📚 Year 13 CIE Physics: Case Study Practice | 案例分析实战演练
In Year 13 CIE A Level Physics, applying theoretical knowledge to solve real-world or exam-style case studies is essential for mastering the subject. This article presents ten carefully selected case studies spanning gravitational fields, electric fields, capacitance, magnetic fields, electromagnetic induction, alternating current, quantum physics, nuclear physics, medical physics and astrophysics. Each case is broken down step by step, with clear explanations and calculations that mirror the type of reasoning expected in CIE assessments.
在 Year 13 CIE A Level 物理中，将理论知识应用于解决实际或考试风格的案例分析是掌握这门学科的关键。本文精选了十个案例，涵盖引力场、电场、电容、磁场、电磁感应、交流电、量子物理、核物理、医学物理和天体物理。每个案例都逐步分解，配有清晰的解释和计算，完美契合 CIE 考试所要求的推理过程。
1. Gravitational Fields: Satellite Orbit Analysis | 引力场：卫星轨道分析
A communications satellite is to be placed in a geostationary orbit above the Earth’s equator. Given the Earth’s mass M = 5.97 × 10²⁴ kg, radius R = 6.37 × 10⁶ m, and the gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻², determine the orbital radius and the altitude of the satellite. The orbital period is T = 24 hours.
一颗通信卫星将要放置在地球赤道上方的地球同步轨道上。已知地球质量 M = 5.97 × 10²⁴ kg，半径 R = 6.37 × 10⁶ m，引力常数 G = 6.67 × 10⁻¹¹ N m² kg⁻²，请计算轨道半径和卫星高度。轨道周期 T = 24 小时。
Step 1: The gravitational force supplies the centripetal force required for circular motion.
步骤1：引力提供了圆周运动所需的向心力。
GMm / r² = m ω² r
where ω = 2π / T.
其中 ω = 2π / T。
Rearranging gives: r³ = (GM T²) / (4π²).
整理得到：r³ = (GM T²) / (4π²)。
Convert T to seconds: T = 24 × 3600 = 86400 s.
将 T 转化为秒：T = 24 × 3600 = 86400 s。
Substitute the values: r³ = [6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (86400)²] / (4π²) ≈ 7.54 × 10²² m³, so r ≈ 4.22 × 10⁷ m.
代入数值：r³ = [6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (86400)²] / (4π²) ≈ 7.54 × 10²² m³，所以 r ≈ 4.22 × 10⁷ m。
The altitude h = r – R = 4.22 × 10⁷ – 6.37 × 10⁶ ≈ 3.58 × 10⁷ m (about 35 800 km).
高度 h = r – R = 4.22 × 10⁷ – 6.37 × 10⁶ ≈ 3.58 × 10⁷ m（约 35 800 km）。
2. Electric Fields: Millikan-type Oil Drop | 电场：密立根式油滴实验
A tiny charged oil drop of mass m = 3.2 × 10⁻¹⁵ kg is held stationary between two horizontal parallel plates. The uniform electric field between the plates is E = 2.0 × 10⁴ N C⁻¹ directed upwards. Calculate the charge on the drop and discuss its quantisation. Take g = 9.81 m s⁻².
一个质量为 m = 3.2 × 10⁻¹⁵ kg 的带电微小油滴静止在两块水平平行板之间。板间匀强电场方向向上，场强 E = 2.0 × 10⁴ N C⁻¹。计算油滴所带电荷并讨论电荷的量子化。取 g = 9.81 m s⁻²。
For the drop to be stationary, the electric force must balance its weight.
为了使油滴静止，电场力必须与其重力平衡。
qE = mg
Hence the charge magnitude is q = mg / E.
因此电荷量的大小为 q = mg / E。
q = (3.2 × 10⁻¹⁵ × 9.81) / (2.0 × 10⁴) ≈ 1.57 × 10⁻¹⁸ C.
q = (3.2 × 10⁻¹⁵ × 9.81) / (2.0 × 10⁴) ≈ 1.57 × 10⁻¹⁸ C。
Since the elementary charge e = 1.60 × 10⁻¹⁹ C, the drop carries n = q/e ≈ 9.8, which is close to 10e. This integer multiple confirms the quantisation of charge, with small errors likely from measurements.
由于元电荷 e = 1.60 × 10⁻¹⁹ C，该油滴携带的电荷为 n = q/e ≈ 9.8，非常接近 10e。这个整数倍证实了电荷的量子化，小偏差可能来自测量误差。
3. Capacitance: Discharge of an RC Circuit | 电容：RC 电路的放电
A 2.0 μF capacitor is charged to 12 V and then discharged through a 1.0 MΩ resistor. Determine the time taken for the voltage across the capacitor to fall to 1.5 V.
一个 2.0 μF 的电容器充电至 12 V，然后通过一个 1.0 MΩ 的电阻放电。求电容器两端电压降至 1.5 V 所需的时间。
During discharge, the voltage follows an exponential decay: V = V₀ e^(-t / RC).
放电过程中，电压遵循指数衰减：V = V₀ e^(-t / RC)。
The time constant τ = RC = 1.0 × 10⁶ Ω × 2.0 × 10⁻⁶ F = 2.0 s.
时间常数 τ = RC = 1.0 × 10⁶ Ω × 2.0 × 10⁻⁶ F = 2.0 s。
Rearrange to find t: t = -RC ln(V / V₀) = -2.0 × ln(1.5 / 12).
整理求 t：t = -RC ln(V / V₀) = -2.0 × ln(1.5 / 12)。
ln(1.5 / 12) = ln(0.125) = -2.079, so t = -2.0 × (-2.079) ≈ 4.16 s.
ln(1.5 / 12) = ln(0.125) = -2.079，所以 t = -2.0 × (-2.079) ≈ 4.16 s。
Thus it takes approximately 4.2 seconds for the voltage to drop to 1.5 V.
因此电压降至 1.5 V 大约需要 4.2 秒。
4. Magnetic Fields: Current Balance Measurement | 磁场：电流天平测量
In a current balance experiment, a straight wire of length 5.0 cm carrying a current of 3.0 A experiences a magnetic force of 1.5 × 10⁻³ N when placed perpendicular to a uniform magnetic field. Determine the magnetic flux density B.
在一个电流天平实验中，一根长 5.0 cm 的直导线通有 3.0 A 电流，当它垂直于匀强磁场放置时受到 1.5 × 10⁻³ N 的磁力。求磁通量密度 B。
For a wire perpendicular to the field, the magnetic force is given by F = BIL.
对于垂直于磁场的导线，磁力由 F = BIL 给出。
B = F / (I L)
Convert length to metres: L = 5.0 cm = 0.050 m.
将长度转换为米：L = 5.0 cm = 0.050 m。
Substitute the values: B = 1.5 × 10⁻³ / (3.0 × 0.050) = 1.5 × 10⁻³ / 0.15 = 1.0 × 10⁻² T.
代入数值：B = 1.5 × 10⁻³ / (3.0 × 0.050) = 1.5 × 10⁻³ / 0.15 = 1.0 × 10⁻² T。
The magnetic flux density is 0.010 T (or 10 mT). This simple principle underpins methods for calibrating magnetic field probes.
磁通量密度为 0.010 T（或 10 mT）。这一简单原理是校准磁场探头方法的基础。
5. Electromagnetic Induction: AC Generator | 电磁感应：交流发电机
A rectangular coil with 200 turns and area 0.020 m² rotates in a uniform magnetic field of 0.50 T at 50 revolutions per second. Calculate the maximum induced emf.
一个矩形线圈有 200 匝，面积 0.020 m²，在 0.50 T 的匀强磁场中以每秒 50 转的转速旋转。计算最大感应电动势。
The magnetic flux linkage is Φ = NBA cos θ, where θ is the angle between the field and the normal to the coil.
磁通匝链数为 Φ = NBA cos θ，其中 θ 是磁场与线圈法线之间的夹角。
The induced emf is ε = -dΦ/dt, and the maximum emf occurs when the rate of change of flux is greatest: ε₀ = NBA ω.
感应电动势 ε = -dΦ/dt，当磁通量变化率最大时出现最大电动势：ε₀ = NBA ω。
The angular frequency ω = 2π f = 2π × 50 = 100π rad s⁻¹.
角频率 ω = 2π f = 2π × 50 = 100π rad s⁻¹。
ε₀ = 200 × 0.50 × 0.020 × 100π ≈ 200 × 0.50 × 0.020 × 314 = 628 V (approximately).
ε₀ = 200 × 0.50 × 0.020 × 100π ≈ 200 × 0.50 × 0.020 × 314 ≈ 628 V。
Therefore the peak output voltage of the generator is about 630 V.
因此发电机的峰值输出电压约为 630 V。
6. Alternating Current: Ideal Transformer | 交流电：理想变压器
A transformer has a primary coil of 500 turns connected to a 240 V AC mains supply. The secondary coil has 100 turns. Assuming an ideal transformer, find the output voltage and the primary current when the secondary delivers a current of 2.0 A.
一个变压器初级线圈有 500 匝，连接到 240 V 交流电源。次级线圈有 100 匝。假设为理想变压器，求输出电压以及当次级输出 2.0 A 电流时的初级电流。
For an ideal transformer, the voltage ratio equals the turns ratio: V_s / V_p = N_s / N_p.
对于理想变压器，电压比等于匝数比：V_s / V_p = N_s / N_p。
V_s = (N_s / N_p) × V_p = (100 / 500) × 240 = 48 V.
V_s = (N_s / N_p) × V_p = (100 / 500) × 240 = 48 V。
Power conservation for an ideal transformer means V_p I_p = V_s I_s.
理想变压器的能量守恒意味着 V_p I_p = V_s I_s。
I_p = V_s I_s / V_p = (48 × 2.0) / 240 = 0.40 A.
I_p = V_s I_s / V_p = (48 × 2.0) / 240 = 0.40 A。
Thus the secondary output is a stepped-down 48 V, and the primary draws 0.40 A.
因此次级输出为 48 V 的降压电压，初级电流为 0.40 A。
7. Quantum Physics: Photoelectric Effect Stopping Voltage | 量子物理：光电效应截止电压
Ultraviolet light of wavelength 300 nm strikes a metal surface with a work function of 2.0 eV. Calculate the maximum kinetic energy of the emitted electrons and the stopping voltage required to reduce the photocurrent to zero. (Planck constant h = 6.63 × 10⁻³⁴ J s, speed of light c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.60 × 10⁻¹⁹ J)
波长为 300 nm 的紫外光照射到功函数为 2.0 eV 的金属表面。计算发射出的电子的最大动能以及将光电流降为零所需的截止电压。（普朗克常数 h = 6.63 × 10⁻³⁴ J s，光速 c = 3.00 × 10⁸ m s⁻¹，1 eV = 1.60 × 10⁻¹⁹ J）
First, find the photon energy E = hf = hc / λ.
首先，计算光子能量 E = hf = hc / λ。
E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (300 × 10⁻⁹) = 6.63 × 10⁻¹⁹ J.
E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (300 × 10⁻⁹) = 6.63 × 10⁻¹⁹ J。
Convert to eV: E (eV) = 6.63 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ ≈ 4.14 eV.
转换为 eV：E (eV) = 6.63 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ ≈ 4.14 eV。
Maximum kinetic energy K_max = hf – φ = 4.14 – 2.0 = 2.14 eV.
最大动能 K_max = hf – φ = 4.14 – 2.0 = 2.14 eV。
The stopping voltage V_s is given by e V_s = K_max, so V_s = 2.14 V.
截止电压 V_s 满足 e V_s = K_max，所以 V_s = 2.14 V。
8. Nuclear Physics: Radiocarbon Dating | 核物理：放射性碳年代测定
A sample of ancient wood has a carbon-14 activity that is 25% of that found in a living tree. The half-life of ¹⁴C is 5730 years. Determine the age of the wood sample.
一块古木样本的碳-14 活度是存活树木的 25%。¹⁴C 的半衰期为 5730 年。判断该木样本的年代。
Radioactive decay follows the exponential law A = A₀ e^{−λt}, where λ = ln 2 / T₁/₂.
放射性衰变遵循指数规律 A = A₀ e^{−λt}，其中 λ = ln 2 / T₁/₂。
Given A / A₀ = 0.25, we can find t without calculating λ: 0.25 = (1/2)², meaning exactly two half-lives have elapsed.
已知 A / A₀ = 0.25，无需计算 λ：0.25 = (1/2)²，意味着恰好经过了两个半衰期。
Thus t = 2 × T₁/₂ = 2 × 5730 = 11460 years.
因此 t = 2 × T₁/₂ = 2 × 5730 = 11460 年。
Alternatively, using the equation t = (T₁/₂ / ln 2) × ln(A₀/A) gives the same result: t = (5730 / 0.693) × ln 4 = 5730 × 2 = 11460 years. The wood is about 11 500 years
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