Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus D4.1 to D4.6考纲 D4.1 至 D4.6PHYSICS HL
Circular coil, $80$ turns, radius $4.0\ \mathrm{cm}$, in $B = 0.25\ \mathrm{T}$; normal at $30^{\circ}$ to the field. (a) flux through one turn; (b) flux linkage and unit.圆线圈 $80$ 匝、半径 $4.0\ \mathrm{cm}$，处于 $B = 0.25\ \mathrm{T}$；法线与场成 $30^{\circ}$。(a) 单匝磁通量；(b) 磁链与单位。
The angle is measured from the normal, so use $\Phi = BA\cos\theta$. The area is $A = \pi r^{2} = \pi(0.040)^{2} = 5.03\times 10^{-3}\ \mathrm{m^{2}}$. (M1)
$$ \Phi = BA\cos\theta = (0.25)(5.03\times 10^{-3})\cos 30^{\circ} \approx 1.1\times 10^{-3}\ \mathrm{Wb}. $$(A1)
Flux linkage is the flux multiplied by the number of turns: $N\Phi = 80 \times 1.089\times 10^{-3}$. (M1)
$$ N\Phi \approx 8.7\times 10^{-2}\ \mathrm{Wb}. $$The unit of flux linkage is the weber (sometimes written weber-turns). (A1)
角度自法线量起，故用 $\Phi = BA\cos\theta$。面积 $A = \pi r^{2} = \pi(0.040)^{2} = 5.03\times 10^{-3}\ \mathrm{m^{2}}$。(M1)
$$ \Phi = BA\cos\theta = (0.25)(5.03\times 10^{-3})\cos 30^{\circ} \approx 1.1\times 10^{-3}\ \mathrm{Wb}. $$(A1)
磁链是磁通量乘以匝数：$N\Phi = 80 \times 1.089\times 10^{-3}$。(M1)
$$ N\Phi \approx 8.7\times 10^{-2}\ \mathrm{Wb}. $$磁链的单位是韦伯（有时写作韦伯-匝）。(A1)
Rod $L = 0.50\ \mathrm{m}$ at $v = 6.0\ \mathrm{m\,s^{-1}}$, $B = 0.40\ \mathrm{T}$ perpendicular, rails closed by $R = 5.0\ \Omega$. (a) emf; (b) current; (c) power in the resistor.棒 $L = 0.50\ \mathrm{m}$、$v = 6.0\ \mathrm{m\,s^{-1}}$、$B = 0.40\ \mathrm{T}$ 垂直，导轨由 $R = 5.0\ \Omega$ 闭合。(a) 电动势；(b) 电流；(c) 电阻功率。
The rod, field and velocity are mutually perpendicular, so use $\varepsilon = BvL$. (M1)
$$ \varepsilon = BvL = (0.40)(6.0)(0.50) = 1.2\ \mathrm{V}. $$(A1)
The rod acts as a cell of emf $1.2\ \mathrm{V}$ across $R = 5.0\ \Omega$: $I = \varepsilon / R$. (M1)
$$ I = \frac{1.2}{5.0} = 0.24\ \mathrm{A}. $$(A1)
$P = \varepsilon I = (1.2)(0.24) = 0.288\ \mathrm{W}$, equivalently $I^{2}R = (0.24)^{2}(5.0)$. (M1)
$$ P \approx 0.29\ \mathrm{W}. $$(A1)
棒、场与速度两两垂直，故用 $\varepsilon = BvL$。(M1)
$$ \varepsilon = BvL = (0.40)(6.0)(0.50) = 1.2\ \mathrm{V}. $$(A1)
棒相当于一个 $1.2\ \mathrm{V}$ 电动势的电池，接在 $R = 5.0\ \Omega$ 上：$I = \varepsilon / R$。(M1)
$$ I = \frac{1.2}{5.0} = 0.24\ \mathrm{A}. $$(A1)
$P = \varepsilon I = (1.2)(0.24) = 0.288\ \mathrm{W}$，亦即 $I^{2}R = (0.24)^{2}(5.0)$。(M1)
$$ P \approx 0.29\ \mathrm{W}. $$(A1)
Coil $400$ turns, $A = 2.5\times 10^{-3}\ \mathrm{m^{2}}$, plane perpendicular to field; $B$ falls from $0.60\ \mathrm{T}$ to $0$ in $0.015\ \mathrm{s}$. (a) flux change per turn; (b) Faraday's law and emf; (c) one change that doubles the emf.线圈 $400$ 匝、$A = 2.5\times 10^{-3}\ \mathrm{m^{2}}$，平面垂直于场；$B$ 在 $0.015\ \mathrm{s}$ 内由 $0.60\ \mathrm{T}$ 降到 $0$。(a) 单匝磁通量变化；(b) 法拉第定律与电动势；(c) 使电动势翻倍的一种改动。
The plane is perpendicular to $B$, so $\theta = 0$ and $\Phi = BA$. Only $B$ changes: $\Delta\Phi = A\,\Delta B = (2.5\times 10^{-3})(0.60)$. (M1)
$$ |\Delta\Phi| = 1.5\times 10^{-3}\ \mathrm{Wb}. $$(A1)
Faraday's law: the induced emf equals the rate of change of flux linkage, $\varepsilon = -N\,\Delta\Phi/\Delta t$. (R1)
Taking magnitudes: $|\varepsilon| = N\,|\Delta\Phi|/\Delta t = 400 \times (1.5\times 10^{-3})/0.015$. (M1)
$$ |\varepsilon| = 400 \times 0.10 = 40\ \mathrm{V}. $$(A1)
Halve the collapse time (from $0.015\ \mathrm{s}$ to $0.0075\ \mathrm{s}$), since $\varepsilon \propto 1/\Delta t$. Doubling the number of turns would do the same. (B1)
平面垂直于 $B$，故 $\theta = 0$、$\Phi = BA$。只有 $B$ 变：$\Delta\Phi = A\,\Delta B = (2.5\times 10^{-3})(0.60)$。(M1)
$$ |\Delta\Phi| = 1.5\times 10^{-3}\ \mathrm{Wb}. $$(A1)
法拉第定律：感应电动势等于磁链变化率，$\varepsilon = -N\,\Delta\Phi/\Delta t$。(R1)
取大小：$|\varepsilon| = N\,|\Delta\Phi|/\Delta t = 400 \times (1.5\times 10^{-3})/0.015$。(M1)
$$ |\varepsilon| = 400 \times 0.10 = 40\ \mathrm{V}. $$(A1)
把衰减时间减半（由 $0.015\ \mathrm{s}$ 到 $0.0075\ \mathrm{s}$），因为 $\varepsilon \propto 1/\Delta t$。把匝数加倍亦可。(B1)
Horizontal ring; field directed vertically downward through it and increasing. (a) state Lenz's law; (b) direction of the induced current seen from above, justified by the flux change; (c) why Lenz's law follows from conservation of energy.水平环；场竖直向下穿过它且正在增大。(a) 陈述楞次定律；(b) 从上方看感应电流方向，并以磁通量变化论证；(c) 楞次定律为何源于能量守恒。
The induced current flows in the direction that opposes the change in magnetic flux that produces it. (B1)
The downward flux through the ring is increasing. (M1)
By Lenz's law the induced current must oppose the increase, so it creates a magnetic field pointing upward inside the ring. (M1)
By the right-hand grip rule, a field pointing up out of the ring requires the current to flow anticlockwise as seen from above. (A1)
If the induced current instead aided the increase, it would reinforce the flux, drive an ever-larger current, and release electrical energy without any work being done. (R1)
That would create energy from nothing, which is forbidden. The opposing direction means work must be done against the induced effect to keep changing the flux, and that work supplies the electrical energy. (R1)
感应电流的方向总是反抗产生它的磁通量变化。(B1)
穿过环的向下磁通量正在增大。(M1)
由楞次定律，感应电流必须反抗这一增大，故在环内产生向上的磁场。(M1)
由右手握拳定则，环内向上、即指向环外的场要求电流从上方看为逆时针。(A1)
若感应电流反而助长增大，它会强化磁通量、驱动越来越大的电流，并在不做功的情况下释放电能。(R1)
那将凭空创造能量，这是不允许的。反抗方向意味着必须克服感应效应做功才能持续改变磁通量，而这份功提供电能。(R1)
Ideal step-up transformer: $N_p = 500$, $V_p = 240\ \mathrm{V}$, $V_s = 12\,000\ \mathrm{V}$; cable $R = 8.0\ \Omega$ carries $6.0\ \mathrm{kW}$. (a) secondary turns; (b) cable current and power lost; (c) loss if sent at $240\ \mathrm{V}$, and what this shows.理想升压变压器：$N_p = 500$、$V_p = 240\ \mathrm{V}$、$V_s = 12\,000\ \mathrm{V}$；输电线 $R = 8.0\ \Omega$ 输送 $6.0\ \mathrm{kW}$。(a) 次级匝数；(b) 线电流与损耗；(c) 以 $240\ \mathrm{V}$ 输送时的损耗及其意义。
For an ideal transformer $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$, so $N_s = N_p \dfrac{V_s}{V_p} = 500 \times \dfrac{12\,000}{240}$. (M1)
$$ N_s = 500 \times 50 = 25\,000\ \text{turns}. $$(A1)
The transmission current follows from $P = VI$ at the transmission voltage: $I = P/V = 6000/12\,000 = 0.50\ \mathrm{A}$. (M1)
Power lost in the cable is $P_{\text{loss}} = I^{2}R = (0.50)^{2}(8.0)$. (M1)
$$ P_{\text{loss}} = 0.25 \times 8.0 = 2.0\ \mathrm{W}. $$(A1)
At $240\ \mathrm{V}$ the current is $I = 6000/240 = 25\ \mathrm{A}$. (M1)
$$ P_{\text{loss}} = I^{2}R = (25)^{2}(8.0) = 5000\ \mathrm{W} = 5.0\ \mathrm{kW}. $$(A1)
The loss is $2500$ times larger, and it would consume most of the $6.0\ \mathrm{kW}$ being sent. Transmitting at high voltage keeps the current, and therefore the $I^{2}R$ loss, very small. (R1)
理想变压器 $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$，故 $N_s = N_p \dfrac{V_s}{V_p} = 500 \times \dfrac{12\,000}{240}$。(M1)
$$ N_s = 500 \times 50 = 25\,000\ \text{匝}. $$(A1)
输电电流由输电电压下的 $P = VI$ 得出：$I = P/V = 6000/12\,000 = 0.50\ \mathrm{A}$。(M1)
线中损耗功率 $P_{\text{loss}} = I^{2}R = (0.50)^{2}(8.0)$。(M1)
$$ P_{\text{loss}} = 0.25 \times 8.0 = 2.0\ \mathrm{W}. $$(A1)
$240\ \mathrm{V}$ 时电流为 $I = 6000/240 = 25\ \mathrm{A}$。(M1)
$$ P_{\text{loss}} = I^{2}R = (25)^{2}(8.0) = 5000\ \mathrm{W} = 5.0\ \mathrm{kW}. $$(A1)
损耗大了 $2500$ 倍，几乎会吞掉所输送的 $6.0\ \mathrm{kW}$。高压输电使电流、从而 $I^{2}R$ 损耗都很小。(R1)
Flux linkage vs time in three straight segments: $0 \to 0.40\ \mathrm{Wb}$ over $0.20\ \mathrm{s}$; constant at $0.40\ \mathrm{Wb}$ for $0.20\ \mathrm{s}$; $0.40 \to 0\ \mathrm{Wb}$ over $0.10\ \mathrm{s}$. (a) how emf comes from the graph; (b) emf in each segment; (c) sketch emf vs time; (d) percentage uncertainty in segment-1 emf.磁链对时间分三段直线：$0.20\ \mathrm{s}$ 内 $0 \to 0.40\ \mathrm{Wb}$；$0.20\ \mathrm{s}$ 内恒为 $0.40\ \mathrm{Wb}$；$0.10\ \mathrm{s}$ 内 $0.40 \to 0\ \mathrm{Wb}$。(a) 电动势如何从图得出；(b) 各段电动势；(c) 画电动势-时间图；(d) 第 1 段电动势的百分比不确定度。
Faraday's law gives $|\varepsilon| = |\Delta(N\Phi)/\Delta t|$, which is the magnitude of the gradient of the flux-linkage-time graph. (M1)
A steeper segment means a larger emf; a flat segment (constant flux) gives zero emf. (A1)
Segment 1: $|\varepsilon| = 0.40/0.20 = 2.0\ \mathrm{V}$. (A1)
Segment 2: flux constant, so gradient $= 0$ and $|\varepsilon| = 0\ \mathrm{V}$. (A1)
Segment 3: $|\varepsilon| = 0.40/0.10 = 4.0\ \mathrm{V}$. (A1)
The emf-time graph is a series of horizontal steps. (M1)
From $0$ to $0.20\ \mathrm{s}$ it sits at $2.0\ \mathrm{V}$; from $0.20$ to $0.40\ \mathrm{s}$ it drops to $0\ \mathrm{V}$. (A1)
From $0.40$ to $0.50\ \mathrm{s}$ it jumps to the opposite sign at magnitude $4.0\ \mathrm{V}$, because the flux is now decreasing rather than increasing. (A1)
The emf is a quotient, so the percentage uncertainties add: $\dfrac{\Delta\varepsilon}{\varepsilon} = \dfrac{\Delta(N\Phi)}{N\Phi} + \dfrac{\Delta t}{t} = \dfrac{0.02}{0.40} + \dfrac{0.01}{0.20}$. (M1)
$$ = 5\% + 5\% = 10\%. $$(A1)
法拉第定律给出 $|\varepsilon| = |\Delta(N\Phi)/\Delta t|$，即磁链-时间图斜率的大小。(M1)
段越陡电动势越大；平直段（磁通量不变）给出零电动势。(A1)
第 1 段：$|\varepsilon| = 0.40/0.20 = 2.0\ \mathrm{V}$。(A1)
第 2 段：磁通量恒定，斜率 $= 0$，$|\varepsilon| = 0\ \mathrm{V}$。(A1)
第 3 段：$|\varepsilon| = 0.40/0.10 = 4.0\ \mathrm{V}$。(A1)
电动势-时间图为一系列水平台阶。(M1)
$0$ 至 $0.20\ \mathrm{s}$ 停在 $2.0\ \mathrm{V}$；$0.20$ 至 $0.40\ \mathrm{s}$ 降到 $0\ \mathrm{V}$。(A1)
$0.40$ 至 $0.50\ \mathrm{s}$ 跳到相反符号、大小 $4.0\ \mathrm{V}$，因为此时磁通量在减小而非增大。(A1)
电动势是商，故百分比不确定度相加：$\dfrac{\Delta\varepsilon}{\varepsilon} = \dfrac{\Delta(N\Phi)}{N\Phi} + \dfrac{\Delta t}{t} = \dfrac{0.02}{0.40} + \dfrac{0.01}{0.20}$。(M1)
$$ = 5\% + 5\% = 10\%. $$(A1)
Coil $200$ turns, $A = 0.015\ \mathrm{m^{2}}$, $f = 60\ \mathrm{Hz}$, $B = 0.080\ \mathrm{T}$, face-on at $t = 0$. (a) max flux linkage; (b) $\omega$ and peak emf; (c) sketch flux and emf vs time, phase difference; (d) percentage uncertainty in peak emf with $\Delta f = \pm 1\ \mathrm{Hz}$, $\Delta B = \pm 0.002\ \mathrm{T}$.线圈 $200$ 匝、$A = 0.015\ \mathrm{m^{2}}$、$f = 60\ \mathrm{Hz}$、$B = 0.080\ \mathrm{T}$，$t = 0$ 正对。(a) 最大磁链；(b) $\omega$ 与峰值电动势；(c) 画磁链与电动势对时间、相位差；(d) 由 $\Delta f = \pm 1\ \mathrm{Hz}$、$\Delta B = \pm 0.002\ \mathrm{T}$ 求峰值电动势百分比不确定度。
The flux linkage is greatest when the coil is face-on, where $N\Phi_{\max} = NBA$. (M1)
$$ N\Phi_{\max} = (200)(0.080)(0.015) = 0.24\ \mathrm{Wb}. $$(A1)
$\omega = 2\pi f = 2\pi(60) \approx 377\ \mathrm{rad\,s^{-1}}$. (M1)
Peak emf $\varepsilon_{0} = NBA\omega = (0.24)(377)$, using $NBA = 0.24\ \mathrm{Wb}$ from (a). (M1)
$$ \varepsilon_{0} \approx 90.5\ \mathrm{V}. $$(A1)
The flux linkage is a cosine, starting at its maximum $0.24\ \mathrm{Wb}$ at $t = 0$ (coil face-on). (M1·A1)
The emf is a sine of amplitude $90.5\ \mathrm{V}$, starting at zero at $t = 0$ and peaking a quarter-period later. (A1)
The emf and flux are $90^{\circ}$ out of phase: the emf peaks exactly where the flux crosses zero. (A1)
$\varepsilon_{0} = NBA(2\pi f)$ is a product, so the percentage uncertainties of $B$ and $f$ add ($N$, $A$ exact): $\dfrac{\Delta\varepsilon_{0}}{\varepsilon_{0}} = \dfrac{\Delta B}{B} + \dfrac{\Delta f}{f} = \dfrac{0.002}{0.080} + \dfrac{1}{60}$. (M1)
$$ = 2.5\% + 1.7\% \approx 4.2\%. $$(A1)
磁链在线圈正对时最大，此时 $N\Phi_{\max} = NBA$。(M1)
$$ N\Phi_{\max} = (200)(0.080)(0.015) = 0.24\ \mathrm{Wb}. $$(A1)
$\omega = 2\pi f = 2\pi(60) \approx 377\ \mathrm{rad\,s^{-1}}$。(M1)
峰值电动势 $\varepsilon_{0} = NBA\omega = (0.24)(377)$，用 (a) 中 $NBA = 0.24\ \mathrm{Wb}$。(M1)
$$ \varepsilon_{0} \approx 90.5\ \mathrm{V}. $$(A1)
磁链为余弦，$t = 0$（线圈正对）时从最大值 $0.24\ \mathrm{Wb}$ 出发。(M1·A1)
电动势为幅值 $90.5\ \mathrm{V}$ 的正弦，$t = 0$ 时为零，四分之一周期后达峰值。(A1)
电动势与磁链相位差 $90^{\circ}$：电动势的峰值恰在磁通量过零处。(A1)
$\varepsilon_{0} = NBA(2\pi f)$ 为乘积，故 $B$ 与 $f$ 的百分比不确定度相加（$N$、$A$ 精确）：$\dfrac{\Delta\varepsilon_{0}}{\varepsilon_{0}} = \dfrac{\Delta B}{B} + \dfrac{\Delta f}{f} = \dfrac{0.002}{0.080} + \dfrac{1}{60}$。(M1)
$$ = 2.5\% + 1.7\% \approx 4.2\%. $$(A1)
AC generator: coil $80$ turns, $A = 0.020\ \mathrm{m^{2}}$, $f = 50\ \mathrm{Hz}$, $B = 0.15\ \mathrm{T}$. (a) principle and law; (b) $\omega$ and peak emf; (c) emf vs time and rms emf; (d) effect of doubling $f$; (e) coil position at peak emf, explained by rate of change of flux.交流发电机：线圈 $80$ 匝、$A = 0.020\ \mathrm{m^{2}}$、$f = 50\ \mathrm{Hz}$、$B = 0.15\ \mathrm{T}$。(a) 原理与定律；(b) $\omega$ 与峰值电动势；(c) 电动势-时间与有效值；(d) $f$ 加倍的影响；(e) 峰值电动势时线圈位置，用磁通量变化率解释。
As the coil rotates, the flux linking it changes continuously, which induces an emf. (M1)
This is Faraday's law of electromagnetic induction: the emf equals the rate of change of flux linkage. (A1)
$\omega = 2\pi f = 2\pi(50) \approx 314\ \mathrm{rad\,s^{-1}}$. (M1)
$\varepsilon_{0} = NBA\omega = (80)(0.15)(0.020)(314)$. (M1)
$$ \varepsilon_{0} = 0.24 \times 314 \approx 75.4\ \mathrm{V}. $$(A1)
With the coil face-on at $t = 0$, the flux is a cosine and the emf is a sine: $\varepsilon = 75.4\sin(314\,t)\ \mathrm{V}$. (A1)
For a sinusoid the rms value is the peak divided by $\sqrt{2}$: $\varepsilon_{\text{rms}} = \varepsilon_{0}/\sqrt{2} = 75.4/\sqrt{2}$. (M1)
$$ \varepsilon_{\text{rms}} \approx 53.3\ \mathrm{V}. $$(A1)
The output frequency doubles, since it equals the rotation frequency. (A1)
The peak emf also doubles, because $\varepsilon_{0} = NBA\omega \propto \omega \propto f$; doubling $\omega$ doubles $\varepsilon_{0}$. (R1)
The emf is greatest when the flux is changing fastest, which is when the flux is passing through zero. That occurs when the coil plane is parallel to the field (the normal at $90^{\circ}$). (M1)
At that instant $\Phi = 0$ but $|\mathrm{d}\Phi/\mathrm{d}t|$ is maximal, so $|\varepsilon|$ is maximal. (A1)
线圈旋转时，交链它的磁通量连续变化，从而感应出电动势。(M1)
这就是法拉第电磁感应定律：电动势等于磁链变化率。(A1)
$\omega = 2\pi f = 2\pi(50) \approx 314\ \mathrm{rad\,s^{-1}}$。(M1)
$\varepsilon_{0} = NBA\omega = (80)(0.15)(0.020)(314)$。(M1)
$$ \varepsilon_{0} = 0.24 \times 314 \approx 75.4\ \mathrm{V}. $$(A1)
$t = 0$ 时线圈正对，磁通量为余弦、电动势为正弦：$\varepsilon = 75.4\sin(314\,t)\ \mathrm{V}$。(A1)
正弦量的有效值为峰值除以 $\sqrt{2}$：$\varepsilon_{\text{rms}} = \varepsilon_{0}/\sqrt{2} = 75.4/\sqrt{2}$。(M1)
$$ \varepsilon_{\text{rms}} \approx 53.3\ \mathrm{V}. $$(A1)
输出频率加倍，因为它等于旋转频率。(A1)
峰值电动势也加倍，因为 $\varepsilon_{0} = NBA\omega \propto \omega \propto f$；$\omega$ 加倍使 $\varepsilon_{0}$ 加倍。(R1)
电动势在磁通量变化最快时最大，即磁通量过零时。这发生在线圈平面平行于场（法线成 $90^{\circ}$）时。(M1)
此刻 $\Phi = 0$ 但 $|\mathrm{d}\Phi/\mathrm{d}t|$ 最大，故 $|\varepsilon|$ 最大。(A1)
Rod $L = 0.25\ \mathrm{m}$ on rails closed by $R = 0.20\ \Omega$; $B = 0.50\ \mathrm{T}$ vertical; pushed at constant $v = 4.0\ \mathrm{m\,s^{-1}}$. (a) emf and current; (b) magnetic force and its direction by Lenz; (c) mechanical power needed; (d) show mechanical power $=$ electrical power and why.棒 $L = 0.25\ \mathrm{m}$，导轨由 $R = 0.20\ \Omega$ 闭合；$B = 0.50\ \mathrm{T}$ 竖直；以恒定 $v = 4.0\ \mathrm{m\,s^{-1}}$ 推动。(a) 电动势与电流；(b) 磁场力及其楞次方向；(c) 所需机械功率；(d) 证明机械功率 $=$ 电功率及原因。
$\varepsilon = BvL = (0.50)(4.0)(0.25) = 0.50\ \mathrm{V}$. (M1·A1)
$I = \varepsilon/R = 0.50/0.20 = 2.5\ \mathrm{A}$. (A1)
The current-carrying rod in the field feels a force $F = BIL = (0.50)(2.5)(0.25)$. (M1)
$$ F = 0.3125 \approx 0.31\ \mathrm{N}. $$(A1)
By Lenz's law the induced current opposes the change in flux, so the force on the rod acts opposite to its velocity, that is, it is a retarding force. (R1)
To keep the rod at constant speed the external force must exactly balance the retarding force, so $F_{\text{ext}} = 0.3125\ \mathrm{N}$ and $P_{\text{mech}} = F_{\text{ext}}\,v = (0.3125)(4.0)$. (M1)
$$ P_{\text{mech}} = 1.25\ \mathrm{W}. $$(A1)
The electrical power dissipated is $P_{\text{elec}} = \varepsilon I = (0.50)(2.5) = 1.25\ \mathrm{W}$ (or $I^{2}R = (2.5)^{2}(0.20) = 1.25\ \mathrm{W}$), equal to $P_{\text{mech}}$. (A1)
At constant speed the kinetic energy is unchanged, so by conservation of energy all the mechanical work done against the magnetic force is converted to electrical energy and dissipated as heat in $R$. (R1)
$\varepsilon = BvL = (0.50)(4.0)(0.25) = 0.50\ \mathrm{V}$。(M1·A1)
$I = \varepsilon/R = 0.50/0.20 = 2.5\ \mathrm{A}$。(A1)
载流棒在场中受力 $F = BIL = (0.50)(2.5)(0.25)$。(M1)
$$ F = 0.3125 \approx 0.31\ \mathrm{N}. $$(A1)
由楞次定律，感应电流反抗磁通量的变化，故棒受的力与其速度反向，即为阻碍力。(R1)
要使棒匀速，外力须恰好平衡阻碍力，故 $F_{\text{ext}} = 0.3125\ \mathrm{N}$，$P_{\text{mech}} = F_{\text{ext}}\,v = (0.3125)(4.0)$。(M1)
$$ P_{\text{mech}} = 1.25\ \mathrm{W}. $$(A1)
耗散的电功率为 $P_{\text{elec}} = \varepsilon I = (0.50)(2.5) = 1.25\ \mathrm{W}$（或 $I^{2}R = (2.5)^{2}(0.20) = 1.25\ \mathrm{W}$），与 $P_{\text{mech}}$ 相等。(A1)
匀速时动能不变，故由能量守恒，克服磁场力所做的全部机械功都转化为电能，并在 $R$ 中以热耗散。(R1)
Generator: $20\ \mathrm{kW}$ at $250\ \mathrm{V}$ AC, sent through a cable $R = 4.0\ \Omega$; ideal step-up transformer raises the voltage to $10\,000\ \mathrm{V}$. (a) turns ratio $N_s : N_p$; (b) cable current and power lost at $10\,000\ \mathrm{V}$; (c) why a transformer needs AC and why the core is laminated.发电机：$250\ \mathrm{V}$ 交流、$20\ \mathrm{kW}$，经 $R = 4.0\ \Omega$ 输电线输送；理想升压变压器把电压升到 $10\,000\ \mathrm{V}$。(a) 匝数比 $N_s : N_p$；(b) $10\,000\ \mathrm{V}$ 下线电流与损耗；(c) 变压器为何需交流、铁芯为何叠片。
For an ideal transformer the turns ratio equals the voltage ratio: $\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p} = \dfrac{10\,000}{250}$. (M1)
$$ \frac{N_s}{N_p} = 40, \quad \text{so } N_s : N_p = 40 : 1. $$(A1)
At the transmission voltage, $I = P/V = 20\,000/10\,000 = 2.0\ \mathrm{A}$. (M1)
Power lost in the cable is $P_{\text{loss}} = I^{2}R = (2.0)^{2}(4.0)$. (M1)
$$ P_{\text{loss}} = 4.0 \times 4.0 = 16\ \mathrm{W}. $$(A1)
A transformer relies on a changing flux to induce an emf in the secondary. AC continuously changes the flux in the core; a steady DC would give constant flux and no induced emf. (R1·R1)
The changing flux also induces eddy currents in the core, which waste energy as heat. Laminating the core (thin insulated sheets) breaks the eddy-current loops into small high-resistance paths, reducing this loss. (B1)
理想变压器匝数比等于电压比：$\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p} = \dfrac{10\,000}{250}$。(M1)
$$ \frac{N_s}{N_p} = 40, \quad \text{即 } N_s : N_p = 40 : 1. $$(A1)
在输电电压下，$I = P/V = 20\,000/10\,000 = 2.0\ \mathrm{A}$。(M1)
线中损耗功率 $P_{\text{loss}} = I^{2}R = (2.0)^{2}(4.0)$。(M1)
$$ P_{\text{loss}} = 4.0 \times 4.0 = 16\ \mathrm{W}. $$(A1)
变压器依赖变化的磁通量在次级感应电动势。交流使铁芯中的磁通量持续变化；恒定直流给出恒定磁通量，不产生感应电动势。(R1·R1)
变化的磁通量还会在铁芯中感应出涡流，以热的形式浪费能量。把铁芯叠片（薄绝缘片）将涡流回路分割成小的高阻路径，减小此损耗。(B1)