📚 ESAT Physics: Electromagnetism | ESAT 物理：电磁学专题
Electromagnetism is one of the most heavily tested topics in the ESAT Physics paper. It combines conceptual understanding with quantitative problem-solving, and questions often link electric fields, circuits, and magnetic induction into a single scenario. Mastering the core laws, their vector directions, and the units involved is essential for scoring high marks.
电磁学是 ESAT 物理试卷中考查最频繁的专题之一。它既考查概念理解，又考查定量计算能力，题目常将电场、电路和磁感应整合在同一情境中。掌握核心定律、矢量方向及相关单位，是取得高分的关键。
1. Coulomb’s Law and Electric Fields | 库仑定律与电场
Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. For charges Q₁ and Q₂ separated by distance r, the force magnitude is given by:
库仑定律指出，两个点电荷之间的静电力与电荷量的乘积成正比，与它们之间距离的平方成反比。对于相距为 r 的电荷 Q₁ 和 Q₂，力的大小为：
F = kQ₁Q₂ / r²
where k = 8.99 × 10⁹ N·m²·C⁻². The force is repulsive for like charges and attractive for unlike charges. In ESAT questions, always check whether forces are vectors — draw a free-body diagram and resolve components when multiple charges are present.
其中 k = 8.99 × 10⁹ N·m²·C⁻²。同号电荷相斥，异号电荷相吸。在 ESAT 题目中，务必注意力是矢量——当存在多个电荷时，应画出受力分析图并分解分量。
The electric field E is defined as the force per unit positive charge:
电场强度 E 定义为每单位正电荷所受的力：
E = F / q = kQ / r²
Field lines point away from positive charges and toward negative charges. The density of field lines indicates the strength of the field. For a uniform field between two parallel plates, the relationship E = V / d applies, where V is the potential difference and d is the plate separation.
电场线从正电荷出发，终止于负电荷。电场线的疏密反映电场的强弱。对于两块平行板之间的匀强电场，适用关系式 E = V / d，其中 V 为电势差，d 为板间距。
2. Electric Potential and Energy | 电势与电势能
Electric potential V at a point is the work done per unit charge in bringing a positive test charge from infinity to that point. For a point charge Q, the potential at distance r is:
电场中某点的电势 V 是指将正的试探电荷从无穷远处移至该点所做的功除以电荷量。对于点电荷 Q，距离 r 处的电势为：
V = kQ / r
Potential is a scalar quantity, so potentials from multiple charges simply add algebraically. This is a common ESAT shortcut — scalar addition is much easier than vector addition of fields.
电势是标量，多个电荷产生的电势可直接代数相加。这是 ESAT 中常用的技巧——标量相加比场强的矢量相加简单得多。
The work done in moving a charge q through a potential difference ΔV is W = qΔV. When a charged particle is accelerated through a potential difference, conservation of energy gives:
将电荷 q 移动通过电势差 ΔV 所做的功为 W = qΔV。当带电粒子经电势差加速时，能量守恒给出：
½mv² = qΔV
This equation frequently appears in ESAT questions involving electron beams or ion acceleration.
该方程常见于 ESAT 中涉及电子束或离子加速的题目。
3. Capacitance | 电容
A capacitor stores charge and electrical energy. Its capacitance is defined as the charge stored per unit potential difference:
电容器储存电荷和电能。其电容定义为储存的电荷量与电势差之比：
C = Q / V
The unit of capacitance is the farad (F). For a parallel-plate capacitor, the capacitance depends on the plate area A, plate separation d, and the permittivity of the material between the plates:
电容的单位是法拉（F）。对于平行板电容器，电容取决于极板面积 A、板间距 d 以及极板间材料的介电常数：
C = ε₀εᵣA / d
where ε₀ = 8.85 × 10⁻¹² F·m⁻¹ is the permittivity of free space and εᵣ is the relative permittivity of the dielectric. Inserting a dielectric increases capacitance — a fact often tested conceptually.
其中 ε₀ = 8.85 × 10⁻¹² F·m⁻¹ 是真空介电常数，εᵣ 是相对介电常数。插入电介质会增大电容——这一概念经常被考查。
The energy stored in a charged capacitor can be expressed in three equivalent forms:
充电电容器中储存的能量有三种等价表达形式：
E = ½QV = ½CV² = Q² / (2C)
Choose the form that uses the quantities given in the question. If two capacitors are connected in parallel, voltages are equal and charges add; in series, charges are equal and voltages add.
根据题目给出的已知量选择最合适的表达式。两个电容器并联时电压相等、电荷相加；串联时电荷相等、电压相加。
4. Current, Resistance and Resistivity | 电流、电阻与电阻率
Electric current is the rate of flow of charge. In a conductor, the current I is related to the number density n of charge carriers, their charge e, the cross-sectional area A, and the drift velocity v:
电流是电荷流动的速率。在导体中，电流 I 与载流子数密度 n、载流子电荷 e、横截面积 A 和漂移速度 v 有关：
I = nAve
Ohm’s law states that the potential difference across a conductor is proportional to the current through it, provided physical conditions remain constant:
欧姆定律指出，在物理条件恒定的情况下，导体两端的电势差与通过它的电流成正比：
V = IR
Resistance depends on the material’s resistivity ρ, length L, and cross-sectional area A:
电阻取决于材料的电阻率 ρ、长度 L 和横截面积 A：
R = ρL / A
For metals, resistivity increases with temperature because lattice vibrations scatter conduction electrons more frequently. For semiconductors, resistivity decreases with temperature. This distinction is a favourite ESAT multiple-choice trap.
金属的电阻率随温度升高而增大，因为晶格振动更频繁地散射导电电子。半导体的电阻率则随温度升高而减小。这一区别是 ESAT 选择题中常见的陷阱。
5. Series and Parallel Circuits | 串联与并联电路
In a series circuit, the current is the same through all components, and the total potential difference is the sum of individual potential differences. The total resistance is:
在串联电路中，通过所有元件的电流相同，总电势差等于各元件电势差之和。总电阻为：
R_total = R₁ + R₂ + R₃ + …
In a parallel circuit, the potential difference is the same across all branches, and the total current divides among the branches. The total resistance is given by:
在并联电路中，各支路两端的电势差相同，总电流在各支路之间分配。总电阻由下式给出：
1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …
For two resistors in parallel, a useful simplification is R_total = R₁R₂ / (R₁ + R₂). The total resistance in parallel is always smaller than the smallest individual resistance — a quick check for numerical answers.
对于两个并联电阻，简化公式为 R_total = R₁R₂ / (R₁ + R₂)。并联总电阻总是小于其中最小的电阻——这是检验数值答案的快捷方法。
The potential divider rule is essential for ESAT: for two resistors in series, the voltage across R₁ is V₁ = V × R₁ / (R₁ + R₂). This appears in sensor circuits, thermistors, and light-dependent resistors.
分压规则对 ESAT 至关重要：对于两个串联电阻，R₁ 两端的电压为 V₁ = V × R₁ / (R₁ + R₂)。这出现在传感器电路、热敏电阻和光敏电阻的题目中。
6. Kirchhoff’s Laws | 基尔霍夫定律
Kirchhoff’s first law (the junction rule) states that the total current entering a junction equals the total current leaving it. This follows from conservation of charge:
基尔霍夫第一定律（节点定律）指出，流入节点的总电流等于流出节点的总电流。这是电荷守恒的体现：
Σ I_in = Σ I_out
Kirchhoff’s second law (the loop rule) states that the sum of the electromotive forces (emfs) in any closed loop equals the sum of the potential differences across the components in that loop. This follows from conservation of energy:
基尔霍夫第二定律（回路定律）指出，在任何闭合回路中，电动势之和等于回路中各元件电势差之和。这是能量守恒的体现：
Σ EMF = Σ IR
When applying the loop rule, assign a direction around the loop and be consistent with signs: a potential rise is positive, a potential drop is negative. Set up simultaneous equations and solve systematically.
应用回路定律时，先设定绕行方向并保持一致的正负号：电势升高为正，电势降低为负。列联立方程后系统求解。
A common ESAT scenario involves a circuit with two batteries and three resistors. Use the junction rule to relate currents, then use two independent loops to obtain enough equations. Always check that your final currents satisfy both laws.
ESAT 常见情境是包含两节电池和三个电阻的电路。用节点定律关联电流，再用两个独立回路获得足够的方程。务必检查最终电流是否同时满足两条定律。
7. Magnetic Fields and Forces | 磁场与磁力
A magnetic field exerts a force on a moving charge or on a current-carrying conductor. For a straight conductor of length L carrying current I in a magnetic field B, the force is:
磁场对运动电荷或载流导体施加力的作用。对于长度为 L、通有电流 I 的直导体，在磁感应强度为 B 的磁场中所受的力为：
F = BIL sinθ
where θ is the angle between the conductor and the magnetic field. When the conductor is perpendicular to the field, sinθ = 1 and F = BIL. The direction of the force is given by Fleming’s left-hand rule: thumb points along the force, first finger along the field, second finger along conventional current.
其中 θ 是导体与磁场方向的夹角。当导体与磁场垂直时，sinθ = 1，F = BIL。力的方向由弗莱明左手定则确定：拇指指向力的方向，食指指向磁场方向，中指指向常规电流方向。
For a single charge q moving with velocity v in a magnetic field, the force is:
对于以速度 v 在磁场中运动的单个电荷 q，所受的力为：
F = qvB sinθ
When the charge moves perpendicular to the field, it follows a circular path. Equating the magnetic force to the centripetal force gives r = mv / (qB). This equation is frequently used in questions about mass spectrometers and particle accelerators.
当电荷垂直于磁场运动时，它沿圆形轨道运动。将磁力与向心力相等，可得 r = mv / (qB)。该方程常用于质谱仪和粒子加速器的题目。
8. Magnetic Fields Created by Currents | 电流产生的磁场
A current-carrying wire generates a magnetic field around itself. For a long straight wire, the magnetic flux density at distance r from the wire is:
载流导线在周围产生磁场。对于长直导线，距离导线 r 处的磁感应强度为：
B = μ₀I / (2πr)
where μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ is the permeability of free space. The field lines form concentric circles around the wire, and the direction is given by the right-hand grip rule: point the thumb of your right hand along the conventional current; your fingers curl in the direction of the field.
其中 μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ 是真空磁导率。磁场线绕导线形成同心圆，方向由右手螺旋定则确定：右手拇指指向常规电流方向，四指弯曲的方向即为磁场方向。
For a solenoid, the magnetic field inside is nearly uniform and is given by:
对于螺线管，其内部磁场近似均匀，由下式给出：
B = μ₀nI
where n is the number of turns per unit length. The field is stronger inside the solenoid and weaker outside, making the solenoid an effective electromagnet. Adding a soft iron core increases B dramatically because iron has a high relative permeability.
其中 n 是单位长度的匝数。螺线管内部磁场较强、外部较弱，因此螺线管是有效的电磁铁。加入软铁芯会显著增大 B，因为铁具有很高的相对磁导率。
When two parallel wires carry currents in the same direction, they attract each other; when currents are opposite, they repel. This can be explained using the right-hand grip rule combined with Fleming’s left-hand rule.
两根平行导线通有同向电流时相互吸引，通有反向电流时相互排斥。这可以用右手螺旋定则结合弗莱明左手定则来解释。
9. Magnetic Flux and Electromagnetic Induction | 磁通量与电磁感应
Magnetic flux Φ through a surface is the product of the magnetic flux density and the area perpendicular to the field:
通过某一表面的磁通量 Φ 等于磁感应强度与垂直于磁场的面积的乘积：
Φ = BA cosθ
where θ is the angle between the magnetic field and the normal to the surface. The unit of flux is the weber (Wb), where 1 Wb = 1 T·m².
其中 θ 是磁场方向与表面法线方向的夹角。磁通量的单位是韦伯（Wb），1 Wb = 1 T·m²。
Electromagnetic induction occurs when the magnetic flux linked with a circuit changes. The induced emf is proportional to the rate of change of flux linkage. For a coil of N turns, the flux linkage is NΦ, and Faraday’s law gives:
当与电路交链的磁通量发生变化时，就会产生电磁感应。感应电动势与磁通链的变化率成正比。对于 N 匝线圈，磁通链为 NΦ，法拉第定律给出：
ε = −N dΦ/dt
Flux can change by moving a magnet relative to a coil, by rotating a coil in a magnetic field, or by changing the current in a nearby coil. ESAT questions often ask students to identify which action produces an induced current — any change in Φ is sufficient.
磁通量可以通过磁体与线圈的相对运动、线圈在磁场中转动或改变附近线圈中的电流来改变。ESAT 题目常要求判断哪种操作会产生感应电流——只要 Φ 发生变化即可。
10. Lenz’s Law and Energy Conservation | 楞次定律与能量守恒
Lenz’s law states that the direction of the induced current opposes the change that produced it. This is a direct consequence of conservation of energy: if the induced current aided the change, energy would be created from nothing.
楞次定律指出，感应电流的方向总是阻碍引起它的磁通量变化。这是能量守恒的直接推论：如果感应电流助长变化，能量就会凭空产生。
Consider a bar magnet falling through a vertical copper tube. The induced currents in the tube create magnetic fields that oppose the magnet’s motion, so the magnet falls more slowly than in free fall. The energy lost by the magnet appears as heat in the tube.
考虑一块条形磁铁从竖直铜管中下落的情形。管中产生的感应电流会形成阻碍磁铁运动的磁场，因此磁铁下落速度比自由落体慢。磁铁损失的能量以热的形式出现在铜管中。
A common ESAT question involves determining the direction of the induced current. Use Lenz’s law as follows: identify whether the flux is increasing or decreasing, determine the direction of the induced field needed to oppose this change, then use the right-hand grip rule to find the current direction.
ESAT 常见题目是判断感应电流的方向。按如下步骤使用楞次定律：判断磁通量是增加还是减小，确定为阻碍这一变化所需的感应磁场方向，再用右手螺旋定则确定电流方向。
The negative sign in Faraday’s law ε = −N dΦ/dt formally expresses Lenz’s law. In numerical problems involving a coil pulled out of a magnetic field, compute the flux change ΔΦ, divide by the time interval Δt, and multiply by N.
法拉第定律 ε = −N dΦ/dt 中的负号正式表达了楞次定律。在涉及线圈拉出磁场的数值问题中，计算磁通量变化 ΔΦ，除以时间间隔 Δt，再乘以匝数 N 即可。
11. AC Circuits and Transformers | 交流电路与变压器
An alternating current varies sinusoidally with time, producing a voltage that alternates direction. The root-mean-square (RMS) value of an alternating current is the equivalent direct current that would dissipate the same power in a resistor:
交流电随时间呈正弦变化，电压方向周期性反转。交流电的均方根（RMS）值是指在电阻上产生相同热功率的等效直流值：
I_RMS = I_peak / √2, V_RMS = V_peak / √2
Mains electricity is rated by its RMS value, so a 230 V supply has a peak voltage of 230 × √2 ≈ 325 V. This distinction is frequently tested in conceptual ESAT questions.
市电以 RMS 值标称，因此 230 V 电源的峰值电压为 230 × √2 ≈ 325 V。这一区别经常在 ESAT 概念题中出现。
A transformer uses electromagnetic induction to change the voltage of an alternating supply. It consists of a primary coil, a secondary coil, and a soft iron core. The voltages and the number of turns are related by:
变压器利用电磁感应改变交流电压。它由初级线圈、次级线圈和软铁芯组成。电压与匝数的关系为：
V_p / V_s = N_p / N_s
For an ideal transformer with 100% efficiency, the power is conserved:
对于效率为 100% 的理想变压器，功率守恒：
V_pI_p = V_sI_s
Transformers only work with AC because a changing current in the primary coil produces a changing magnetic flux, which induces an emf in the secondary coil. A direct current produces a constant flux and no induction. The soft iron core confines the magnetic flux and increases the flux linkage.
变压器只能使用交流电，因为初级线圈中变化的电流产生变化的磁通量，进而在次级线圈中感应出电动势。直流电产生恒定磁通，不会产生感应。软铁芯约束磁通量并增大磁通链。
12. Electromagnetic Waves | 电磁波
Maxwell’s great synthesis showed that changing electric and magnetic fields generate each other, producing self-propagating electromagnetic waves. In a vacuum, all electromagnetic waves travel at the speed of light:
麦克斯韦的伟大综合表明，变化的电场和磁场相互激发，产生自传播的电磁波。在真空中，所有电磁波都以光速传播：
c = 3.00 × 10⁸ m·s⁻¹
The electromagnetic spectrum, in order of increasing frequency, includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. All obey the wave equation c = fλ, where f is frequency and λ is wavelength.
电磁波谱按频率从低到高依次包括无线电波、微波、红外线、可见光、紫外线、X 射线和伽马射线。它们都满足波动方程 c = fλ，其中 f 是频率，λ 是波长。
Electric and magnetic fields in an electromagnetic wave oscillate perpendicular to each other and perpendicular to the direction of propagation. The ratio of the electric field strength to the magnetic field strength in a vacuum equals the speed of light.
电磁波中的电场和磁场相互垂直，且都垂直于传播方向。真空中电场强度与磁场强度之比等于光速。
For ESAT, remember that electromagnetic waves transfer energy without needing a medium. This is why light from the Sun reaches Earth through empty space, and why radio communication works between spacecraft and Earth.
对于 ESAT，请记住电磁波传播能量不需要介质。这就是太阳光能穿过真空到达地球，以及航天器与地球之间能进行无线电通信的原因。
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