PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. State whether a quoted angle is measured from the normal or from the plane before using $\Phi = BA\cos\theta$. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。使用 $\Phi = BA\cos\theta$ 前先写明所给角度是相对法线还是相对平面。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1HL ONLYflux and flux linkage geometry磁通量与磁链几何[4 marks]
A circular coil of $80$ turns and radius $4.0\ \mathrm{cm}$ is placed in a uniform magnetic field of $0.25\ \mathrm{T}$. The coil is tilted so that the normal to its plane makes an angle of $30^{\circ}$ with the field.一个 $80$ 匝、半径 $4.0\ \mathrm{cm}$ 的圆线圈置于 $0.25\ \mathrm{T}$ 的均匀磁场中。线圈倾斜，使其平面法线与场成 $30^{\circ}$。
(a)Calculate the magnetic flux through one turn of the coil.计算穿过线圈单匝的磁通量。[2]
(b)Hence determine the flux linkage of the whole coil, and state the unit.由此求整个线圈的磁链，并写出单位。[2]
Q2MEDIUMPaper 1HL ONLYmotional emf on rails导轨上的动生电动势[6 marks]
A straight conducting rod of length $0.50\ \mathrm{m}$ slides at a constant $6.0\ \mathrm{m\,s^{-1}}$ along frictionless horizontal rails. A uniform field of $0.40\ \mathrm{T}$ is directed vertically, perpendicular to the plane of the rails. The rails are joined at one end by a $5.0\ \Omega$ resistor.一根长 $0.50\ \mathrm{m}$ 的直导体棒以恒定 $6.0\ \mathrm{m\,s^{-1}}$ 沿无摩擦水平导轨滑动。$0.40\ \mathrm{T}$ 的均匀场竖直方向，垂直于导轨平面。导轨一端由 $5.0\ \Omega$ 电阻连接。
(a)Calculate the emf induced across the rod.计算棒两端感应的电动势。[2]
(b)Calculate the current in the circuit.计算电路中的电流。[2]
(c)Calculate the power dissipated in the resistor.计算电阻中耗散的功率。[2]
A coil of $400$ turns and cross-sectional area $2.5\times 10^{-3}\ \mathrm{m^{2}}$ has its plane perpendicular to a uniform magnetic field. The field falls uniformly from $0.60\ \mathrm{T}$ to zero in a time of $0.015\ \mathrm{s}$.一个 $400$ 匝、横截面积 $2.5\times 10^{-3}\ \mathrm{m^{2}}$ 的线圈，其平面垂直于均匀磁场。场在 $0.015\ \mathrm{s}$ 内由 $0.60\ \mathrm{T}$ 匀减到零。
(a)Calculate the change in flux through one turn of the coil.计算穿过线圈单匝的磁通量变化。[2]
(b)State Faraday's law and use it to calculate the magnitude of the induced emf.陈述法拉第定律，并用它计算感应电动势的大小。[3]
(c)State one change to the experiment that would double the induced emf without changing the field.写出在不改变磁场的前提下使感应电动势翻倍的一种改动。[1]
Q4HARDPaper 1HL ONLYLenz's law and energy楞次定律与能量[6 marks]
A horizontal circular conducting ring lies in a region where a uniform magnetic field is directed vertically downward through it. The strength of the field is steadily increasing.一个水平圆形导电环位于某区域，均匀磁场竖直向下穿过它。场强正在稳定增大。
(a)State Lenz's law.陈述楞次定律。[1]
(b)Determine the direction of the induced current in the ring, as seen from above, justifying your answer by reference to the change in flux.求从上方看环中感应电流的方向，并结合磁通量的变化加以论证。[3]
(c)Explain why Lenz's law is a consequence of the conservation of energy.解释楞次定律为何是能量守恒的结果。[2]
An ideal step-up transformer has a primary of $500$ turns connected to a $240\ \mathrm{V}$ AC supply, and delivers an output of $12\,000\ \mathrm{V}$. The output feeds a transmission cable of total resistance $8.0\ \Omega$ that carries $6.0\ \mathrm{kW}$ of power.一台理想升压变压器初级为 $500$ 匝，接 $240\ \mathrm{V}$ 交流电源，输出 $12\,000\ \mathrm{V}$。输出端接一条总电阻 $8.0\ \Omega$ 的输电线，输送 $6.0\ \mathrm{kW}$ 的功率。
(a)Calculate the number of turns on the secondary coil.计算次级线圈的匝数。[2]
(b)Calculate the current in the transmission cable and hence the power dissipated in it.计算输电线中的电流，并由此求其耗散功率。[3]
(c)The same $6.0\ \mathrm{kW}$ could in principle be sent along the cable at $240\ \mathrm{V}$ instead. Calculate the power that would then be dissipated, and state what this shows about high-voltage transmission.原则上同样的 $6.0\ \mathrm{kW}$ 也可以 $240\ \mathrm{V}$ 沿该线输送。计算此时的耗散功率，并说明这对高压输电意味着什么。[3]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward reading the emf as the gradient of a flux-linkage graph and correct handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查把电动势读作磁链图斜率，以及对不确定度的正确处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q6HARDPaper 1BHL ONLYflux-time graph to emf磁通量-时间图求电动势[10 marks]
The flux linkage $N\Phi$ through a coil is recorded against time $t$. The graph consists of three straight-line segments, summarised in the table below.记录穿过某线圈的磁链 $N\Phi$ 随时间 $t$ 的变化。图线由三段直线组成，汇总于下表。
Segment区段
$t\ /\ \mathrm{s}$
$N\Phi\ /\ \mathrm{Wb}$
1
$0.00 \to 0.20$
$0.00 \to 0.40$
2
$0.20 \to 0.40$
$0.40 \to 0.40$
3
$0.40 \to 0.50$
$0.40 \to 0.00$
(a)Explain how the magnitude of the induced emf can be obtained from a graph of flux linkage against time.解释如何从磁链-时间图得到感应电动势的大小。[2]
(b)Calculate the magnitude of the induced emf during each of the three segments.计算三段中各段感应电动势的大小。[3]
(c)Sketch the graph of induced emf against time for the whole interval $0$ to $0.50\ \mathrm{s}$, marking the value of the emf on each segment.画出整个 $0$ 至 $0.50\ \mathrm{s}$ 区间感应电动势对时间的图，并在每段标出电动势的数值。[3]
(d)In segment 1 the flux-linkage change is $0.40\ \mathrm{Wb}$ with an absolute uncertainty of $\pm 0.02\ \mathrm{Wb}$, and the time interval is $0.20\ \mathrm{s}$ with an uncertainty of $\pm 0.01\ \mathrm{s}$. Calculate the percentage uncertainty in the segment-1 emf.在第 1 段中，磁链变化为 $0.40\ \mathrm{Wb}$，绝对不确定度 $\pm 0.02\ \mathrm{Wb}$；时间间隔为 $0.20\ \mathrm{s}$，不确定度 $\pm 0.01\ \mathrm{s}$。计算第 1 段电动势的百分比不确定度。[2]
Q7HARDPaper 1BHL ONLYrotating coil, phase and uncertainty旋转线圈、相位与不确定度[12 marks]
A coil of $200$ turns and area $0.015\ \mathrm{m^{2}}$ rotates at a steady frequency $f = 60\ \mathrm{Hz}$ in a uniform field of $0.080\ \mathrm{T}$. Take the coil to be face-on to the field (flux a maximum) at $t = 0$.一个 $200$ 匝、面积 $0.015\ \mathrm{m^{2}}$ 的线圈在 $0.080\ \mathrm{T}$ 的均匀场中以稳定频率 $f = 60\ \mathrm{Hz}$ 旋转。取 $t = 0$ 时线圈正对场（磁通量最大）。
(a)Calculate the maximum flux linkage of the coil.计算线圈的最大磁链。[2]
(b)Calculate the angular frequency of rotation and hence the peak emf, $\varepsilon_{0} = NBA\omega$.计算旋转角频率，由此求峰值电动势 $\varepsilon_{0} = NBA\omega$。[3]
(c)On the same axes, sketch the flux linkage and the induced emf against time for one full rotation, and state the phase difference between them.在同一坐标上画出一整圈内磁链与感应电动势随时间的曲线，并写出两者的相位差。[4]
(d)The frequency is known to $\pm 1\ \mathrm{Hz}$ and the field to $\pm 0.002\ \mathrm{T}$, while $N$ and $A$ are exact. Calculate the percentage uncertainty in the peak emf.已知频率为 $\pm 1\ \mathrm{Hz}$、场为 $\pm 0.002\ \mathrm{T}$，而 $N$ 与 $A$ 为精确值。计算峰值电动势的百分比不确定度。[3]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem with a clear diagram and a stated sign convention. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先画清晰的示意图并写明正负号约定。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
A simple AC generator consists of a coil of $80$ turns and area $0.020\ \mathrm{m^{2}}$ rotating at a steady frequency of $50\ \mathrm{Hz}$ in a uniform magnetic field of $0.15\ \mathrm{T}$.一台简单交流发电机由 $80$ 匝、面积 $0.020\ \mathrm{m^{2}}$ 的线圈构成，在 $0.15\ \mathrm{T}$ 的均匀磁场中以稳定频率 $50\ \mathrm{Hz}$ 旋转。
(a)State the principle by which the rotating coil generates an emf, naming the law involved.写出旋转线圈产生电动势所依据的原理，并指明所涉及的定律。[2]
(b)Calculate the angular frequency and the peak emf of the generator.计算发电机的角频率与峰值电动势。[3]
(c)Write the emf as a function of time, and calculate the root-mean-square value of the output emf.把电动势写成时间的函数，并计算输出电动势的有效值（均方根值）。[3]
(d)The rotation frequency is now doubled to $100\ \mathrm{Hz}$. State, with reasons, how the peak emf and the output frequency change.现在把旋转频率加倍至 $100\ \mathrm{Hz}$。说明峰值电动势与输出频率如何变化，并给出理由。[2]
(e)State the position of the coil, relative to the field, at the instant the emf is a maximum, and explain this in terms of the rate of change of flux.写出电动势取最大值瞬间线圈相对场的位置，并用磁通量变化率加以解释。[2]
Q9HARDPaper 2HL ONLYrod on rails, Lenz, energy balance导轨上的棒、楞次定律与能量平衡[10 marks]
A conducting rod of length $0.25\ \mathrm{m}$ rests on two horizontal rails joined by a $0.20\ \Omega$ resistor. A uniform field of $0.50\ \mathrm{T}$ points vertically downward through the plane of the rails. An external force pushes the rod at a constant $4.0\ \mathrm{m\,s^{-1}}$. The rails and rod have negligible resistance.一根长 $0.25\ \mathrm{m}$ 的导体棒搁在两条由 $0.20\ \Omega$ 电阻连接的水平导轨上。$0.50\ \mathrm{T}$ 的均匀场竖直向下穿过导轨平面。外力推动棒以恒定 $4.0\ \mathrm{m\,s^{-1}}$ 运动。导轨与棒的电阻可忽略。
(a)Calculate the emf induced in the rod and the current in the circuit.计算棒中感应的电动势与电路中的电流。[3]
(b)Calculate the magnetic force on the rod, and state its direction relative to the motion, with reference to Lenz's law.计算棒所受的磁场力，并结合楞次定律说明其方向相对运动如何。[3]
(c)Calculate the mechanical power that the external force must supply to keep the rod moving at constant speed.计算外力为使棒保持匀速运动而必须提供的机械功率。[2]
(d)Show that the mechanical power supplied equals the electrical power dissipated, and explain why this must be so.证明所提供的机械功率等于耗散的电功率，并解释为何必然如此。[2]
A generator produces $20\ \mathrm{kW}$ of electrical power at $250\ \mathrm{V}$ AC. The power is to be sent to a town through a cable of total resistance $4.0\ \Omega$. An ideal step-up transformer raises the voltage to $10\,000\ \mathrm{V}$ before transmission.一台发电机以 $250\ \mathrm{V}$ 交流产生 $20\ \mathrm{kW}$ 电功率。该功率经一条总电阻 $4.0\ \Omega$ 的输电线送往某城镇。输电前用理想升压变压器把电压升到 $10\,000\ \mathrm{V}$。
(a)State the turns ratio $N_s : N_p$ of the step-up transformer.写出升压变压器的匝数比 $N_s : N_p$。[2]
(b)Calculate the current in the cable and the power lost in it during transmission at $10\,000\ \mathrm{V}$.计算以 $10\,000\ \mathrm{V}$ 输电时线中的电流及其损耗功率。[3]
(c)Explain why a transformer works only with alternating current, and state why its core is laminated.解释变压器为何只能用于交流，并说明其铁芯为何叠片。[3]