of total resistance = sum of reciprocals. Voltage is the same across each branch; current divides inversely to resistance.
Parallel: reciprocal of total resistance = sum of reciprocals. Voltage is the same across each branch; current divides inversely to resistance.
📐 串并联电阻公式的推导 / Derivation of Series & Parallel Resistance
串联推导：
串联电路中，电流 $I$ 处处相等。总电压等于各电阻两端电压之和：
$V_{\text{总}} = V_1 + V_2 + V_3 + \ldots$
由欧姆定律 $V = IR$：
$IR_s = IR_1 + IR_2 + IR_3 + \ldots$
两边除以 $I$：
$R_s = R_1 + R_2 + R_3 + \ldots$
并联推导：
并联电路中，各支路两端电压相等。总电流等于各支路电流之和：
$I_{\text{总}} = I_1 + I_2 + I_3 + \ldots$
由欧姆定律 $I = V/R$：
$V/R_p = V/R_1 + V/R_2 + V/R_3 + \ldots$
两边除以 $V$：
$1/R_p = 1/R_1 + 1/R_2 + 1/R_3 + \ldots$
关键区别：
• 串联：电流相同，电压相加 → 电阻相加
• 并联：电压相同，电流相加 → 电导（$1/R$）相加
串联电路中，电流 $I$ 处处相等。总电压等于各电阻两端电压之和：
$V_{\text{总}} = V_1 + V_2 + V_3 + \ldots$
由欧姆定律 $V = IR$：
$IR_s = IR_1 + IR_2 + IR_3 + \ldots$
两边除以 $I$：
$R_s = R_1 + R_2 + R_3 + \ldots$
并联推导：
并联电路中，各支路两端电压相等。总电流等于各支路电流之和：
$I_{\text{总}} = I_1 + I_2 + I_3 + \ldots$
由欧姆定律 $I = V/R$：
$V/R_p = V/R_1 + V/R_2 + V/R_3 + \ldots$
两边除以 $V$：
$1/R_p = 1/R_1 + 1/R_2 + 1/R_3 + \ldots$
关键区别：
• 串联：电流相同，电压相加 → 电阻相加
• 并联：电压相同，电流相加 → 电导（$1/R$）相加
Series derivation:
Current $I$ is the same everywhere. Total voltage = sum of individual voltages:
$V_{\text{total}} = V_1 + V_2 + V_3 + \ldots$
$IR_s = IR_1 + IR_2 + IR_3 + \ldots$
Dividing by $I$: $R_s = R_1 + R_2 + R_3 + \ldots$
Parallel derivation:
Voltage $V$ is the same across each branch. Total current = sum of branch currents:
$I_{\text{total}} = I_1 + I_2 + I_3 + \ldots$
$V/R_p = V/R_1 + V/R_2 + V/R_3 + \ldots$
Dividing by $V$: $1/R_p = 1/R_1 + 1/R_2 + 1/R_3 + \ldots$
Key difference: Series = same current, add voltages → add resistances. Parallel = same voltage, add currents → add conductances.
Current $I$ is the same everywhere. Total voltage = sum of individual voltages:
$V_{\text{total}} = V_1 + V_2 + V_3 + \ldots$
$IR_s = IR_1 + IR_2 + IR_3 + \ldots$
Dividing by $I$: $R_s = R_1 + R_2 + R_3 + \ldots$
Parallel derivation:
Voltage $V$ is the same across each branch. Total current = sum of branch currents:
$I_{\text{total}} = I_1 + I_2 + I_3 + \ldots$
$V/R_p = V/R_1 + V/R_2 + V/R_3 + \ldots$
Dividing by $V$: $1/R_p = 1/R_1 + 1/R_2 + 1/R_3 + \ldots$
Key difference: Series = same current, add voltages → add resistances. Parallel = same voltage, add currents → add conductances.
基尔霍夫定律 / Kirchhoff's Laws
$$\sum I_{\text{进}} = \sum I_{\text{出}} \quad \text{（节点电流定律 KCL）} \qquad \sum V = 0 \quad \text{（回路电压定律 KVL）}$$
基尔霍夫电流定律（Kirchhoff's Current Law, KCL）：流入电路中任意节点的总电流等于流出该节点的总电流——电荷守恒。
基尔霍夫电压定律（Kirchhoff's Voltage Law, KVL）：沿任意闭合回路，各元件上的电势降的代数和为零——能量守恒。
基尔霍夫电压定律（Kirchhoff's Voltage Law, KVL）：沿任意闭合回路，各元件上的电势降的代数和为零——能量守恒。
Kirchhoff's Current Law (KCL): The sum of currents entering any junction equals the sum of currents leaving it — conservation of charge.
Kirchhoff's Voltage Law (KVL): The sum of potential differences around any closed loop is zero — conservation of energy.
Kirchhoff's Voltage Law (KVL): The sum of potential differences around any closed loop is zero — conservation of energy.
📐 从守恒定律推导基尔霍夫定律 / Deriving Kirchhoff's Laws from Conservation
KCL — 电荷守恒推导：
电荷既不能创造也不能消灭。在一个电路节点处，电荷不能积累（直流稳态条件下）。因此单位时间内流入节点的总电荷等于流出节点的总电荷：
$I_{\text{进}} \Delta t = I_{\text{出}} \Delta t$
除以 $\Delta t$ 即得：$\sum I_{\text{进}} = \sum I_{\text{出}}$
KVL — 能量守恒推导：
静电力是保守力，电场力移动电荷做的功与路径无关。将单位正电荷沿闭合回路移动一周，电场力做的总功为零。
沿闭合回路一周，电势差的代数和必须为零，否则就意味着电荷获得或损失了净能量——这违反能量守恒。
数学表达：$\oint \vec{E} \cdot d\vec{l} = 0$，即静电场的环量为零。在电路中等价于：
$\sum V = 0$（沿任何闭合回路）
解题应用：
1. 标出各支路电流方向（可假设）
2. 对每个节点列出 KCL 方程
3. 选择独立的回路，应用 KVL
4. 解方程组
电荷既不能创造也不能消灭。在一个电路节点处，电荷不能积累（直流稳态条件下）。因此单位时间内流入节点的总电荷等于流出节点的总电荷：
$I_{\text{进}} \Delta t = I_{\text{出}} \Delta t$
除以 $\Delta t$ 即得：$\sum I_{\text{进}} = \sum I_{\text{出}}$
KVL — 能量守恒推导：
静电力是保守力，电场力移动电荷做的功与路径无关。将单位正电荷沿闭合回路移动一周，电场力做的总功为零。
沿闭合回路一周，电势差的代数和必须为零，否则就意味着电荷获得或损失了净能量——这违反能量守恒。
数学表达：$\oint \vec{E} \cdot d\vec{l} = 0$，即静电场的环量为零。在电路中等价于：
$\sum V = 0$（沿任何闭合回路）
解题应用：
1. 标出各支路电流方向（可假设）
2. 对每个节点列出 KCL 方程
3. 选择独立的回路，应用 KVL
4. 解方程组
KCL — from conservation of charge:
Charge cannot be created or destroyed. At a junction, charge cannot accumulate (DC steady state). So charge entering per unit time = charge leaving per unit time: $\sum I_{\text{in}} = \sum I_{\text{out}}$.
KVL — from conservation of energy:
The electrostatic force is conservative. Moving a unit charge around any closed loop does zero net work. The sum of potential differences around any closed loop must be zero — otherwise charge would gain or lose net energy, violating energy conservation.
Mathematically: $\oint \vec{E} \cdot d\vec{l} = 0$ — the circulation of the electrostatic field is zero.
Charge cannot be created or destroyed. At a junction, charge cannot accumulate (DC steady state). So charge entering per unit time = charge leaving per unit time: $\sum I_{\text{in}} = \sum I_{\text{out}}$.
KVL — from conservation of energy:
The electrostatic force is conservative. Moving a unit charge around any closed loop does zero net work. The sum of potential differences around any closed loop must be zero — otherwise charge would gain or lose net energy, violating energy conservation.
Mathematically: $\oint \vec{E} \cdot d\vec{l} = 0$ — the circulation of the electrostatic field is zero.
内阻 / Internal Resistance
$$\mathcal{E} = I(R + r) \qquad V = \mathcal{E} - Ir$$
所有实际电源都有内阻（internal resistance）$r$。电源的电动势（electromotive force, emf）$\mathcal{E}$ 是电源开路时的端电压。当电源输出电流 $I$ 时，电源内部的电阻 $r$ 引起电压降 $Ir$，所以路端电压（terminal voltage）$V = \mathcal{E} - Ir$。
当外电阻 $R \gg r$ 时，$V \approx \mathcal{E}$（近似理想电源）。当外电阻 $R \ll r$ 时，大部分电压降在内阻上，电源效率很低。
当外电阻 $R \gg r$ 时，$V \approx \mathcal{E}$（近似理想电源）。当外电阻 $R \ll r$ 时，大部分电压降在内阻上，电源效率很低。
All real power sources have internal resistance $r$. The electromotive force (emf) $\mathcal{E}$ is the open-circuit terminal voltage. When current $I$ flows, internal resistance causes a voltage drop $Ir$, so terminal voltage $V = \mathcal{E} - Ir$.
When $R \gg r$, $V \approx \mathcal{E}$ (near-ideal). When $R \ll r$, most voltage drops across the internal resistance — very inefficient.
When $R \gg r$, $V \approx \mathcal{E}$ (near-ideal). When $R \ll r$, most voltage drops across the internal resistance — very inefficient.
📐 内阻与最大功率传输 / Internal Resistance & Maximum Power Transfer
推导电源的输出功率：
电源的输出功率（传递给外电阻 $R$ 的功率）：
$P = I^2 R = \left(\frac{\mathcal{E}}{R + r}\right)^2 R$
为求 $R$ 为何值时 $P$ 最大，对 $R$ 求导并令导数为零：
$\frac{dP}{dR} = \mathcal{E}^2 \frac{(R+r)^2 - 2R(R+r)}{(R+r)^4} = 0$
$(R+r)^2 - 2R(R+r) = 0$
$(R+r)(R+r - 2R) = (R+r)(r - R) = 0$
所以 $R = r$
最大功率传输定理：当外电阻等于电源内阻时，电源输出功率最大。
最大功率：$P_{\max} = \frac{\mathcal{E}^2}{4r}$
此时效率只有 50%（一半功率消耗在内阻上）。实际电路中往往根据需求权衡效率和功率。
电源的输出功率（传递给外电阻 $R$ 的功率）：
$P = I^2 R = \left(\frac{\mathcal{E}}{R + r}\right)^2 R$
为求 $R$ 为何值时 $P$ 最大，对 $R$ 求导并令导数为零：
$\frac{dP}{dR} = \mathcal{E}^2 \frac{(R+r)^2 - 2R(R+r)}{(R+r)^4} = 0$
$(R+r)^2 - 2R(R+r) = 0$
$(R+r)(R+r - 2R) = (R+r)(r - R) = 0$
所以 $R = r$
最大功率传输定理：当外电阻等于电源内阻时，电源输出功率最大。
最大功率：$P_{\max} = \frac{\mathcal{E}^2}{4r}$
此时效率只有 50%（一半功率消耗在内阻上）。实际电路中往往根据需求权衡效率和功率。
Derivation of output power:
Power delivered to load $R$: $P = I^2 R = \left(\frac{\mathcal{E}}{R + r}\right)^2 R$
To maximize $P$, differentiate w.r.t. $R$ and set to zero:
$\frac{dP}{dR} = 0$ → $R = r$
Maximum Power Transfer Theorem: Maximum power is delivered when the load resistance equals the internal resistance.
Maximum power: $P_{\max} = \frac{\mathcal{E}^2}{4r}$
Efficiency at this point is only 50%.
Power delivered to load $R$: $P = I^2 R = \left(\frac{\mathcal{E}}{R + r}\right)^2 R$
To maximize $P$, differentiate w.r.t. $R$ and set to zero:
$\frac{dP}{dR} = 0$ → $R = r$
Maximum Power Transfer Theorem: Maximum power is delivered when the load resistance equals the internal resistance.
Maximum power: $P_{\max} = \frac{\mathcal{E}^2}{4r}$
Efficiency at this point is only 50%.
分压器 / Potential Divider
$$V_{\text{out}} = V_{\text{in}} \frac{R_2}{R_1 + R_2} \quad \text{（分压公式）}$$
分压器（potential divider / voltage divider）由两个串联的电阻组成，将输入电压 $V_{\text{in}}$ 按电阻比例分配。$R_2$ 两端的输出电压 $V_{\text{out}} = V_{\text{in}} \times R_2/(R_1+R_2)$。分压器广泛应用于传感器电路（如光敏电阻 LDR、热敏电阻 thermistor 与固定电阻串联构成分压）。
A potential divider uses two series resistors to divide input voltage proportionally. Output voltage across $R_2$: $V_{\text{out}} = V_{\text{in}} \times R_2/(R_1+R_2)$. Widely used in sensor circuits (LDR, thermistor with fixed resistor).
电功率 / Electric Power
$$P = IV = I^2 R = \frac{V^2}{R}$$
电功率（electric power）：电流通过电阻时，电能转化为内能（焦耳热）。$P = IV$ 适用于所有电路元件，$P = I^2 R = V^2/R$ 仅适用于纯电阻（焦耳定律 Joule's Law）。
Electric power: Current passing through a resistor converts electrical energy to heat (Joule heating). $P = IV$ applies to any circuit element; $P = I^2 R = V^2/R$ applies to purely resistive loads (Joule's Law).
🎯 IB 考试要点：① 基尔霍夫定律是分析复杂电路的基本工具——KCL 用于节点，KVL 用于回路；② 内阻是常考题型，注意区分 $\mathcal{E}$（电动势）和 $V$（路端电压）；③ 分压器公式要熟记，常与传感器结合出题；④ 最大功率传输定理 $R = r$ 是 HL 考点；⑤ 功率公式 $P = IV$ 是基本公式，$P = I^2 R$ 用于计算电阻发热。
🎯 IB Exam Tips: ① Kirchhoff's Laws are essential for complex circuit analysis — KCL for junctions, KVL for loops; ② Internal resistance is a common topic — distinguish $\mathcal{E}$ (emf) from $V$ (terminal voltage); ③ Memorize the potential divider formula — often combined with sensors; ④ Maximum power transfer $R = r$ is an HL topic; ⑤ $P = IV$ is fundamental, $P = I^2 R$ for resistive heating.
例 5.4 / Example 5.4
一个电池的电动势 $\mathcal{E} = 12.0$ V，内阻 $r = 0.50\ \Omega$，外接电阻 $R = 5.50\ \Omega$。求：(a) 电路电流；(b) 路端电压；(c) 电源输出功率；(d) 内阻消耗功率。
Battery: $\mathcal{E} = 12.0$ V, $r = 0.50\ \Omega$, load $R = 5.50\ \Omega$. Find: (a) current; (b) terminal voltage; (c) output power; (d) internal power loss.
解题过程 / Solution
(a) $I = \mathcal{E}/(R + r) = 12.0/(5.50 + 0.50) = 12.0/6.00 = 2.00$ A
(b) $V = \mathcal{E} - Ir = 12.0 - 2.00 \times 0.50 = 12.0 - 1.00 = 11.0$ V
也可以 $V = IR = 2.00 \times 5.50 = 11.0$ V
(c) $P_{\text{out}} = IV = 2.00 \times 11.0 = 22.0$ W
或 $P_{\text{out}} = I^2 R = 4.00 \times 5.50 = 22.0$ W
(d) $P_{\text{内}} = I^2 r = 4.00 \times 0.50 = 2.00$ W
验证：$\mathcal{E}I = 12.0 \times 2.00 = 24.0$ W = $P_{\text{out}} + P_{\text{内}} = 22.0 + 2.00 = 24.0$ W ✓
(b) $V = \mathcal{E} - Ir = 12.0 - 2.00 \times 0.50 = 12.0 - 1.00 = 11.0$ V
也可以 $V = IR = 2.00 \times 5.50 = 11.0$ V
(c) $P_{\text{out}} = IV = 2.00 \times 11.0 = 22.0$ W
或 $P_{\text{out}} = I^2 R = 4.00 \times 5.50 = 22.0$ W
(d) $P_{\text{内}} = I^2 r = 4.00 \times 0.50 = 2.00$ W
验证：$\mathcal{E}I = 12.0 \times 2.00 = 24.0$ W = $P_{\text{out}} + P_{\text{内}} = 22.0 + 2.00 = 24.0$ W ✓
(a) $I = \mathcal{E}/(R + r) = 12.0/6.00 = 2.00$ A
(b) $V = \mathcal{E} - Ir = 12.0 - 1.00 = 11.0$ V
(c) $P_{\text{out}} = IV = 2.00 \times 11.0 = 22.0$ W
(d) $P_{\text{int}} = I^2 r = 4.00 \times 0.50 = 2.00$ W
Check: $\mathcal{E}I = 24.0$ W = $P_{\text{out}} + P_{\text{int}}$ ✓
(b) $V = \mathcal{E} - Ir = 12.0 - 1.00 = 11.0$ V
(c) $P_{\text{out}} = IV = 2.00 \times 11.0 = 22.0$ W
(d) $P_{\text{int}} = I^2 r = 4.00 \times 0.50 = 2.00$ W
Check: $\mathcal{E}I = 24.0$ W = $P_{\text{out}} + P_{\text{int}}$ ✓
例 5.5 / Example 5.5
三个电阻 $R_1 = 4.0\ \Omega, R_2 = 6.0\ \Omega, R_3 = 3.0\ \Omega$，$R_1$ 和 $R_2$ 并联后再与 $R_3$ 串联，接在 12 V 电源上（内阻忽略）。求总电阻、总电流和各电阻的电流、电压。
Three resistors: $R_1 = 4.0\ \Omega$, $R_2 = 6.0\ \Omega$ in parallel, then series with $R_3 = 3.0\ \Omega$. $V = 12$ V (neglect internal resistance). Find total resistance, total current, and current/voltage for each.
解题过程 / Solution
$R_1$ 和 $R_2$ 并联：$1/R_{12} = 1/4.0 + 1/6.0 = 0.250 + 0.167 = 0.417$
$R_{12} = 1/0.417 = 2.40\ \Omega$
串联总电阻：$R_{\text{总}} = R_{12} + R_3 = 2.40 + 3.00 = 5.40\ \Omega$
总电流：$I_{\text{总}} = V/R_{\text{总}} = 12.0/5.40 = 2.22$ A
$R_3$ 电流 = $I_{\text{总}} = 2.22$ A
$V_{R3} = IR_3 = 2.22 \times 3.0 = 6.67$ V
$R_{12}$ 两端电压：$V_{12} = I_{\text{总}} \times R_{12} = 2.22 \times 2.40 = 5.33$ V
（验证：$V_{12} + V_{R3} = 5.33 + 6.67 = 12.0$ V ✓）
$R_1$ 电流：$I_1 = V_{12}/R_1 = 5.33/4.0 = 1.33$ A
$R_2$ 电流：$I_2 = V_{12}/R_2 = 5.33/6.0 = 0.889$ A
（验证：$I_1 + I_2 = 1.33 + 0.889 = 2.22$ A = $I_{\text{总}}$ ✓）
$R_{12} = 1/0.417 = 2.40\ \Omega$
串联总电阻：$R_{\text{总}} = R_{12} + R_3 = 2.40 + 3.00 = 5.40\ \Omega$
总电流：$I_{\text{总}} = V/R_{\text{总}} = 12.0/5.40 = 2.22$ A
$R_3$ 电流 = $I_{\text{总}} = 2.22$ A
$V_{R3} = IR_3 = 2.22 \times 3.0 = 6.67$ V
$R_{12}$ 两端电压：$V_{12} = I_{\text{总}} \times R_{12} = 2.22 \times 2.40 = 5.33$ V
（验证：$V_{12} + V_{R3} = 5.33 + 6.67 = 12.0$ V ✓）
$R_1$ 电流：$I_1 = V_{12}/R_1 = 5.33/4.0 = 1.33$ A
$R_2$ 电流：$I_2 = V_{12}/R_2 = 5.33/6.0 = 0.889$ A
（验证：$I_1 + I_2 = 1.33 + 0.889 = 2.22$ A = $I_{\text{总}}$ ✓）
$R_1 \parallel R_2$: $R_{12} = 1/(1/4.0 + 1/6.0) = 2.40\ \Omega$
Total: $R_{\text{total}} = 2.40 + 3.00 = 5.40\ \Omega$
$I_{\text{total}} = 12.0/5.40 = 2.22$ A
Across $R_3$: $I = 2.22$ A, $V = 6.67$ V
Across $R_{12}$: $V = 5.33$ V
$I_1 = 1.33$ A, $I_2 = 0.889$ A
Total: $R_{\text{total}} = 2.40 + 3.00 = 5.40\ \Omega$
$I_{\text{total}} = 12.0/5.40 = 2.22$ A
Across $R_3$: $I = 2.22$ A, $V = 6.67$ V
Across $R_{12}$: $V = 5.33$ V
$I_1 = 1.33$ A, $I_2 = 0.889$ A
例 5.6 / Example 5.6
用 KVL 分析：一个 6.0 V 电池（内阻 0.2 $\Omega$）连接两个并联电阻 $R_1 = 10\ \Omega$ 和 $R_2 = 15\ \Omega$。求各支路电流和路端电压。
KVL analysis: 6.0 V battery ($r = 0.2\ \Omega$), parallel resistors $R_1 = 10\ \Omega$, $R_2 = 15\ \Omega$. Find branch currents and terminal voltage.
解题过程 / Solution
先求并联等效电阻：$1/R_p = 1/10 + 1/15 = 0.100 + 0.0667 = 0.1667$
$R_p = 6.0\ \Omega$
总电阻（含内阻）：$R_{\text{总}} = R_p + r = 6.0 + 0.2 = 6.2\ \Omega$
总电流：$I_{\text{总}} = \mathcal{E}/R_{\text{总}} = 6.0/6.2 = 0.968$ A
路端电压：$V = \mathcal{E} - Ir = 6.0 - 0.968 \times 0.2 = 6.0 - 0.194 = 5.81$ V
（也等于 $V = I_{\text{总}} \times R_p = 0.968 \times 6.0 = 5.81$ V ✓）
$R_1$ 电流：$I_1 = V/R_1 = 5.81/10 = 0.581$ A
$R_2$ 电流：$I_2 = V/R_2 = 5.81/15 = 0.387$ A
验证：$I_1 + I_2 = 0.581 + 0.387 = 0.968$ A = $I_{\text{总}}$ ✓
$R_p = 6.0\ \Omega$
总电阻（含内阻）：$R_{\text{总}} = R_p + r = 6.0 + 0.2 = 6.2\ \Omega$
总电流：$I_{\text{总}} = \mathcal{E}/R_{\text{总}} = 6.0/6.2 = 0.968$ A
路端电压：$V = \mathcal{E} - Ir = 6.0 - 0.968 \times 0.2 = 6.0 - 0.194 = 5.81$ V
（也等于 $V = I_{\text{总}} \times R_p = 0.968 \times 6.0 = 5.81$ V ✓）
$R_1$ 电流：$I_1 = V/R_1 = 5.81/10 = 0.581$ A
$R_2$ 电流：$I_2 = V/R_2 = 5.81/15 = 0.387$ A
验证：$I_1 + I_2 = 0.581 + 0.387 = 0.968$ A = $I_{\text{总}}$ ✓
Parallel equivalent: $R_p = 1/(1/10 + 1/15) = 6.0\ \Omega$
Total: $R_{\text{total}} = 6.0 + 0.2 = 6.2\ \Omega$
$I_{\text{total}} = 6.0/6.2 = 0.968$ A
$V_{\text{terminal}} = \mathcal{E} - Ir = 5.81$ V
$I_1 = 0.581$ A, $I_2 = 0.387$ A
Total: $R_{\text{total}} = 6.0 + 0.2 = 6.2\ \Omega$
$I_{\text{total}} = 6.0/6.2 = 0.968$ A
$V_{\text{terminal}} = \mathcal{E} - Ir = 5.81$ V
$I_1 = 0.581$ A, $I_2 = 0.387$ A
5.3 磁场 / Magnetic Fields
IB 5.3SL & HL 核心洛伦兹力 + 安培力
磁场的基本概念 / Basic Concepts of Magnetic Fields
磁场（magnetic field）由磁体、电流或运动电荷产生。磁场用磁感应强度（magnetic flux density / magnetic field strength）$B$ 描述，单位是特斯拉（T, Tesla）。
磁场线的特点：
• 磁感线是闭合曲线——从 N 极出发进入 S 极，在磁体内部从 S 极回到 N 极
• 磁感线的疏密表示磁场的强弱
• 电流产生的磁场方向用右手螺旋定则（right-hand grip rule / right-hand screw rule）判断
磁场线的特点：
• 磁感线是闭合曲线——从 N 极出发进入 S 极，在磁体内部从 S 极回到 N 极
• 磁感线的疏密表示磁场的强弱
• 电流产生的磁场方向用右手螺旋定则（right-hand grip rule / right-hand screw rule）判断
Magnetic fields are produced by magnets, currents, and moving charges. The magnetic field (flux density) $B$ is measured in tesla (T).
Magnetic field lines:
• Form closed loops — leave N, enter S, continue inside from S to N
• Density indicates field strength
• Direction determined by the right-hand grip rule
Magnetic field lines:
• Form closed loops — leave N, enter S, continue inside from S to N
• Density indicates field strength
• Direction determined by the right-hand grip rule
运动电荷在磁场中的受力 / Magnetic Force on Moving Charge
$$F = qvB\sin\theta$$
洛伦兹力（Lorentz force）：电荷量为 $q$、速度为 $v$ 的带电粒子在磁感应强度为 $B$ 的磁场中受到的力。力的大小为 $F = qvB\sin\theta$，其中 $\theta$ 是速度 $v$ 与磁场 $B$ 之间的夹角。当 $v \perp B$（$\theta = 90^\circ$）时力最大 $F = qvB$；当 $v \parallel B$（$\theta = 0^\circ$）时力为零。
方向由右手定则（right-hand rule）或左手定则（left-hand rule / Fleming's left-hand rule）判断。注意：正电荷和负电荷的偏转方向相反。
方向由右手定则（right-hand rule）或左手定则（left-hand rule / Fleming's left-hand rule）判断。注意：正电荷和负电荷的偏转方向相反。
Lorentz force: A charged particle $q$ moving at velocity $v$ in a magnetic field $B$ experiences a force: $F = qvB\sin\theta$, where $\theta$ is the angle between $v$ and $B$. Maximum when $v \perp B$; zero when $v \parallel B$.
Direction determined by Fleming's left-hand rule. Positive and negative charges deflect in opposite directions.
Direction determined by Fleming's left-hand rule. Positive and negative charges deflect in opposite directions.
📐 洛伦兹力公式的推导 / Derivation of Lorentz Force
从安培力推导洛伦兹力：
安培力（载流导线在磁场中受力）$F = BIL\sin\theta$
电流 $I$ 与电荷运动的关系：$I = nAqv_d$，其中：
• $n$ = 单位体积的载流子数
• $A$ = 导线横截面积
• $q$ = 每个载流子的电荷量
• $v_d$ = 载流子的漂移速度
长度为 $L$ 的导线中总载流子数 $N = nAL$
安培力：$F = BIL\sin\theta = B(nAqv_d)L\sin\theta = (nAL) \times qv_d B\sin\theta = N \times qv_d B\sin\theta$
所以每个载流子受到的力：
$F_{\text{单个}} = F/N = qv_d B\sin\theta$
即为洛伦兹力 $F = qvB\sin\theta$。
完整的洛伦兹力（含电场力）：$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
安培力（载流导线在磁场中受力）$F = BIL\sin\theta$
电流 $I$ 与电荷运动的关系：$I = nAqv_d$，其中：
• $n$ = 单位体积的载流子数
• $A$ = 导线横截面积
• $q$ = 每个载流子的电荷量
• $v_d$ = 载流子的漂移速度
长度为 $L$ 的导线中总载流子数 $N = nAL$
安培力：$F = BIL\sin\theta = B(nAqv_d)L\sin\theta = (nAL) \times qv_d B\sin\theta = N \times qv_d B\sin\theta$
所以每个载流子受到的力：
$F_{\text{单个}} = F/N = qv_d B\sin\theta$
即为洛伦兹力 $F = qvB\sin\theta$。
完整的洛伦兹力（含电场力）：$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
Deriving Lorentz force from Ampere force:
Ampere force on a current-carrying wire: $F = BIL\sin\theta$
Current and charge motion: $I = nAqv_d$
• $n$ = charge carrier density
• $A$ = cross-sectional area
• $q$ = charge per carrier
• $v_d$ = drift velocity
Number of carriers in length $L$: $N = nAL$
Ampere force: $F = BIL\sin\theta = B(nAqv_d)L\sin\theta = N \times qv_d B\sin\theta$
Force per carrier: $F_{\text{single}} = F/N = qv_d B\sin\theta$
This is the Lorentz force $F = qvB\sin\theta$.
Full Lorentz force (including electric force): $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
Ampere force on a current-carrying wire: $F = BIL\sin\theta$
Current and charge motion: $I = nAqv_d$
• $n$ = charge carrier density
• $A$ = cross-sectional area
• $q$ = charge per carrier
• $v_d$ = drift velocity
Number of carriers in length $L$: $N = nAL$
Ampere force: $F = BIL\sin\theta = B(nAqv_d)L\sin\theta = N \times qv_d B\sin\theta$
Force per carrier: $F_{\text{single}} = F/N = qv_d B\sin\theta$
This is the Lorentz force $F = qvB\sin\theta$.
Full Lorentz force (including electric force): $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
载流导线在磁场中的受力 / Force on a Current-Carrying Wire
$$F = BIL\sin\theta$$
安培力（Ampere force）：长度为 $L$、电流为 $I$ 的直导线在匀强磁场 $B$ 中受到的力 $F = BIL\sin\theta$，$\theta$ 是电流方向与磁场方向的夹角。方向由左手定则（Fleming's left-hand rule）判断：左手拇指为力的方向（F）、食指为磁场方向（B）、中指为电流方向（I）。
Ampere force: A straight wire of length $L$ carrying current $I$ in a uniform field $B$ experiences force $F = BIL\sin\theta$, where $\theta$ is the angle between current and field direction. Direction is given by Fleming's left-hand rule: thumb = Force, index finger = Field, middle finger = Current.
带电粒子在匀强磁场中的运动 / Charged Particle in Uniform B Field
$$r = \frac{mv}{qB} \qquad T = \frac{2\pi m}{qB}$$
当带电粒子以速度 $v$ 垂直进入匀强磁场 $B$ 时，洛伦兹力提供向心力：$qvB = mv^2/r$，由此得回旋半径（cyclotron radius / gyroradius）$r = mv/qB$，回旋周期 $T = 2\pi m/qB$（周期与速度无关！）。
这是回旋加速器（cyclotron）和质谱仪（mass spectrometer）的基本原理。
这是回旋加速器（cyclotron）和质谱仪（mass spectrometer）的基本原理。
When a charged particle enters a uniform $B$ field perpendicularly, the Lorentz force provides centripetal force: $qvB = mv^2/r$. The cyclotron radius $r = mv/qB$ and cyclotron period $T = 2\pi m/qB$ (period is independent of speed!).
This is the principle behind cyclotrons and mass spectrometers.
This is the principle behind cyclotrons and mass spectrometers.
📐 回旋半径与周期的推导 / Derivation of Cyclotron Radius & Period
回旋半径：
粒子以速度 $v$ 垂直进入匀强磁场 $B$，洛伦兹力提供向心力：
$F = qvB = \frac{mv^2}{r}$
解出 $r$：$r = \frac{mv}{qB}$
可见半径 $r$ 与动量 $p = mv$ 成正比，与磁场 $B$ 成反比。
回旋周期：
圆周运动周期：$T = \frac{2\pi r}{v} = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$
周期与速度 $v$ 无关！这意味着在均匀磁场中，所有同种粒子（相同 $m/q$）都有相同的回旋周期，与速度无关——这是回旋加速器能够工作的关键。
回旋频率（cyclotron frequency）：$f = \frac{1}{T} = \frac{qB}{2\pi m}$
粒子以速度 $v$ 垂直进入匀强磁场 $B$，洛伦兹力提供向心力：
$F = qvB = \frac{mv^2}{r}$
解出 $r$：$r = \frac{mv}{qB}$
可见半径 $r$ 与动量 $p = mv$ 成正比，与磁场 $B$ 成反比。
回旋周期：
圆周运动周期：$T = \frac{2\pi r}{v} = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$
周期与速度 $v$ 无关！这意味着在均匀磁场中，所有同种粒子（相同 $m/q$）都有相同的回旋周期，与速度无关——这是回旋加速器能够工作的关键。
回旋频率（cyclotron frequency）：$f = \frac{1}{T} = \frac{qB}{2\pi m}$
Cyclotron radius:
$F = qvB = mv^2/r$ → $r = mv/qB$
Radius is proportional to momentum $p = mv$ and inversely proportional to $B$.
Cyclotron period:
$T = 2\pi r/v = 2\pi m/qB$
Period is independent of speed! All particles with the same $m/q$ have the same period — this is why cyclotrons work.
Cyclotron frequency: $f = 1/T = qB/2\pi m$
$F = qvB = mv^2/r$ → $r = mv/qB$
Radius is proportional to momentum $p = mv$ and inversely proportional to $B$.
Cyclotron period:
$T = 2\pi r/v = 2\pi m/qB$
Period is independent of speed! All particles with the same $m/q$ have the same period — this is why cyclotrons work.
Cyclotron frequency: $f = 1/T = qB/2\pi m$
应用：回旋加速器 / Application: Cyclotron
回旋加速器（cyclotron）利用交变电场和恒定磁场加速带电粒子。结构：两个 D 形盒（dees）置于匀强磁场中，D 形盒之间加交变电压。
工作原理：
① 粒子从中心附近注入，在磁场中做半圆周运动
② 到达缝隙时，交变电场的方向恰好改变，给粒子加速
③ 加速后粒子速度增大、回旋半径增大（周期不变）
④ 重复此过程，粒子沿螺旋轨道运动到边缘后射出
粒子最终能量：$E_k = \frac{q^2 B^2 R^2}{2m}$，其中 $R$ 是 D 形盒的最大半径。
工作原理：
① 粒子从中心附近注入，在磁场中做半圆周运动
② 到达缝隙时，交变电场的方向恰好改变，给粒子加速
③ 加速后粒子速度增大、回旋半径增大（周期不变）
④ 重复此过程，粒子沿螺旋轨道运动到边缘后射出
粒子最终能量：$E_k = \frac{q^2 B^2 R^2}{2m}$，其中 $R$ 是 D 形盒的最大半径。
A cyclotron uses alternating electric fields and a constant magnetic field to accelerate charged particles. Structure: two D-shaped hollow electrodes (dees) in a uniform magnetic field with alternating voltage between them.
Working principle:
① Particles are injected near the center and move in semicircles
② At the gap, the alternating field accelerates them
③ Speed increases → radius increases (period stays constant)
④ Particles spiral outward and exit at the edge
Final energy: $E_k = \frac{q^2 B^2 R^2}{2m}$, where $R$ is the maximum radius.
Working principle:
① Particles are injected near the center and move in semicircles
② At the gap, the alternating field accelerates them
③ Speed increases → radius increases (period stays constant)
④ Particles spiral outward and exit at the edge
Final energy: $E_k = \frac{q^2 B^2 R^2}{2m}$, where $R$ is the maximum radius.
应用：直流电动机 / Application: DC Motor
直流电动机（DC motor）将电能转化为机械能。工作原理：
• 载流线圈在磁场中受到安培力产生力矩，使线圈转动
• 线圈转动半圈后，换向器（commutator）改变电流方向，使力矩方向保持不变
• 电动机的扭矩 $\tau = NBIA\cos\theta$，其中 $N$ 为线圈匝数，$A$ 为线圈面积
• 转速由电压和负载决定
• 载流线圈在磁场中受到安培力产生力矩，使线圈转动
• 线圈转动半圈后，换向器（commutator）改变电流方向，使力矩方向保持不变
• 电动机的扭矩 $\tau = NBIA\cos\theta$，其中 $N$ 为线圈匝数，$A$ 为线圈面积
• 转速由电压和负载决定
A DC motor converts electrical energy to mechanical energy. How it works:
• Current-carrying coil in a magnetic field experiences torque
• After half a rotation, the commutator reverses current direction so torque continues in the same direction
• Torque: $\tau = NBIA\cos\theta$, where $N$ = turns, $A$ = area
• Speed depends on voltage and load
• Current-carrying coil in a magnetic field experiences torque
• After half a rotation, the commutator reverses current direction so torque continues in the same direction
• Torque: $\tau = NBIA\cos\theta$, where $N$ = turns, $A$ = area
• Speed depends on voltage and load
🎯 IB 考试要点：① 用左手定则判断安培力和洛伦兹力的方向——拇指 F，食指 B，中指 I（正电荷运动方向）；② 负电荷受力方向与正电荷相反；③ 粒子垂直进入匀强磁场做匀速圆周运动 $r = mv/qB$；④ 速度方向与磁场方向平行时不受力；⑤ 回旋加速器中周期与速度无关是理解加速器工作的关键。
🎯 IB Exam Tips: ① Fleming's left-hand rule: thumb = F, index = B, middle = I (or v for positive charge); ② Negative charges deflect opposite to positive charges; ③ Perpendicular entry into uniform B field → circular motion $r = mv/qB$; ④ Zero force when v ∥ B; ⑤ Cyclotron period independent of speed is key to understanding accelerators.
例 5.7 / Example 5.7
一个质子（$q = 1.60 \times 10^{-19}$ C, $m = 1.67 \times 10^{-27}$ kg）以 $2.0 \times 10^6$ m/s 垂直进入 $B = 0.50$ T 的匀强磁场。求：(a) 洛伦兹力大小；(b) 回旋半径；(c) 回旋周期。
A proton enters a uniform $B = 0.50$ T field perpendicularly at $v = 2.0 \times 10^6$ m/s. Find: (a) Lorentz force; (b) cyclotron radius; (c) period.
解题过程 / Solution
(a) $F = qvB = 1.60 \times 10^{-19} \times 2.0 \times 10^6 \times 0.50 = 1.60 \times 10^{-13}$ N
(b) $r = mv/qB = (1.67 \times 10^{-27} \times 2.0 \times 10^6) / (1.60 \times 10^{-19} \times 0.50)$
$r = 3.34 \times 10^{-21} / 8.0 \times 10^{-20} = 0.0418$ m = 4.18 cm
(c) $T = 2\pi m/qB = 2\pi \times 1.67 \times 10^{-27} / (1.60 \times 10^{-19} \times 0.50)$
$T = 1.049 \times 10^{-26} / 8.0 \times 10^{-20} = 1.31 \times 10^{-7}$ s = 0.131 $\mu$s
(b) $r = mv/qB = (1.67 \times 10^{-27} \times 2.0 \times 10^6) / (1.60 \times 10^{-19} \times 0.50)$
$r = 3.34 \times 10^{-21} / 8.0 \times 10^{-20} = 0.0418$ m = 4.18 cm
(c) $T = 2\pi m/qB = 2\pi \times 1.67 \times 10^{-27} / (1.60 \times 10^{-19} \times 0.50)$
$T = 1.049 \times 10^{-26} / 8.0 \times 10^{-20} = 1.31 \times 10^{-7}$ s = 0.131 $\mu$s
(a) $F = qvB = 1.60 \times 10^{-13}$ N
(b) $r = mv/qB = 0.0418$ m = 4.18 cm
(c) $T = 2\pi m/qB = 1.31 \times 10^{-7}$ s = 0.131 $\mu$s
(b) $r = mv/qB = 0.0418$ m = 4.18 cm
(c) $T = 2\pi m/qB = 1.31 \times 10^{-7}$ s = 0.131 $\mu$s
例 5.8 / Example 5.8
一根长 0.20 m 的直导线载有 5.0 A 电流，与 0.30 T 的匀强磁场成 30° 角。求导线所受安培力的大小和方向。
A 0.20 m wire carries 5.0 A at 30° to a 0.30 T uniform field. Find the Ampere force magnitude.
解题过程 / Solution
$F = BIL\sin\theta = 0.30 \times 5.0 \times 0.20 \times \sin 30^\circ$
$F = 0.30 \times 5.0 \times 0.20 \times 0.5 = 0.15$ N
方向用左手定则判断：让磁感线垂直穿入手心（B），四指指向电流方向（I），拇指所指方向即为力的方向。
$F = 0.30 \times 5.0 \times 0.20 \times 0.5 = 0.15$ N
方向用左手定则判断：让磁感线垂直穿入手心（B），四指指向电流方向（I），拇指所指方向即为力的方向。
$F = BIL\sin\theta = 0.30 \times 5.0 \times 0.20 \times \sin 30^\circ = 0.15$ N
Use Fleming's left-hand rule to determine direction.
Use Fleming's left-hand rule to determine direction.
5.4 电磁感应 / Electromagnetic Induction
IB 5.4SL & HL 核心法拉第定律 + 楞次定律 + 变压器
磁通量 / Magnetic Flux
$$\Phi = BA\cos\theta$$
磁通量（magnetic flux）$\Phi$ 穿过面积为 $A$ 的平面的磁感线总数，$\Phi = BA\cos\theta$，其中 $\theta$ 是平面法线与磁场 $B$ 的夹角。磁通量的单位是韦伯（Wb, Weber），$1\ \text{Wb} = 1\ \text{T·m}^2$。
当平面垂直于磁场（$\theta = 0$）时 $\Phi = BA$（最大）；当平面平行于磁场（$\theta = 90^\circ$）时 $\Phi = 0$。
当平面垂直于磁场（$\theta = 0$）时 $\Phi = BA$（最大）；当平面平行于磁场（$\theta = 90^\circ$）时 $\Phi = 0$。
Magnetic flux $\Phi$ is the number of magnetic field lines passing through an area $A$: $\Phi = BA\cos\theta$, where $\theta$ is the angle between the normal to the surface and $B$. Unit: weber (Wb), $1\ \text{Wb} = 1\ \text{T·m}^2$.
Maximum when surface is perpendicular to $B$ ($\theta = 0^\circ$, $\Phi = BA$); zero when parallel ($\theta = 90^\circ$).
Maximum when surface is perpendicular to $B$ ($\theta = 0^\circ$, $\Phi = BA$); zero when parallel ($\theta = 90^\circ$).
法拉第电磁感应定律 / Faraday's Law
$$\mathcal{E} = -N\frac{\Delta\Phi}{\Delta t}$$
法拉第电磁感应定律（Faraday's Law of Induction）：闭合回路中产生的感应电动势 $\mathcal{E}$ 与穿过回路的磁通量变化率成正比。$N$ 为线圈匝数。
感应电动势产生的条件：磁通量发生变化——可以通过改变磁场 $B$、面积 $A$ 或角度 $\theta$ 来实现。
感应电动势产生的条件：磁通量发生变化——可以通过改变磁场 $B$、面积 $A$ 或角度 $\theta$ 来实现。
Faraday's Law of Induction: The induced emf in a closed loop is proportional to the rate of change of magnetic flux through the loop. $\mathcal{E} = -N \Delta\Phi/\Delta t$, where $N$ is the number of turns.
Induced emf requires changing flux — achieved by changing $B$, $A$, or $\theta$.
Induced emf requires changing flux — achieved by changing $B$, $A$, or $\theta$.
楞次定律 / Lenz's Law
$$\mathcal{E} = -N\frac{\Delta\Phi}{\Delta t} \quad \text{（公式中的负号 = 楞次定律）}$$
楞次定律（Lenz's Law）：感应电流的方向总是抵抗（oppose）引起它的磁通量变化。换句话说，感应电流产生的磁场总是阻碍原磁场的变化。
法拉第定律中的负号就是楞次定律的数学体现。楞次定律本质上是能量守恒的体现——如果感应电流不抵抗变化，就会产生永动机。
法拉第定律中的负号就是楞次定律的数学体现。楞次定律本质上是能量守恒的体现——如果感应电流不抵抗变化，就会产生永动机。
Lenz's Law: The direction of induced current is such that it opposes the change in magnetic flux that produced it. In other words, the induced magnetic field always acts to counteract the change in the original field.
The negative sign in Faraday's Law is the mathematical expression of Lenz's Law. Lenz's Law is a consequence of conservation of energy — if induced current didn't oppose the change, perpetual motion would be possible.
The negative sign in Faraday's Law is the mathematical expression of Lenz's Law. Lenz's Law is a consequence of conservation of energy — if induced current didn't oppose the change, perpetual motion would be possible.
📐 法拉第定律的推导 / Derivation of Faraday's Law
从洛伦兹力推导法拉第定律（导线切割磁感线的情况）：
考虑一根长度为 $L$ 的直导线以速度 $v$ 垂直于磁场 $B$ 运动。导线中的自由电子随着导线运动，受到洛伦兹力：
$F = qvB$（方向沿导线方向）
这个力使电子移动到导线的一端，产生感应电动势。电动势等于移动单位正电荷所做的功：
$\mathcal{E} = \frac{W}{q} = \frac{F \cdot L}{q} = \frac{qvBL}{q} = BLv$
另一方面，磁通量变化率：
在时间 $\Delta t$ 内，导线扫过的面积 $\Delta A = L \cdot v\Delta t$
磁通量变化 $\Delta\Phi = B\Delta A = BLv\Delta t$
所以 $\Delta\Phi/\Delta t = BLv$
因此 $\mathcal{E} = BLv = \Delta\Phi/\Delta t$
加上楞次定律的负号：$\mathcal{E} = -\Delta\Phi/\Delta t$
对于 $N$ 匝线圈：$\mathcal{E} = -N\Delta\Phi/\Delta t$
这就从洛伦兹力推导出了法拉第电磁感应定律。
考虑一根长度为 $L$ 的直导线以速度 $v$ 垂直于磁场 $B$ 运动。导线中的自由电子随着导线运动，受到洛伦兹力：
$F = qvB$（方向沿导线方向）
这个力使电子移动到导线的一端，产生感应电动势。电动势等于移动单位正电荷所做的功：
$\mathcal{E} = \frac{W}{q} = \frac{F \cdot L}{q} = \frac{qvBL}{q} = BLv$
另一方面，磁通量变化率：
在时间 $\Delta t$ 内，导线扫过的面积 $\Delta A = L \cdot v\Delta t$
磁通量变化 $\Delta\Phi = B\Delta A = BLv\Delta t$
所以 $\Delta\Phi/\Delta t = BLv$
因此 $\mathcal{E} = BLv = \Delta\Phi/\Delta t$
加上楞次定律的负号：$\mathcal{E} = -\Delta\Phi/\Delta t$
对于 $N$ 匝线圈：$\mathcal{E} = -N\Delta\Phi/\Delta t$
这就从洛伦兹力推导出了法拉第电磁感应定律。
Deriving Faraday's Law from Lorentz force (wire cutting field lines):
Consider a straight wire of length $L$ moving at $v$ perpendicular to $B$. Free electrons in the wire experience Lorentz force: $F = qvB$ (along the wire direction).
This force moves electrons to one end, creating an induced emf. Emf = work per unit charge:
$\mathcal{E} = F \cdot L / q = qvBL/q = BLv$
Flux change: in time $\Delta t$, area swept $\Delta A = Lv\Delta t$
$\Delta\Phi = B\Delta A = BLv\Delta t$
$\Delta\Phi/\Delta t = BLv$
Therefore $\mathcal{E} = BLv = \Delta\Phi/\Delta t$
With Lenz's Law sign: $\mathcal{E} = -N\Delta\Phi/\Delta t$
Consider a straight wire of length $L$ moving at $v$ perpendicular to $B$. Free electrons in the wire experience Lorentz force: $F = qvB$ (along the wire direction).
This force moves electrons to one end, creating an induced emf. Emf = work per unit charge:
$\mathcal{E} = F \cdot L / q = qvBL/q = BLv$
Flux change: in time $\Delta t$, area swept $\Delta A = Lv\Delta t$
$\Delta\Phi = B\Delta A = BLv\Delta t$
$\Delta\Phi/\Delta t = BLv$
Therefore $\mathcal{E} = BLv = \Delta\Phi/\Delta t$
With Lenz's Law sign: $\mathcal{E} = -N\Delta\Phi/\Delta t$
交流发电机 / AC Generator
$$\mathcal{E} = \mathcal{E}_0 \sin(\omega t) \qquad \mathcal{E}_0 = NBA\omega$$
交流发电机（AC generator / alternator）将机械能转化为电能。工作原理：
• 线圈在匀强磁场中以角速度 $\omega$ 匀速转动
• 穿过线圈的磁通量 $\Phi = BA\cos\theta = BA\cos(\omega t)$ 随时间正弦变化
• 感应电动势 $\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
• 峰值电动势 $\mathcal{E}_0 = NBA\omega$
• 产生正弦交流电
• 线圈在匀强磁场中以角速度 $\omega$ 匀速转动
• 穿过线圈的磁通量 $\Phi = BA\cos\theta = BA\cos(\omega t)$ 随时间正弦变化
• 感应电动势 $\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
• 峰值电动势 $\mathcal{E}_0 = NBA\omega$
• 产生正弦交流电
An AC generator (alternator) converts mechanical energy to electrical energy:
• A coil rotates at angular speed $\omega$ in a uniform field $B$
• Flux through the coil: $\Phi = BA\cos(\omega t)$
• Induced emf: $\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
• Peak emf: $\mathcal{E}_0 = NBA\omega$
• Produces sinusoidal alternating current
• A coil rotates at angular speed $\omega$ in a uniform field $B$
• Flux through the coil: $\Phi = BA\cos(\omega t)$
• Induced emf: $\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
• Peak emf: $\mathcal{E}_0 = NBA\omega$
• Produces sinusoidal alternating current
📐 交流发电机电动势的推导 / Derivation of AC Generator Emf
推导交流发电机的感应电动势：
设线圈面积为 $A$，在匀强磁场 $B$ 中以角速度 $\omega$ 匀速转动。$t = 0$ 时线圈平面垂直于磁场。
任意时刻 $t$，线圈平面法线与磁场方向的夹角 $\theta = \omega t$。
穿过线圈的磁通量：$\Phi = BA\cos\theta = BA\cos(\omega t)$
由法拉第定律，感应电动势：
$\mathcal{E} = -N\frac{d\Phi}{dt} = -N\frac{d}{dt}[BA\cos(\omega t)]$
$\mathcal{E} = -NBA \cdot [-\omega\sin(\omega t)]$
$\mathcal{E} = NBA\omega\sin(\omega t)$
峰值电动势：$\mathcal{E}_0 = NBA\omega$
当线圈平面平行于磁场（$\theta = 90^\circ$）时，$\sin(\omega t) = 1$，电动势最大。此时磁通量变化率最大。
当线圈平面垂直于磁场（$\theta = 0^\circ$）时，$\sin(\omega t) = 0$，电动势为零。此时磁通量最大但变化率为零。
设线圈面积为 $A$，在匀强磁场 $B$ 中以角速度 $\omega$ 匀速转动。$t = 0$ 时线圈平面垂直于磁场。
任意时刻 $t$，线圈平面法线与磁场方向的夹角 $\theta = \omega t$。
穿过线圈的磁通量：$\Phi = BA\cos\theta = BA\cos(\omega t)$
由法拉第定律，感应电动势：
$\mathcal{E} = -N\frac{d\Phi}{dt} = -N\frac{d}{dt}[BA\cos(\omega t)]$
$\mathcal{E} = -NBA \cdot [-\omega\sin(\omega t)]$
$\mathcal{E} = NBA\omega\sin(\omega t)$
峰值电动势：$\mathcal{E}_0 = NBA\omega$
当线圈平面平行于磁场（$\theta = 90^\circ$）时，$\sin(\omega t) = 1$，电动势最大。此时磁通量变化率最大。
当线圈平面垂直于磁场（$\theta = 0^\circ$）时，$\sin(\omega t) = 0$，电动势为零。此时磁通量最大但变化率为零。
Deriving AC generator emf:
Coil area $A$ rotates at $\omega$ in uniform $B$. At $t=0$, coil plane is perpendicular to $B$.
Angle at time $t$: $\theta = \omega t$
Flux: $\Phi = BA\cos(\omega t)$
By Faraday's Law:
$\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
Peak emf: $\mathcal{E}_0 = NBA\omega$
Maximum emf when coil is parallel to $B$ ($\theta = 90^\circ$, max rate of flux change).
Zero emf when coil is perpendicular to $B$ ($\theta = 0^\circ$, flux is maximum but rate of change is zero).
Coil area $A$ rotates at $\omega$ in uniform $B$. At $t=0$, coil plane is perpendicular to $B$.
Angle at time $t$: $\theta = \omega t$
Flux: $\Phi = BA\cos(\omega t)$
By Faraday's Law:
$\mathcal{E} = -N d\Phi/dt = NBA\omega\sin(\omega t)$
Peak emf: $\mathcal{E}_0 = NBA\omega$
Maximum emf when coil is parallel to $B$ ($\theta = 90^\circ$, max rate of flux change).
Zero emf when coil is perpendicular to $B$ ($\theta = 0^\circ$, flux is maximum but rate of change is zero).
变压器 / Transformers
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad V_p I_p = V_s I_s \quad \text{（理想变压器，忽略损耗）}$$
变压器（transformer）利用电磁感应原理改变交流电压。结构：两个线圈（初级 $N_p$ 匝，次级 $N_s$ 匝）绕在同一铁芯上。
工作原理：初级线圈通交流电 → 铁芯中产生交变磁场 → 次级线圈中产生感应电动势。
• 电压比：$V_s/V_p = N_s/N_p$（$N_s > N_p$ → 升压 step-up；$N_s < N_p$ → 降压 step-down）
• 理想变压器：输入功率 = 输出功率，$V_p I_p = V_s I_s$
• $N_s > N_p$：电压升高，电流降低（用于远距离输电减少热损耗）
• 变压器不能改变直流电压（需要变化的磁通量）
工作原理：初级线圈通交流电 → 铁芯中产生交变磁场 → 次级线圈中产生感应电动势。
• 电压比：$V_s/V_p = N_s/N_p$（$N_s > N_p$ → 升压 step-up；$N_s < N_p$ → 降压 step-down）
• 理想变压器：输入功率 = 输出功率，$V_p I_p = V_s I_s$
• $N_s > N_p$：电压升高，电流降低（用于远距离输电减少热损耗）
• 变压器不能改变直流电压（需要变化的磁通量）
A transformer uses electromagnetic induction to change AC voltage. Structure: two coils (primary $N_p$, secondary $N_s$) wound on a common iron core.
How it works: AC in primary → changing magnetic field in core → induced emf in secondary.
• Voltage ratio: $V_s/V_p = N_s/N_p$
• Ideal transformer: input power = output power, $V_p I_p = V_s I_s$
• Step-up ($N_s > N_p$): voltage up, current down (used for power transmission)
• Transformers only work with AC (requires changing flux)
How it works: AC in primary → changing magnetic field in core → induced emf in secondary.
• Voltage ratio: $V_s/V_p = N_s/N_p$
• Ideal transformer: input power = output power, $V_p I_p = V_s I_s$
• Step-up ($N_s > N_p$): voltage up, current down (used for power transmission)
• Transformers only work with AC (requires changing flux)
📐 变压器原理的推导 / Derivation of Transformer Equation
推导变压器的电压比：
设铁芯中的磁通量为 $\Phi(t)$（随时间变化）。根据法拉第定律：
初级线圈中的感应电动势：$\mathcal{E}_p = -N_p \frac{d\Phi}{dt}$
次级线圈中的感应电动势：$\mathcal{E}_s = -N_s \frac{d\Phi}{dt}$
因为铁芯中的磁通量 $\Phi$ 对两个线圈是相同的（理想情况下无漏磁），所以：
$\frac{\mathcal{E}_s}{\mathcal{E}_p} = \frac{N_s}{N_p}$
对于理想变压器，初级线圈的端电压 $V_p \approx \mathcal{E}_p$（忽略初级线圈电阻），次级端电压 $V_s = \mathcal{E}_s$（开路）：
$\frac{V_s}{V_p} \approx \frac{N_s}{N_p}$
推导电流关系（能量守恒）：
理想变压器无能量损失，输入功率 = 输出功率：
$V_p I_p = V_s I_s$
所以电流比是电压比的倒数：$\frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}$
实际变压器有损耗：铜损（线圈电阻发热）、铁损（涡流 eddy current 发热）、磁滞损耗（hysteresis）。
设铁芯中的磁通量为 $\Phi(t)$（随时间变化）。根据法拉第定律：
初级线圈中的感应电动势：$\mathcal{E}_p = -N_p \frac{d\Phi}{dt}$
次级线圈中的感应电动势：$\mathcal{E}_s = -N_s \frac{d\Phi}{dt}$
因为铁芯中的磁通量 $\Phi$ 对两个线圈是相同的（理想情况下无漏磁），所以：
$\frac{\mathcal{E}_s}{\mathcal{E}_p} = \frac{N_s}{N_p}$
对于理想变压器，初级线圈的端电压 $V_p \approx \mathcal{E}_p$（忽略初级线圈电阻），次级端电压 $V_s = \mathcal{E}_s$（开路）：
$\frac{V_s}{V_p} \approx \frac{N_s}{N_p}$
推导电流关系（能量守恒）：
理想变压器无能量损失，输入功率 = 输出功率：
$V_p I_p = V_s I_s$
所以电流比是电压比的倒数：$\frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}$
实际变压器有损耗：铜损（线圈电阻发热）、铁损（涡流 eddy current 发热）、磁滞损耗（hysteresis）。
Deriving transformer voltage ratio:
Flux in core: $\Phi(t)$. By Faraday's Law:
$\mathcal{E}_p = -N_p d\Phi/dt$, $\mathcal{E}_s = -N_s d\Phi/dt$
Since the same flux passes through both coils (ideal, no leakage):
$\mathcal{E}_s/\mathcal{E}_p = N_s/N_p$
For an ideal transformer, $V_p \approx \mathcal{E}_p$ and $V_s = \mathcal{E}_s$:
$V_s/V_p \approx N_s/N_p$
Current relation (energy conservation):
$V_p I_p = V_s I_s$ → $I_s/I_p = N_p/N_s$
Flux in core: $\Phi(t)$. By Faraday's Law:
$\mathcal{E}_p = -N_p d\Phi/dt$, $\mathcal{E}_s = -N_s d\Phi/dt$
Since the same flux passes through both coils (ideal, no leakage):
$\mathcal{E}_s/\mathcal{E}_p = N_s/N_p$
For an ideal transformer, $V_p \approx \mathcal{E}_p$ and $V_s = \mathcal{E}_s$:
$V_s/V_p \approx N_s/N_p$
Current relation (energy conservation):
$V_p I_p = V_s I_s$ → $I_s/I_p = N_p/N_s$
应用：电动机与发电机的对比 / Motor vs Generator
电动机 vs 发电机——本质上是同一装置的不同工作模式：
• 电动机（Motor）：输入电能，输出机械能。电流通过磁场中的线圈产生力矩使其转动。
• 发电机（Generator / Dynamo）：输入机械能，输出电能。外力使线圈在磁场中转动产生感应电流。
同一个装置既可以做电动机也可以做发电机（可逆性）。发电机中的
反电动势（back emf）：电动机转动时，线圈切割磁感线也产生感应电动势，方向与外加电压相反——这叫反电动势，它限制了电动机的电流。
• 电动机（Motor）：输入电能，输出机械能。电流通过磁场中的线圈产生力矩使其转动。
• 发电机（Generator / Dynamo）：输入机械能，输出电能。外力使线圈在磁场中转动产生感应电流。
同一个装置既可以做电动机也可以做发电机（可逆性）。发电机中的
反电动势（back emf）：电动机转动时，线圈切割磁感线也产生感应电动势，方向与外加电压相反——这叫反电动势，它限制了电动机的电流。
Motor vs Generator — the same device operating in reverse:
• Motor: electrical energy → mechanical energy. Current in a coil in a magnetic field produces torque.
• Generator: mechanical energy → electrical energy. External force rotates a coil in a magnetic field, inducing current.
The same device can function as either (reversibility). In a motor, the rotating coil also generates a back emf opposing the applied voltage, limiting the current.
• Motor: electrical energy → mechanical energy. Current in a coil in a magnetic field produces torque.
• Generator: mechanical energy → electrical energy. External force rotates a coil in a magnetic field, inducing current.
The same device can function as either (reversibility). In a motor, the rotating coil also generates a back emf opposing the applied voltage, limiting the current.
应用：粒子加速器中的电磁应用 / Application: Particle Accelerators
电磁学在粒子加速器中有着核心应用：
• 回旋加速器（Cyclotron）：磁场使带电粒子做圆周运动，电场加速。$r = mv/qB$，$T = 2\pi m/qB$
• 同步加速器（Synchrotron）：粒子在环形轨道中运动，磁场随粒子能量增加而增强以保持轨道半径恒定
• 直线加速器（Linac）：一系列电极排列成直线，交变电场依次加速粒子
• LHC（大型强子对撞机）：使用超导磁体产生 8.3 T 的强磁场，使质子沿 27 km 的环形隧道运动，能量达到 TeV 量级
• 回旋加速器（Cyclotron）：磁场使带电粒子做圆周运动，电场加速。$r = mv/qB$，$T = 2\pi m/qB$
• 同步加速器（Synchrotron）：粒子在环形轨道中运动，磁场随粒子能量增加而增强以保持轨道半径恒定
• 直线加速器（Linac）：一系列电极排列成直线，交变电场依次加速粒子
• LHC（大型强子对撞机）：使用超导磁体产生 8.3 T 的强磁场，使质子沿 27 km 的环形隧道运动，能量达到 TeV 量级
Electromagnetism is central to particle accelerators:
• Cyclotron: magnetic field bends particles into circles, electric fields accelerate them. $r = mv/qB$, $T = 2\pi m/qB$
• Synchrotron: particles in a ring; $B$ increases with energy to keep radius constant
• Linac: linear array of electrodes with alternating fields
• LHC (Large Hadron Collider): superconducting magnets produce 8.3 T fields, guiding protons along a 27 km ring to TeV energies
• Cyclotron: magnetic field bends particles into circles, electric fields accelerate them. $r = mv/qB$, $T = 2\pi m/qB$
• Synchrotron: particles in a ring; $B$ increases with energy to keep radius constant
• Linac: linear array of electrodes with alternating fields
• LHC (Large Hadron Collider): superconducting magnets produce 8.3 T fields, guiding protons along a 27 km ring to TeV energies
🎯 IB 考试要点：① 楞次定律判断感应电流方向：感应电流的磁场总"反抗"原磁场的变化——"增反减同"；② 法拉第定律 $\mathcal{E} = -N\Delta\Phi/\Delta t$ 计算感应电动势的大小；③ 变压器只适用于交流电（需要变化的磁通量）；④ 发电机峰值电动势 $\mathcal{E}_0 = NBA\omega$；⑤ 远距离输电用升压变压器减少电流，从而减少 $I^2 R$ 损耗。
🎯 IB Exam Tips: ① Lenz's Law: induced current opposes the change in flux; ② Faraday's Law $\mathcal{E} = -N\Delta\Phi/\Delta t$ gives induced emf magnitude; ③ Transformers work only with AC; ④ Generator peak emf $\mathcal{E}_0 = NBA\omega$; ⑤ High-voltage transmission reduces $I^2 R$ losses.
例 5.9 / Example 5.9
一个 200 匝的线圈，面积 $4.0 \times 10^{-3}$ m$^2$，在 0.50 T 的匀强磁场中以角速度 100 rad/s 匀速转动。求：(a) 最大感应电动势；(b) $t = 0.010$ s 时的感应电动势（设 $t=0$ 时线圈平面垂直于磁场）。
200-turn coil, area $4.0 \times 10^{-3}$ m$^2$, $B = 0.50$ T, $\omega = 100$ rad/s. Find: (a) peak emf; (b) emf at $t = 0.010$ s ($t=0$ when coil plane $\perp$ field).
解题过程 / Solution
(a) $\mathcal{E}_0 = NBA\omega = 200 \times 0.50 \times 4.0 \times 10^{-3} \times 100$
$\mathcal{E}_0 = 200 \times 0.50 \times 0.0040 \times 100 = 40$ V
(b) $\mathcal{E}(t) = \mathcal{E}_0 \sin(\omega t) = 40 \times \sin(100 \times 0.010)$
$\mathcal{E} = 40 \times \sin(1.0\ \text{rad}) = 40 \times 0.841 = 33.7$ V
$\mathcal{E}_0 = 200 \times 0.50 \times 0.0040 \times 100 = 40$ V
(b) $\mathcal{E}(t) = \mathcal{E}_0 \sin(\omega t) = 40 \times \sin(100 \times 0.010)$
$\mathcal{E} = 40 \times \sin(1.0\ \text{rad}) = 40 \times 0.841 = 33.7$ V
(a) $\mathcal{E}_0 = NBA\omega = 200 \times 0.50 \times 4.0 \times 10^{-3} \times 100 = 40$ V
(b) $\mathcal{E} = \mathcal{E}_0 \sin(\omega t) = 40 \times \sin(1.0) = 33.7$ V
(b) $\mathcal{E} = \mathcal{E}_0 \sin(\omega t) = 40 \times \sin(1.0) = 33.7$ V
例 5.10 / Example 5.10
一个理想变压器，初级线圈 1000 匝，输入 240 V 交流电。次级输出电压 12 V。求：(a) 次级线圈匝数；(b) 若次级接 6.0 $\Omega$ 负载，求初、次级电流。
Ideal transformer: $N_p = 1000$, $V_p = 240$ V AC, $V_s = 12$ V. Find: (a) $N_s$; (b) primary and secondary currents if secondary load = 6.0 $\Omega$.
解题过程 / Solution
(a) $V_s/V_p = N_s/N_p$
$12/240 = N_s/1000$
$N_s = 1000 \times 12/240 = 50$ 匝
(b) 次级电流：$I_s = V_s/R = 12/6.0 = 2.0$ A
理想变压器功率守恒：$V_p I_p = V_s I_s$
$240 \times I_p = 12 \times 2.0$
$I_p = 24/240 = 0.10$ A
验证电流比：$I_s/I_p = 2.0/0.10 = 20$，$N_p/N_s = 1000/50 = 20$，一致。
$12/240 = N_s/1000$
$N_s = 1000 \times 12/240 = 50$ 匝
(b) 次级电流：$I_s = V_s/R = 12/6.0 = 2.0$ A
理想变压器功率守恒：$V_p I_p = V_s I_s$
$240 \times I_p = 12 \times 2.0$
$I_p = 24/240 = 0.10$ A
验证电流比：$I_s/I_p = 2.0/0.10 = 20$，$N_p/N_s = 1000/50 = 20$，一致。
(a) $N_s = N_p \times V_s/V_p = 1000 \times 12/240 = 50$ turns
(b) $I_s = V_s/R = 12/6.0 = 2.0$ A
$V_p I_p = V_s I_s$ → $I_p = 12 \times 2.0/240 = 0.10$ A
(b) $I_s = V_s/R = 12/6.0 = 2.0$ A
$V_p I_p = V_s I_s$ → $I_p = 12 \times 2.0/240 = 0.10$ A
例 5.11 / Example 5.11
一个 100 匝的线圈中的磁通量在 0.050 s 内从 0.020 Wb 均匀变化到 0.080 Wb。求感应电动势的大小。
A 100-turn coil: flux changes uniformly from 0.020 Wb to 0.080 Wb in 0.050 s. Find the induced emf.
解题过程 / Solution
$\Delta\Phi = 0.080 - 0.020 = 0.060$ Wb
$\Delta t = 0.050$ s
$\mathcal{E} = N|\Delta\Phi/\Delta t| = 100 \times 0.060/0.050 = 100 \times 1.20 = 120$ V
方向由楞次定律确定：磁通量在增加，感应电流产生的磁场方向与原磁场相反（抵抗增加）。
$\Delta t = 0.050$ s
$\mathcal{E} = N|\Delta\Phi/\Delta t| = 100 \times 0.060/0.050 = 100 \times 1.20 = 120$ V
方向由楞次定律确定：磁通量在增加，感应电流产生的磁场方向与原磁场相反（抵抗增加）。
$\Delta\Phi = 0.080 - 0.020 = 0.060$ Wb, $\Delta t = 0.050$ s
$\mathcal{E} = N|\Delta\Phi/\Delta t| = 100 \times 0.060/0.050 = 120$ V
Direction: Lenz's Law — increasing flux → induced field opposes the original field.
$\mathcal{E} = N|\Delta\Phi/\Delta t| = 100 \times 0.060/0.050 = 120$ V
Direction: Lenz's Law — increasing flux → induced field opposes the original field.
🏆 挑战题 / Challenge Problems
多概念综合 / Multi-concept难度 / Hard
以下题目需要综合运用本章的多个知识点。建议先复习所有内容后再尝试。
These problems require integrating multiple concepts from this chapter. Review all sections before attempting.
挑战 1 / Challenge 1 电场与磁场综合 / Electric and Magnetic Fields Combined
一个质子（$q = 1.60 \times 10^{-19}$ C, $m = 1.67 \times 10^{-27}$ kg）以 $5.0 \times 10^5$ m/s 垂直射入正交的匀强电场（$E = 3.0 \times 10^4$ N/C）和匀强磁场区域。要使质子沿直线运动（速度选择器），求磁感应强度 $B$ 的大小。
A proton enters perpendicular crossed uniform $E$ and $B$ fields: $E = 3.0 \times 10^4$ N/C, $v = 5.0 \times 10^5$ m/s. Find $B$ for straight-line motion (velocity selector).
解题过程 / Solution
速度选择器原理：电场力和洛伦兹力平衡，合力为零。
电场力：$F_E = qE$（方向与电场方向一致）
洛伦兹力：$F_B = qvB$（方向与电场力相反）
平衡条件：$qE = qvB$
$E = vB$
$B = E/v = 3.0 \times 10^4 / 5.0 \times 10^5 = 0.060$ T
只有速度恰好为 $v = E/B$ 的粒子才能沿直线通过速度选择器（速度与 $m$ 和 $q$ 无关）。速度不同的粒子会偏转。
电场力：$F_E = qE$（方向与电场方向一致）
洛伦兹力：$F_B = qvB$（方向与电场力相反）
平衡条件：$qE = qvB$
$E = vB$
$B = E/v = 3.0 \times 10^4 / 5.0 \times 10^5 = 0.060$ T
只有速度恰好为 $v = E/B$ 的粒子才能沿直线通过速度选择器（速度与 $m$ 和 $q$ 无关）。速度不同的粒子会偏转。
Velocity selector: electric force balances magnetic force.
$qE = qvB$ → $B = E/v = 3.0 \times 10^4 / 5.0 \times 10^5 = 0.060$ T
Only particles with $v = E/B$ pass straight through (independent of $m$ and $q$).
$qE = qvB$ → $B = E/v = 3.0 \times 10^4 / 5.0 \times 10^5 = 0.060$ T
Only particles with $v = E/B$ pass straight through (independent of $m$ and $q$).
挑战 2 / Challenge 2 质谱仪 / Mass Spectrometer
一个未知离子（电荷量 $q = +1.60 \times 10^{-19}$ C）被 2000 V 的电压从静止加速后，进入 $B = 0.40$ T 的匀强磁场做半圆周运动，打在探测器上，测得轨道半径 $r = 0.050$ m。求离子的质量。
An unknown ion ($q = +1.60 \times 10^{-19}$ C) is accelerated through 2000 V, then enters $B = 0.40$ T field, radius $r = 0.050$ m. Find the ion mass.
解题过程 / Solution
第一步：电场加速，能量守恒
$qV = \frac12 mv^2$
$v = \sqrt{2qV/m}$
第二步：磁场中做圆周运动
$qvB = mv^2/r$
$r = mv/qB$
代入第一步的速度表达式：
$r = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
解出质量：
$m = \frac{qB^2 r^2}{2V} = \frac{1.60 \times 10^{-19} \times (0.40)^2 \times (0.050)^2}{2 \times 2000}$
$m = \frac{1.60 \times 10^{-19} \times 0.16 \times 0.0025}{4000}$
$m = \frac{6.40 \times 10^{-23}}{4000} = 1.60 \times 10^{-26}$ kg
$qV = \frac12 mv^2$
$v = \sqrt{2qV/m}$
第二步：磁场中做圆周运动
$qvB = mv^2/r$
$r = mv/qB$
代入第一步的速度表达式：
$r = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
解出质量：
$m = \frac{qB^2 r^2}{2V} = \frac{1.60 \times 10^{-19} \times (0.40)^2 \times (0.050)^2}{2 \times 2000}$
$m = \frac{1.60 \times 10^{-19} \times 0.16 \times 0.0025}{4000}$
$m = \frac{6.40 \times 10^{-23}}{4000} = 1.60 \times 10^{-26}$ kg
Step 1: $qV = \frac12 mv^2$ → $v = \sqrt{2qV/m}$
Step 2: $qvB = mv^2/r$ → $r = mv/qB$
Combined: $r = \frac{1}{B}\sqrt{2mV/q}$ → $m = qB^2 r^2/(2V)$
$m = 1.60 \times 10^{-19} \times 0.16 \times 0.0025 / 4000 = 1.60 \times 10^{-26}$ kg
Step 2: $qvB = mv^2/r$ → $r = mv/qB$
Combined: $r = \frac{1}{B}\sqrt{2mV/q}$ → $m = qB^2 r^2/(2V)$
$m = 1.60 \times 10^{-19} \times 0.16 \times 0.0025 / 4000 = 1.60 \times 10^{-26}$ kg
挑战 3 / Challenge 3 复杂电路分析 / Complex Circuit Analysis
一个电路：电池 $\mathcal{E} = 20$ V（内阻 $r = 1.0\ \Omega$），$R_1 = 4.0\ \Omega$，$R_2 = 6.0\ \Omega$，$R_3 = 8.0\ \Omega$。$R_1$ 和 $R_2$ 并联后再与 $R_3$ 串联。求：(a) 总电流；(b) $R_3$ 两端的电压；(c) $R_1$ 通过的电流；(d) 当外电路短路时的短路电流。
Circuit: $\mathcal{E} = 20$ V ($r = 1.0\ \Omega$), $R_1 = 4.0\ \Omega$, $R_2 = 6.0\ \Omega$ in parallel, series with $R_3 = 8.0\ \Omega$. Find: (a) total current; (b) $V_{R3}$; (c) $I_1$; (d) short-circuit current.
解题过程 / Solution
(a) $R_{12} = (1/4.0 + 1/6.0)^{-1} = (0.25 + 0.1667)^{-1} = (0.4167)^{-1} = 2.40\ \Omega$
$R_{\text{外}} = R_{12} + R_3 = 2.40 + 8.0 = 10.40\ \Omega$
$R_{\text{总}} = R_{\text{外}} + r = 10.40 + 1.0 = 11.40\ \Omega$
$I_{\text{总}} = \mathcal{E}/R_{\text{总}} = 20/11.40 = 1.754$ A
(b) $V_{R3} = I_{\text{总}} \times R_3 = 1.754 \times 8.0 = 14.04$ V
(c) $V_{12} = I_{\text{总}} \times R_{12} = 1.754 \times 2.40 = 4.21$ V
$I_1 = V_{12}/R_1 = 4.21/4.0 = 1.053$ A
(d) 短路时外电阻为零：$I_{\text{短路}} = \mathcal{E}/r = 20/1.0 = 20$ A
短路电流很大——这就是为什么短路很危险。
$R_{\text{外}} = R_{12} + R_3 = 2.40 + 8.0 = 10.40\ \Omega$
$R_{\text{总}} = R_{\text{外}} + r = 10.40 + 1.0 = 11.40\ \Omega$
$I_{\text{总}} = \mathcal{E}/R_{\text{总}} = 20/11.40 = 1.754$ A
(b) $V_{R3} = I_{\text{总}} \times R_3 = 1.754 \times 8.0 = 14.04$ V
(c) $V_{12} = I_{\text{总}} \times R_{12} = 1.754 \times 2.40 = 4.21$ V
$I_1 = V_{12}/R_1 = 4.21/4.0 = 1.053$ A
(d) 短路时外电阻为零：$I_{\text{短路}} = \mathcal{E}/r = 20/1.0 = 20$ A
短路电流很大——这就是为什么短路很危险。
(a) $R_{\text{total}} = 11.40\ \Omega$, $I_{\text{total}} = 1.754$ A
(b) $V_{R3} = 14.04$ V
(c) $I_1 = 1.053$ A
(d) $I_{\text{short}} = \mathcal{E}/r = 20$ A — very dangerous!
(b) $V_{R3} = 14.04$ V
(c) $I_1 = 1.053$ A
(d) $I_{\text{short}} = \mathcal{E}/r = 20$ A — very dangerous!
挑战 4 / Challenge 4 电磁感应与能量守恒 / Induction and Energy Conservation
一根 $L = 0.50$ m 的金属棒在两根平行金属导轨上以 $v = 4.0$ m/s 向右匀速滑动。导轨间距与棒长相同，磁场 $B = 0.60$ T 垂直纸面向里。导轨右端接有 $R = 3.0\ \Omega$ 的电阻（忽略其他电阻）。求：(a) 感应电动势；(b) 感应电流的大小和方向；(c) 保持棒匀速运动所需的外力大小和方向；(d) 外力做功的功率。
A 0.50 m metal rod slides right at 4.0 m/s on parallel rails. $B = 0.60$ T into page, $R = 3.0\ \Omega$. Find: (a) induced emf; (b) current magnitude and direction; (c) external force to maintain constant speed; (d) power of external force.
解题过程 / Solution
(a) $\mathcal{E} = BLv = 0.60 \times 0.50 \times 4.0 = 1.20$ V
(b) $I = \mathcal{E}/R = 1.20/3.0 = 0.40$ A
方向：由楞次定律，棒向右运动→回路面积增大→磁通量增加→感应电流产生的磁场抵抗增加→用右手定则判断电流方向为顺时针（从棒上看，电流向上）。
(c) 载流导体在磁场中受力：$F_B = BIL = 0.60 \times 0.40 \times 0.50 = 0.120$ N
左手定则：$F_B$ 方向向左（抵抗运动）。为使棒匀速运动，外力必须向右，大小等于 $F_B = 0.120$ N。
(d) $P_{\text{外}} = F_{\text{外}} \times v = 0.120 \times 4.0 = 0.480$ W
验证：电功率 $P_{\text{电}} = I^2 R = (0.40)^2 \times 3.0 = 0.480$ W ✓
机械能完全转化为电能——能量守恒的完美体现！
(b) $I = \mathcal{E}/R = 1.20/3.0 = 0.40$ A
方向：由楞次定律，棒向右运动→回路面积增大→磁通量增加→感应电流产生的磁场抵抗增加→用右手定则判断电流方向为顺时针（从棒上看，电流向上）。
(c) 载流导体在磁场中受力：$F_B = BIL = 0.60 \times 0.40 \times 0.50 = 0.120$ N
左手定则：$F_B$ 方向向左（抵抗运动）。为使棒匀速运动，外力必须向右，大小等于 $F_B = 0.120$ N。
(d) $P_{\text{外}} = F_{\text{外}} \times v = 0.120 \times 4.0 = 0.480$ W
验证：电功率 $P_{\text{电}} = I^2 R = (0.40)^2 \times 3.0 = 0.480$ W ✓
机械能完全转化为电能——能量守恒的完美体现！
(a) $\mathcal{E} = BLv = 1.20$ V
(b) $I = \mathcal{E}/R = 0.40$ A, clockwise
(c) $F_B = BIL = 0.120$ N left → $F_{\text{ext}} = 0.120$ N right
(d) $P_{\text{ext}} = Fv = 0.480$ W = $I^2 R$ — mechanical energy converted to electrical energy!
(b) $I = \mathcal{E}/R = 0.40$ A, clockwise
(c) $F_B = BIL = 0.120$ N left → $F_{\text{ext}} = 0.120$ N right
(d) $P_{\text{ext}} = Fv = 0.480$ W = $I^2 R$ — mechanical energy converted to electrical energy!
挑战 5 / Challenge 5 多概念综合 / Multi-Concept Integration
一个电子（$m_e = 9.11 \times 10^{-31}$ kg, $e = 1.60 \times 10^{-19}$ C）在 $B = 0.020$ T 的匀强磁场中做圆周运动，轨道半径 $r = 0.010$ m。求：(a) 电子的速度；(b) 回旋周期；(c) 若电子同时受到一个向外的径向电场作用，恰好平衡洛伦兹力，求电场强度 $E$ 的大小；(d) 电子的动能（单位 eV）。
An electron moves in a circle in $B = 0.020$ T, radius $r = 0.010$ m. Find: (a) speed; (b) period; (c) $E$ for radial balance; (d) kinetic energy in eV.
解题过程 / Solution
(a) $qvB = mv^2/r$ → $v = qBr/m$
$v = (1.60 \times 10^{-19} \times 0.020 \times 0.010) / 9.11 \times 10^{-31}$
$v = 3.20 \times 10^{-23} / 9.11 \times 10^{-31} = 3.51 \times 10^7$ m/s
(b) $T = 2\pi m/qB = 2\pi \times 9.11 \times 10^{-31} / (1.60 \times 10^{-19} \times 0.020)$
$T = 5.72 \times 10^{-30} / 3.20 \times 10^{-21} = 1.79 \times 10^{-9}$ s = 1.79 ns
(c) 平衡条件：$qE = qvB$ → $E = vB = 3.51 \times 10^7 \times 0.020 = 7.02 \times 10^5$ N/C
(d) $E_k = \frac12 mv^2 = 0.5 \times 9.11 \times 10^{-31} \times (3.51 \times 10^7)^2$
$E_k = 0.5 \times 9.11 \times 10^{-31} \times 1.232 \times 10^{15} = 5.61 \times 10^{-16}$ J
$E_k = 5.61 \times 10^{-16} / 1.60 \times 10^{-19} = 3506$ eV ≈ 3.51 keV
$v = (1.60 \times 10^{-19} \times 0.020 \times 0.010) / 9.11 \times 10^{-31}$
$v = 3.20 \times 10^{-23} / 9.11 \times 10^{-31} = 3.51 \times 10^7$ m/s
(b) $T = 2\pi m/qB = 2\pi \times 9.11 \times 10^{-31} / (1.60 \times 10^{-19} \times 0.020)$
$T = 5.72 \times 10^{-30} / 3.20 \times 10^{-21} = 1.79 \times 10^{-9}$ s = 1.79 ns
(c) 平衡条件：$qE = qvB$ → $E = vB = 3.51 \times 10^7 \times 0.020 = 7.02 \times 10^5$ N/C
(d) $E_k = \frac12 mv^2 = 0.5 \times 9.11 \times 10^{-31} \times (3.51 \times 10^7)^2$
$E_k = 0.5 \times 9.11 \times 10^{-31} \times 1.232 \times 10^{15} = 5.61 \times 10^{-16}$ J
$E_k = 5.61 \times 10^{-16} / 1.60 \times 10^{-19} = 3506$ eV ≈ 3.51 keV
$v = qBr/m = 3.51 \times 10^7$ m/s
$T = 2\pi m/qB = 1.79$ ns
$E = vB = 7.02 \times 10^5$ N/C
$E_k = \frac12 mv^2 = 5.61 \times 10^{-16}$ J = 3.51 keV
$T = 2\pi m/qB = 1.79$ ns
$E = vB = 7.02 \times 10^5$ N/C
$E_k = \frac12 mv^2 = 5.61 \times 10^{-16}$ J = 3.51 keV
附录：核心公式速查 / Formula Reference
| 知识点 / Topic | 公式 / Formula | 说明 / Notes |
|---|---|---|
| 库仑定律 / Coulomb's Law | $F = kq_1q_2/r^2$ | $k = 8.99 \times 10^9$ N·m$^2$/C$^2$ |
| 电场强度 / Electric Field | $E = F/q$ | $E = kQ/r^2$（点电荷） |
| 电势 / Potential | $V = kQ/r$ | $E = V/d$（匀强电场） |
| 电势能 / Potential Energy | $E_p = kq_1q_2/r$ | $W = qV$（电场力做功） |
| 欧姆定律 / Ohm's Law | $V = IR$ | $R = \rho L/A$ |
| 串联电阻 / Series | $R_s = R_1 + R_2 + \ldots$ | 电流相同 |
| 并联电阻 / Parallel | $1/R_p = 1/R_1 + 1/R_2 + \ldots$ | 电压相同 |
| 基尔霍夫 KCL / KCL | $\sum I_{\text{进}} = \sum I_{\text{出}}$ | 节点 / Junction |
| 基尔霍夫 KVL / KVL | $\sum V = 0$ | 回路 / Loop |
| 内阻 / Internal Res. | $V = \mathcal{E} - Ir$ | $\mathcal{E}$ 为电动势 |
| 电功率 / Power | $P = IV = I^2 R = V^2/R$ | 焦耳定律 |
| 分压器 / Pot. Divider | $V_{\text{out}} = V_{\text{in}}R_2/(R_1+R_2)$ | 串联分压 |
| 安培力 / Ampere Force | $F = BIL\sin\theta$ | 左手定则判断方向 |
| 洛伦兹力 / Lorentz Force | $F = qvB\sin\theta$ | $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ |
| 回旋半径 / Cyclotron R. | $r = mv/qB$ | 匀速圆周运动 |
| 回旋周期 / Cyclotron T. | $T = 2\pi m/qB$ | 与速度无关 |
| 磁通量 / Magnetic Flux | $\Phi = BA\cos\theta$ | 单位 Wb |
| 法拉第定律 / Faraday | $\mathcal{E} = -N\Delta\Phi/\Delta t$ | 感应电动势 |
| 发电机 / Generator | $\mathcal{E}_0 = NBA\omega$ | 峰值电动势 |
| 变压器 / Transformer | $V_s/V_p = N_s/N_p$ | $V_p I_p = V_s I_s$（理想） |
📝 分节练习 / Section Practice
按知识点逐节练习，每题即时反馈。完成一节后自动记录进度。
Practice section by section with instant feedback. Progress is automatically saved.
由易到难 / Easy→Hard即时反馈 / Instant记录进度 / Progress Saved
🎯 随机测试 / Random Quiz
从题库随机抽取题目，模拟考试环境。提交后统一判分。
Randomly drawn from the question bank. Simulates exam conditions. Graded after submission.
测试设置 / Quiz Settings
📊 成绩分析 / Score Analysis
📭
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