AP 물리학 C: 전기 및 자기

Electric charge 电荷 comes in positive and negative; like charges repel, opposites attract. Charge is quantized 量子化 – every charge is a whole-number multiple of the elementary charge 基本电荷 $e=1.6\times10^{-19}\ \text{C}$ – and obeys conservation of charge 电荷守恒: charge is never created or destroyed, only moved. The force between two point charges 点电荷 is Coulomb's law 库仑定律:

an inverse-square 平方反比 force along the line joining them, with $\dfrac{1}{4\pi\varepsilon_0}=k=9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$. For several charges (AP uses four or fewer, unless symmetry helps), add the force vectors on the charge you care about – superposition 叠加.

Charge is a fundamental property 基本属性 of matter, and a point charge is a model that ignores the size of a charged object. The constant $\varepsilon_0$ is the permittivity of free space 真空介电常数 – a material property of the vacuum. The permittivity of matter differs from $\varepsilon_0$, set by how easily the material polarizes, which is why a dielectric between charges weakens the force. Note too that although the electric force is far stronger than gravity, gravity dominates at astronomical scales because large bodies are nearly neutral.

Worked example. Two $+2.0\ \mu\text{C}$ charges sit $0.30\ \text{m}$ apart: $F=\dfrac{kq_1q_2}{r^2}=\dfrac{9.0\times10^{9}(2.0\times10^{-6})^2}{(0.30)^2}=0.40\ \text{N}$, repulsive.

Worked example (vectors). Charges $+3.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$ sit at two corners of a right angle, each $0.30\ \text{m}$ from a $+1.0\ \mu\text{C}$ charge at the corner. Each exerts $F=\dfrac{(9.0\times10^{9})(3.0\times10^{-6})(1.0\times10^{-6})}{0.09}=0.30\ \text{N}$ – one a push, one a pull, at right angles to each other. The net force is $F_{\text{net}}=\sqrt{0.30^2+0.30^2}=0.42\ \text{N}$, pointing between them. Never add magnitudes blindly: components first.

A conductor 导体 lets charge move freely; an insulator 绝缘体 holds it in place. Objects gain net charge three ways:

• Charging by friction 摩擦起电: rubbing transfers electrons from one surface to the other.
• Charging by conduction 传导起电: touching a charged object shares charge of the same sign.
• Charging by induction 感应起电: a nearby charge pushes the conductor's charge apart; ground 接地 the far side, remove the ground wire first, and the conductor is left with the opposite sign – all without contact.

An insulator cannot pass charge, but it can develop polarization 极化: its molecules stretch or turn so one face is slightly positive and the other slightly negative. That is why a charged comb picks up neutral paper scraps. An electroscope 验电器 shows net charge by its leaves repelling; on any isolated conductor the excess charge sits entirely on the outer surface.

The electric field 电场 at a point is the force per unit charge that a small positive test charge 检验电荷 would feel there:

A charge placed in a field feels $\vec{F}=q\vec{E}$ – positive charges along the field, negative charges against it. Fields from several sources add as vectors (superposition again). Field lines 电场线 make the field visible: they point away from positive charge and toward negative, their density shows the strength, and they never cross.

In a uniform 均匀 field a charge feels a constant force, so it accelerates uniformly – launched sideways, it follows a parabola, just like a projectile in gravity.

For a continuous charge distribution 连续电荷分布, cut the object into pieces $dq$ and integrate their point-charge fields:

Write $dq$ using the right density: $\lambda\,dl$ with the linear charge density 线电荷密度 (a line or ring), $\sigma\,dA$ with the surface charge density 面电荷密度, or $\rho\,dV$ with the volume charge density 体电荷密度. Then use symmetry 对称性 to cancel components before integrating – that sentence is usually a scored step. AP's calculus cases: a finite line (collinear point or perpendicular bisector), an infinite wire or cylinder, a ring on its axis, and an arc at its centre.

Worked example (ring, on axis). A ring of radius $R$ carries charge $Q$. At a point $z$ along the axis, each element is a distance $\sqrt{R^2+z^2}$ away, and by symmetry the sideways components cancel, leaving only the axial part ($\cos\alpha=z/\sqrt{R^2+z^2}$):

Check the limits: $E=0$ at the centre ($z=0$), and far away ($z\gg R$) it becomes $kQ/z^2$ – a point charge, as it must.

Electric flux 电通量 measures how much field passes through a surface:

For a uniform field through a flat area, $\Phi_E=EA\cos\theta$, where the area vector 面积矢量 is perpendicular to the surface. Flux is positive where field lines exit a closed surface and negative where they enter – only the perpendicular component of $\vec{E}$ counts.

Gauss's law 高斯定律 – the first of Maxwell's equations – relates the flux through any closed surface to the charge inside it:

It is true for any closed surface, but it solves for $E$ only when symmetry lets you choose a Gaussian surface 高斯面 on which $E$ is constant and perpendicular (or parallel, contributing zero). The three symmetries AP tests: spherical symmetry 球对称 (concentric sphere), cylindrical symmetry 柱对称 (coaxial cylinder), and planar symmetry 平面对称 (a straddling "pillbox").

Worked example (sphere). Outside a sphere of total charge $Q$, a spherical Gaussian surface of radius $r$ gives $E(4\pi r^2)=Q/\varepsilon_0$, so $E=\dfrac{kQ}{r^2}$ – identical to a point charge at the centre. Inside a uniformly charged solid sphere of radius $R$, the surface encloses only $Q_{\text{enc}}=Q\,r^3/R^3$, so $E=\dfrac{kQr}{R^3}$: zero at the centre, growing linearly to the surface.

Worked example (wire). For an infinite wire with charge per length $\lambda$, take a coaxial cylinder of radius $r$ and length $L$. The ends contribute nothing ($\vec{E}\perp d\vec{A}$), so $E(2\pi rL)=\dfrac{\lambda L}{\varepsilon_0}$ and $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. The same pillbox method gives an infinite sheet's field, $E=\dfrac{\sigma}{2\varepsilon_0}$, uniform on both sides.

Exam skill. A full-credit Gauss's-law answer has four parts: name the surface, state the symmetry argument (why $E$ is constant and perpendicular on it), count $Q_{\text{enc}}$, then solve. Skipping the symmetry sentence loses the reasoning point – and remember Gauss's law also explains why $E=0$ inside any conductor in equilibrium.

• Use Coulomb's law $F=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r^2}$ and superpose forces as vectors (components, not magnitudes).
• For a continuous charge, integrate $d\vec E$ over $dq=\lambda\,dl,\ \sigma\,dA,\ \rho\,dV$; use symmetry to cancel components.
• The field points away from positive and toward negative charge; draw field lines correctly.
• Distinguish the force on a charge ($\vec F=q\vec E$) from the field the charge creates.
• Keep the constant $k=\tfrac{1}{4\pi\varepsilon_0}$ straight and check units.

When you push two like charges together, you do work 功 against the electric force. That work is stored by the pair as electric potential energy 电势能:

$U_E$ is defined as the work an external force must do to bring the charges from infinitely far away to a distance $r$ apart. The signs carry the physics:

• Like charges: $U_E>0$. An outside push was needed to bring them together. If released, they fly apart, and the stored energy becomes kinetic energy 动能.
• Opposite charges: $U_E<0$. The pair is bound. You must supply $|U_E|$ of work to pull them apart to infinity.

The electric force is a conservative force 保守力: the work it does depends only on the start and end points, not the path, and equals $-\Delta U_E$.

For a system 系统 of several charges, add the potential energy of every distinct pair:

Worked example. Three $+2.0\ \mu\text{C}$ charges sit at the corners of an equilateral triangle with sides $0.30\ \text{m}$. Each pair stores $U=\dfrac{kq^2}{r}=\dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.30}=0.12\ \text{J}$. There are three pairs, so $U_{\text{total}}=3\times0.12\ \text{J}=0.36\ \text{J}$. This is the work needed to assemble the triangle from infinity – and the kinetic energy the charges would share if all three were released.

Electric potential 电势 is electric potential energy per unit charge at a point in space. It is measured in volts ($1\ \text{V}=1\ \text{J/C}$):

Potential is a scalar 标量: it has a sign but no direction. This makes $V$ far easier to work with than the field $\vec{E}$. For several point charges 点电荷, use scalar superposition 叠加 – add the potentials with their signs:

Worked example. The potential $0.10\ \text{m}$ from a $+3.0\ \text{nC}$ point charge is $V=\dfrac{kq}{r}=\dfrac{9.0\times10^{9}(3.0\times10^{-9})}{0.10}=270\ \text{V}$. Now add a $-3.0\ \text{nC}$ charge $0.30\ \text{m}$ from the same point: it contributes $\dfrac{9.0\times10^{9}(-3.0\times10^{-9})}{0.30}=-90\ \text{V}$, so the total is $270-90=180\ \text{V}$. Just a signed sum – no vector components to resolve.

Continuous charge distributions

For a continuous charge distribution 连续电荷分布, cut it into small pieces $dq$, treat each piece as a point charge, and integrate 积分:

This is a scalar integral, so it is usually much easier than the field integral. AP expects you to handle four shapes with calculus: a thin ring (at a point on its axis), an arc (at its centre), a finite line or wire (collinear, or on its perpendicular bisector), and an infinitely long wire or cylinder (at a distance from its axis).

Worked example (ring of charge). A thin ring of radius $R$ carries total charge $Q$. For a point $P$ on the axis a distance $z$ from the centre, every element $dq$ sits at the same distance $r=\sqrt{R^2+z^2}$ from $P$. The distance is constant, so it comes out of the integral:

On an FRQ, say that $r$ is the same for every element – that statement is the marked step. The same idea gives $V=kQ/R$ at the centre of any arc, whatever its angle.

From potential to field

Potential and field contain the same information, linked by a derivative in each direction and by a path integral:

The field is the negative gradient 梯度 of the potential: $\vec{E}$ points from high potential to low potential, "downhill". Where $V$ changes quickly, the field is strong.

Worked example. Along the $x$-axis, $V(x)=3x^{2}-2x$ (volts, with $x$ in metres). Then $E_x=-\dfrac{dV}{dx}=2-6x\ \text{V/m}$. At $x=0.50\ \text{m}$, $E_x=-1.0\ \text{V/m}$: the field there points in the $-x$ direction.

The potential difference 电势差 between two points is the change in potential energy per unit charge moved between them: $\Delta V=\Delta U_E/q$. A battery makes a potential difference chemically: reactions inside separate positive from negative charge and hold the terminals a fixed $\Delta V$ apart.

Equipotential maps

An equipotential 等势面 line (also called an isoline 等值线) joins points that share the same potential. Four rules let you read any map:

• Isolines are perpendicular 垂直 to field lines 电场线 everywhere they cross.
• $\vec{E}$ points from high $V$ to low $V$ – never along an isoline. Moving a charge along an isoline takes no work.
• Closely spaced isolines mean a strong field: $E\approx-\Delta V/\Delta x$.
• You can sketch the field map from an isoline map, and the isoline map from a field map.

Exam skill. Given a map with isolines every $10\ \text{V}$ spaced about $2\ \text{cm}$ apart, estimate $E\approx\dfrac{10}{0.02}=500\ \text{V/m}$, pointing from the higher-value line to the lower one. The work an external agent does moving a charge $q$ slowly from $A$ to $B$ is $W=q(V_B-V_A)$ – the path taken does not matter.

When a charge $q$ moves between two points that differ in potential by $\Delta V$, the potential energy of the charge–field system changes by

If only the electric force acts, total energy is conserved, so the kinetic energy changes by the opposite amount:

Watch the two signs together: a positive charge speeds up when it moves to lower potential, while a negative charge speeds up when it moves to higher potential. Both are just the system trading potential energy for kinetic energy. This is how particle accelerators give charged particles their energy, and it underlies energy analysis in circuits.

Worked example. A proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) starts at rest and is accelerated through a potential drop of $500\ \text{V}$ (it moves to lower potential, so $\Delta V=-500\ \text{V}$ and $\Delta K=-q\,\Delta V=+8.0\times10^{-17}\ \text{J}$). Setting $\Delta K=\tfrac12 mv^2$:

Exam skill. Choose energy methods, not kinematics, whenever the field is non-uniform or the path curves: only the end-point potentials matter. A typical FRQ chain is $q\,\Delta V \to \Delta K \to v$, with one line of justification: "the electric force is conservative, so energy is conserved."

• Relate field and potential by $V=-\int \vec E\cdot d\vec l$ and $\vec E=-\nabla V$ (in 1-D, $E_x=-\tfrac{dV}{dx}$).
• Potential is a scalar — add contributions with signs, no vector components needed.
• Potential energy of a pair is $U=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r}$; use energy conservation for a charge's speed.
• Know that no work is done moving along an equipotential, which is perpendicular to $\vec E$.
• Choose the zero of potential (usually infinity) and state it.

In an ideal conductor 导体, electrons move freely. Put extra charge on one and the charges repel each other until, almost instantly, they settle into electrostatic equilibrium 静电平衡. In that state the conductor has four properties you must be able to state and use:

• The electric field inside the conductor is zero. If it were not, free electrons would still be moving.
• All excess charge sits on the surface. (Negative net charge = extra electrons on the surface; positive = a shortage of electrons there.)
• The field just outside is perpendicular 垂直 to the surface, with magnitude $E=\sigma/\varepsilon_0$. Any parallel component would push charge sideways along the surface.
• The whole conductor is one equipotential surface 等势面: every point, inside and on the surface, is at the same potential.

Charge density is largest where the surface curves sharply – at points and edges – so the outside field is strongest there. In an external field a conductor polarizes: charge shifts on its surface so that the interior stays field-free and the body stays an equipotential. Surrounding a region with a closed conducting shell keeps outside fields out entirely – electrostatic shielding 静电屏蔽, the idea behind the Faraday cage 法拉第笼.

When two conductors touch (or are wired together), charge flows between them until both surfaces reach the same potential – that is the stopping condition, not "equal charges". A larger sphere holds more charge at the same potential ($V=kQ/R$), so it takes the larger share.

Worked example. A small sphere of radius $R$ carrying $+6.0\ \mu\text{C}$ touches a distant sphere of radius $2R$, then they separate. Equal potentials require $\dfrac{kq_1}{R}=\dfrac{kq_2}{2R}$, so $q_2=2q_1$. With $q_1+q_2=6.0\ \mu\text{C}$: $q_1=2.0\ \mu\text{C}$ and $q_2=4.0\ \mu\text{C}$.

Ground 接地 is an idealized reference at zero potential that can absorb or supply any amount of charge. Grounding a conductor while an external charge is nearby leaves the conductor with a net induced charge 感应电荷: the external field pushes charge of one sign to ground, and cutting the ground wire before removing the external charge traps the rest.

A capacitor 电容器 stores charge on two conductors separated by a gap: $+Q$ on one plate, $-Q$ on the other. Its capacitance 电容 relates the stored charge to the potential difference between the plates:

Capacitance depends only on geometry and the material in the gap – never on $Q$ or $\Delta V$ themselves.

Worked derivation (parallel plates). For a parallel-plate capacitor 平行板电容器 with plate area $A$ and small gap $d$: Gauss's law plus superposition 叠加 gives a uniform 均匀 field between the plates, $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$ (each plate alone contributes $\sigma/2\varepsilon_0$; between the plates the two add, outside they cancel). A uniform field means $\Delta V=Ed=\dfrac{Qd}{\varepsilon_0 A}$, so

This $E\to\Delta V\to C$ chain is a standard FRQ derivation – learn it as three steps, and quote each one. The same method handles the other two shapes AP expects: concentric spheres ($C=4\pi\varepsilon_0\dfrac{ab}{b-a}$) and a coaxial cylinder of length $L$, where $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ gives $\Delta V=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$ and so $C=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.

Worked example. Plates of area $A=0.020\ \text{m}^2$ and gap $d=1.0\ \text{mm}$: $C=\dfrac{8.85\times10^{-12}(0.020)}{1.0\times10^{-3}}=1.8\times10^{-10}\ \text{F}$. Charged to $100\ \text{V}$ it holds $Q=CV=1.8\times10^{-8}\ \text{C}$.

Because the field between the plates is uniform, a charged particle there feels a constant force, so it moves with constant acceleration – exactly like projectile motion 抛体运动 in gravity: constant speed across, uniform acceleration towards a plate.

Worked example. An electron enters midway between plates at $v_0=2.0\times10^{7}\ \text{m/s}$, parallel to them. The field is $E=1.0\times10^{3}\ \text{N/C}$ and the plates are $4.0\ \text{cm}$ long. Acceleration: $a=\dfrac{eE}{m}=\dfrac{(1.6\times10^{-19})(1.0\times10^{3})}{9.11\times10^{-31}}=1.8\times10^{14}\ \text{m/s}^2$. Time between the plates: $t=\dfrac{0.040}{2.0\times10^{7}}=2.0\times10^{-9}\ \text{s}$. Deflection: $y=\tfrac12at^{2}=\tfrac12(1.8\times10^{14})(2.0\times10^{-9})^{2}\approx3.5\times10^{-4}\ \text{m}$ – about $0.35\ \text{mm}$ towards the positive plate.

Storing charge takes work 功: an external force must move each bit of charge against the field of the charge already there. The total work ends up as stored potential energy,

The factor $\tfrac12$ is the area of the triangle under the $Q$–$\Delta V$ line: the first charge moved across costs almost nothing, the last costs the full $\Delta V$.

Capacitors combine oppositely to resistors: in parallel 并联 the capacitances add ($C_{\text{eq}}=C_1+C_2$, same $\Delta V$, charges add), while in series 串联 the reciprocals add ($\tfrac{1}{C_{\text{eq}}}=\tfrac{1}{C_1}+\tfrac{1}{C_2}$, same $Q$, voltages add).

Worked example. A $2\ \mu\text{F}$ and a $4\ \mu\text{F}$ capacitor in series: $\tfrac{1}{C}=\tfrac12+\tfrac14$, so $C=\tfrac43\ \mu\text{F}$. The same pair in parallel: $6\ \mu\text{F}$.

A dielectric 电介质 is an insulating material. Its charges cannot travel, but in an external field each molecule stretches or turns slightly – the material becomes polarized 极化. The lined-up molecules create their own small field opposite to the applied one, so the net field inside the material drops:

where $\kappa$ is the dielectric constant 介电常数, the ratio of the material's permittivity to the permittivity of free space 真空介电常数.

Filling a capacitor with dielectric multiplies its capacitance:

What happens next depends on what stays fixed – a favourite exam trap:

Worked example. A $100\ \text{pF}$ capacitor is charged to $12\ \text{V}$, disconnected, then filled with a $\kappa=3$ dielectric. $Q$ is trapped, so $\Delta V$ falls to $4\ \text{V}$ and the stored energy falls to one-third – the missing energy went into pulling the dielectric in. Reconnected to the $12\ \text{V}$ battery instead, the capacitor would hold three times the charge and three times the energy.

Exam skill. Always begin dielectric questions by writing down which quantity is held fixed ($Q$ or $\Delta V$), then let $C=\kappa C_0$ drive everything else through $Q=C\Delta V$ and $U=\tfrac12 C(\Delta V)^2$.

• In electrostatic equilibrium a conductor has $\vec E=0$ inside and all excess charge on the surface.
• Apply Gauss's law $\oint \vec E\cdot d\vec A=\tfrac{q_{enc}}{\varepsilon_0}$ with a symmetry-matched Gaussian surface (sphere, cylinder, pillbox).
• The whole conductor is one equipotential, and the surface field is perpendicular to it.
• For a capacitor use $C=\tfrac{Q}{V}$, energy $U=\tfrac12 CV^2$, and how a dielectric raises $C$.
• Pick the Gaussian surface so $\vec E$ is constant and parallel (or zero) on each part.

A magnetic field 磁场 $\vec{B}$ is a vector field 矢量场 that surrounds magnets, moving charges, and currents; it determines the magnetic force on any moving charge placed in it. Magnetic field lines 磁感线 must form closed loops: they leave the north pole, return to the south pole, and continue through the magnet. Denser lines mean a stronger field.

Closed loops are the content of Gauss's law for magnetism, the second of Maxwell's equations 麦克斯韦方程组:

The net magnetic flux through any closed surface is zero – as many field lines leave as enter. That is exactly the statement that isolated magnetic monopoles 磁单极子 do not exist. Every source of magnetism is a magnetic dipole 磁偶极子 (a north–south pair), made by circulating charge – in materials, the motion of electrons. Break a bar magnet in half and you get two smaller dipoles, never a lone pole. Like poles repel, opposite poles attract, and a free dipole – a compass 指南针 – rotates to line up with the local field. Earth's own field is approximately a dipole, which is why a compass works. Dipole fields weaken with distance.

How a material responds to an external field depends on how its internal dipoles behave:

• Ferromagnetic 铁磁性 (iron, nickel, cobalt): an external field aligns whole magnetic domains 磁畴, and the alignment survives after the field is removed – a permanent magnet.
• Paramagnetic 顺磁性 (aluminum, titanium): dipoles align weakly with the field but relax when it is removed.
• Diamagnetic 抗磁性 (all materials): the electronic structure produces a usually weak alignment opposite to the field.

The strength of a material's response is its magnetic permeability 磁导率. Free space has the constant value $\mu_0$ (the vacuum permeability); the permeability of matter differs from $\mu_0$ and is not even constant – it varies with temperature, orientation, and field strength.

A moving charge does two things: it creates a magnetic field, and it feels a force in an external one. The field it creates at a point is perpendicular 垂直 to both its velocity and the position vector from charge to point (right-hand rule again), and is largest where those two are perpendicular – zero directly ahead of or behind the motion.

The force on a charge moving in a field $\vec{B}$ is the cross product

perpendicular to both $\vec{v}$ and $\vec{B}$ – point the fingers of your right hand along $\vec{v}$, curl toward $\vec{B}$, and the thumb gives the force on a positive charge (reverse it for negative). Because $\vec{F}_B\perp\vec{v}$, the magnetic force does no work on the charge: it changes direction, never speed.

A charge moving perpendicular to a uniform field therefore travels in a circle, the magnetic force supplying the centripetal force 向心力:

Notice the period $T$ does not depend on the speed – faster particles ride larger circles in the same time.

Worked example. A proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) enters a $0.50\ \text{T}$ field at $2.0\times10^{5}\ \text{m/s}$, perpendicular to it. The magnetic force $F=qvB=1.6\times10^{-14}\ \text{N}$ bends it into a circle of radius $r=\dfrac{mv}{qB}=\dfrac{1.67\times10^{-27}(2.0\times10^{5})}{1.6\times10^{-19}(0.50)}=4.2\times10^{-3}\ \text{m}$.

In a region with both an electric and a magnetic field, the two forces act independently and add as vectors. Balancing them makes a velocity selector 速度选择器: with $\vec{E}$ and $\vec{B}$ crossed, only charges with $qE=qvB$, i.e. $v=E/B$, pass straight through. The Hall effect 霍尔效应 is the same physics inside a conductor: a field with a component perpendicular to the current pushes the moving carriers sideways, charge builds up on one face, and a measurable potential difference appears across the conductor – its sign reveals whether the carriers are positive or negative.

A current is a stream of moving charges, so it creates a magnetic field. The Biot–Savart law 毕奥-萨伐尔定律 adds up the field of each current element:

Around any straight segment the field vectors are tangent 相切 to concentric circles 同心圆 centred on the wire – no component toward, away from, or along the wire. Curl your right hand around the wire with the thumb along the current: your fingers give the field direction.

The Biot–Savart integrals AP expects: a long straight wire ($B=\dfrac{\mu_0 I}{2\pi r}$, or a point on the perpendicular bisector of a finite wire), and the centre of a circular loop,

where every element $d\vec{l}$ is perpendicular to $\hat{r}$ and equally distant $R$ – say that in an FRQ derivation. An arc that is a fraction of a full circle contributes that same fraction of $\mu_0 I/2R$ at its centre.

A field also pushes on a current-carrying wire, element by element:

Worked example. Two long parallel wires a distance $d=0.10\ \text{m}$ apart each carry $5.0\ \text{A}$ in the same direction. Wire 1's field at wire 2 is $B=\dfrac{\mu_0 I}{2\pi d}=1.0\times10^{-5}\ \text{T}$, so wire 2 feels $\dfrac{F}{L}=I B=5.0\times10^{-5}\ \text{N/m}$, pulled toward wire 1. Same-direction currents attract; opposite currents repel.

For symmetric current distributions, Ampère's law 安培定律 finds the field far faster than Biot–Savart:

Choose an Amperian loop 安培环路 that matches the symmetry, so that $B$ is constant along the loop (and parallel to it) and comes out of the integral. AP applies it to long straight wires, long solenoids, and conductors carrying a current density 电流密度 (slabs and cylinders); combinations are handled by superposition 叠加.

Worked example. Find the field $0.10\ \text{m}$ from a long wire carrying $5.0\ \text{A}$. A circular loop of radius $r$ shares the field's symmetry, so $B(2\pi r)=\mu_0 I$ and

A long solenoid 螺线管 (assumed: uniform field inside, negligible field outside) is the other classic. Take a rectangular loop with one side of length $L$ inside, parallel to the axis: only that side contributes to the integral, so $BL=\mu_0 (nL) I$ and

with $n$ the turns per metre.

Worked example (cylinder). A solid cylindrical conductor of radius $R$ carries current $I$ spread uniformly over its cross-section. Inside ($r), a circular loop encloses $I_{\text{enc}}=I\dfrac{r^2}{R^2}$, so $B=\dfrac{\mu_0 I r}{2\pi R^2}$ – the field grows linearly to the surface, then falls off as $1/r$ outside.

Exam skill. Every Ampère's-law answer earns its marks in the setup: name the loop, state why $B$ is constant and parallel to it (symmetry), and count $I_{\text{enc}}$ carefully before you solve. Maxwell's fourth equation adds one more idea you should know qualitatively: a changing electric field also creates a magnetic field – but AP will not ask you to compute with that term.

• The magnetic force $\vec F=q\vec v\times\vec B$ is perpendicular to velocity — use the right-hand rule and note it does no work.
• A charge in a uniform field moves in a circle with $r=\tfrac{mv}{qB}$.
• Find the field of currents with Biot–Savart $d\vec B=\tfrac{\mu_0}{4\pi}\tfrac{I\,d\vec l\times\hat r}{r^2}$ or Ampère's law $\oint \vec B\cdot d\vec l=\mu_0 I_{enc}$ when there is symmetry.
• Force on a wire is $\vec F=I\vec L\times\vec B$; keep the cross-product direction straight.
• Distinguish the force on a charge from the field the current creates.

Magnetic flux 磁通量 measures how much magnetic field passes through a surface. For a uniform field through a flat loop it is the dot product $\Phi_B=\vec{B}\cdot\vec{A}=BA\cos\theta$; in general it is the surface integral 面积分

The area vector 面积矢量 is perpendicular to the surface (outward from a closed one), and the sign of the flux comes from the dot product. Flux changes if $B$ changes, the area changes, or the loop turns – keep all three routes in mind, because each one is an exam question.

A changing flux induces an emf 电动势 – Faraday's law 法拉第定律:

With constant area, $\varepsilon=-A\,\dfrac{dB_\perp}{dt}$; with constant field, $\varepsilon=-B\,\dfrac{dA_\perp}{dt}$. A coil of $N$ turns multiplies the single-loop emf: $|\varepsilon_{\text{sol}}|=N\left|\dfrac{d\Phi_B}{dt}\right|$.

Lenz's law 楞次定律 is the minus sign: the induced current 感应电流 flows so that its own magnetic field opposes the change in flux that created it. Push a magnet toward a loop and the loop pushes back; pull it away and the loop pulls it in. Use the right-hand rule 右手定则 to turn "oppose the change" into a current direction.

A rod of length $L$ sliding at speed $v$ across a field is the special case worth memorising – the motional emf 动生电动势 $\varepsilon=BLv$.

Worked example. A $0.20\ \text{m}$ rod slides at $3.0\ \text{m/s}$ across a $0.50\ \text{T}$ field: $\varepsilon=BLv=0.50(0.20)(3.0)=0.30\ \text{V}$. Equivalently, if a single loop's flux drops from $0.020\ \text{Wb}$ to $0.008\ \text{Wb}$ in $0.030\ \text{s}$, the average emf is $\varepsilon=\dfrac{0.012}{0.030}=0.40\ \text{V}$.

Faraday's law is also the third of Maxwell's equations 麦克斯韦方程组, in a deeper form: a changing magnetic flux creates a circulating electric field, $\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_B}{dt}$ – that field is what pushes the charges around the loop. Together, Maxwell's equations predict electromagnetic waves 电磁波 travelling at $c=1/\sqrt{\varepsilon_0\mu_0}$ (you should know this connection, but AP will not ask you to derive it).

Once an induced current flows, the external field exerts forces on it ($\vec{F}=\int I\,d\vec{l}\times\vec{B}$) – and by Lenz's law those forces always resist the motion that causes the induction. Only the segments of the loop actually inside the field feel a force, which can make a loop accelerate, rotate, or brake. This is the origin of eddy currents 涡电流 braking, and you can apply Newton's second law to a moving loop or rod like any other mechanics problem.

The classic setup is a rod sliding on conducting rails connected by a resistor:

Worked example (the full chain). Rails $L=0.20\ \text{m}$ apart with resistance $R=0.60\ \Omega$ sit in a $0.50\ \text{T}$ field into the page. The rod is pushed at a constant $3.0\ \text{m/s}$. Then: $\varepsilon=BLv=0.30\ \text{V}$; $I=\varepsilon/R=0.50\ \text{A}$; the field pushes back on the rod with $F=BIL=0.50(0.50)(0.20)=0.050\ \text{N}$. The pushing force does work at $P=Fv=0.15\ \text{W}$ – exactly the $P=I^2R=0.15\ \text{W}$ dissipated in the resistor. Mechanical work becomes electrical energy: that is a generator 发电机, and energy is conserved. Released with no push, the rod slows exponentially: $ma=-\dfrac{B^2L^2}{R}v$.

Exam skill. FRQs walk this exact chain: flux $\to$ emf $\to$ current $\to$ force $\to$ Newton's second law. Write each link separately and check the direction with Lenz's law at the end.

Inductance 电感 is a conductor's tendency to oppose a change in its own current: changing current changes its own flux, which self-induces an emf. From Faraday's law,

An inductor 电感器 is a circuit element built to have large inductance – usually a solenoid 螺线管, where geometry gives

with $N$ total turns, area $A$, length $\ell$, and the magnetic permeability 磁导率 of the core. (Straight wires are modelled as having zero inductance.) An inductor carrying current $I$ stores energy in its magnetic field:

which can later be dissipated in a resistor or moved into a capacitor – conservation of energy applies as usual.

In an RL circuit RL电路, Kirchhoff's loop rule for a battery $\varepsilon$, resistor $R$, and inductor $L$ in series gives a differential equation 微分方程:

The time constant 时间常数 $\tau$ sets the pace: after one $\tau$ a rising current reaches about $63\%$ of its final value (a decaying one falls to $37\%$); it is also how long the change would take at the initial rate. Learn the two limits – they answer most conceptual questions:

• Just after the switch closes ($t=0$): current cannot jump, so the inductor momentarily blocks it, self-inducing an emf equal and opposite to the applied potential difference.
• Long after ($t\gg\tau$): the current is steady, $dI/dt=0$, and the inductor behaves as a plain wire – the exact opposite of a capacitor.

Worked example. $\varepsilon=12\ \text{V}$, $R=6.0\ \Omega$, $L=3.0\ \text{H}$: final current $\varepsilon/R=2.0\ \text{A}$, $\tau=L/R=0.50\ \text{s}$. At $t=0.50\ \text{s}$ the current is $2.0(1-e^{-1})\approx1.3\ \text{A}$. At $t=0$ the inductor's potential difference is the full $12\ \text{V}$; as $t\to\infty$ it falls to zero. Current, inductor voltage, and stored energy are all exponential in time.

An LC circuit LC电路 has no resistance, so nothing dissipates energy: it oscillates 振荡, sloshing energy between the capacitor's electric field and the inductor's magnetic field. The loop rule gives

the same equation as a mass on a spring – simple harmonic motion 简谐运动 with charge playing the role of displacement and

The total energy $\dfrac{q^2}{2C}+\tfrac12LI^2$ stays constant: all in the capacitor at maximum charge, all in the inductor at maximum current.

Worked example. $L=2.0\ \text{H}$, $C=8.0\ \mu\text{F}$: $\omega=\dfrac{1}{\sqrt{2.0(8.0\times10^{-6})}}=250\ \text{rad/s}$, period $T=2\pi/\omega\approx0.025\ \text{s}$. If the capacitor starts charged to $12\ \text{V}$, then $U=\tfrac12CV^2=5.8\times10^{-4}\ \text{J}$, and the maximum current follows from $\tfrac12LI_{\max}^2=U$: $I_{\max}=\sqrt{2U/L}=0.024\ \text{A}$ – conservation of energy 能量守恒, no calculus needed.

• Compute flux $\Phi_B=\int \vec B\cdot d\vec A$ and get the induced EMF from Faraday's law $\varepsilon=-\tfrac{d\Phi_B}{dt}$.
• Use Lenz's law (the minus sign) to fix the direction: the induced current opposes the change in flux.
• Flux changes three ways — changing $B$, changing area, or changing angle — identify which and differentiate.
• For a rod of length $L$ moving at speed $v$, the motional EMF is $BLv$.
• An inductor stores energy $\tfrac12 LI^2$ and resists changes in current (RL time constant $\tau=L/R$).