Тема 7: Магнетизм и электромагнетизм
AQA · GCSE · Physics Тема 7 19:53 Английское озвучивание · Английский + китайские субтитры (встроенные)
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Follow the energy transfer and distinguish a potential difference from current in a complete circuit.
沿能量传递过程区分电动机效应与发电机效应。 感应电压并不总产生电流，电流需要闭合回路。
Sheets 7.1–7.3 build the field models first.
练习7.1至7.3先建立磁场模型。
These devices use related magnetic effects but are not interchangeable explanations.
这些装置利用相关的磁效应，但不能相互替代地解释。
Initially unmagnetised magnetic material is attracted by induction.
原来未磁化的磁性材料受到感应而被吸引。
Like permanent poles can repel.
同名的永久磁极可以相斥。
A permanent magnet produces its own field.
永磁体产生自己的磁场。
Do not turn the induced-magnet model into a claim that steel can never retain magnetism.
不要把感应磁体模型错误地理解成钢永远不能保留磁性。
The north-seeking compass tip follows the field.
指南针的指北端沿磁场方向。
Lines do not cross; closer spacing represents stronger field.
磁感线不相交，线越密集表示磁场越强。
A single compass reading gives one local direction, not the whole field pattern.
一次指南针读数只给出一个位置的方向，不能表示整个磁场。
Keep the magnet fixed.
保持磁铁位置不变。
Do not connect unrelated compass readings into crossing field lines.
不要把无关的读数连接成相交的磁感线。
A compass aligns with the resultant local field.
指南针沿当地合磁场方向。
Local interference adds to the Earth’s field.
局部干扰磁场与地磁场叠加。
Do not claim a compass gives an accurate geographic-north bearing everywhere, especially near interference or the magnetic poles.
不要声称指南针在所有地方都能准确指出地理北方，尤其是在干扰源或磁极附近。
An outward current gives an anticlockwise wire field.
出纸面的电流产生逆时针磁场。
In the solenoid view, the shown top and bottom currents give an internal field from left to right.
在所示螺线管中，上下方电流共同产生从左向右的内部磁场。
Compare the same compass position at different currents.
在同一位置比较不同电流的效果。
Changing current and distance together would confound the comparison.
若同时改变电流和距离，就无法公平比较。
The wire’s distance dependence does not disappear at larger current.
电流增大并不会消除磁场随距离减弱的规律。
Follow current, magnetisation, attraction and contact movement.
按电流、磁化、吸引和触点移动的顺序解释。
The magnetic action operates the contact without electrically joining the circuits.
磁作用驱动触点，但不把两个电路直接连接。
Use the arrangement shown by a device diagram; do not assume every electromagnetic device has the same contact mechanism.
应根据装置图所示结构解释，不要假定所有电磁装置都有相同的触点机构。
Thumb is force, first finger field, second finger conventional current.
拇指表示力，食指表示磁场，中指表示传统电流方向。
All three directions are mutually perpendicular.
三个方向互相垂直。
Apply the rule afresh rather than assuming every reversal changes force.
每次重新应用左手定则，不要假定任何反转都会改变力的方向。
Use F = BIl for the perpendicular length in the stated GCSE arrangement.
在本考纲给定的垂直放置条件下使用F = BIl。
Distinguish increasing each factor separately from doubling two factors at once.
区分分别增大一个因素与同时把两个因素加倍。
The length is the active conductor length in the field, not a total connecting-wire length.
长度是位于磁场中的有效导体长度，不是全部连接导线长度。
The field is nought point three zero tesla, the current two point zero amperes, the length of rod in the field nought point nought seven zero metres, and the rod's mass five point zero grams.
磁感应强度为零点三零特斯拉，电流为二点零安，磁场中导体棒的长度为零点零七零米，导体棒质量为五点零克。
Calculate the motor force with F equals B I l, convert the mass to kilograms, then use F equals m a to find the initial acceleration.
先用F等于BIl算出电动机力，把质量换算成千克，再用F等于ma求出初始加速度。
Pause and try it.
暂停并试着计算。
Use the initial resultant along the rails in the paper’s model.
使用原题沿轨道方向的初始合力模型。
A force calculation alone is not the acceleration answer.
只算出力还没有回答加速度。
The end view shows current out on the left and into the page on the right.
端视图左侧电流出纸面，右侧入纸面。
With field right, the forces are up and down respectively.
磁场向右时，两侧的力分别向上和向下。
Use the actual split-ring contact model.
使用实际的分裂环触点模型。
A generic statement about forces alone misses the continuation mechanism.
只泛泛描述受力，会漏掉持续旋转的原因。
The interacting fields produce forces on the sides of the coil, pushing them in opposite directions.
磁场相互作用，在线圈两边产生方向相反的力。
Every half turn, the coil's ends swap contacts, so the current and the force on each side reverse.
每转半圈，线圈两端交换接触点，所以每一边的电流和受力都反向。
That reversal keeps the coil turning the same way after its sides change places.
这种反向使两边交换位置后，线圈仍向同一方向转动。
At the insulating gaps there is no current, so no magnetic force and no turning effect; the coil's inertia carries it through.
在绝缘间隙处没有电流，所以没有磁力，也没有转动效果；线圈靠惯性转过去。
Name the motor effect, attachment to the cone and pressure variations.
应说明电动机效应、线圈与纸盆连接以及压强变化。
“Electricity becomes sound” is incomplete.
只说电变成声音是不完整的。
A moving-coil microphone reverses the conversion and uses induction, not motor force as the output mechanism.
动圈式麦克风进行相反的转换，其输出机制是电磁感应。
Motion must change magnetic linkage or cut field lines.
运动必须改变磁场连接情况或切割磁感线。
A stationary magnet is not a continuous current source.
静止磁铁不是持续电流源。
The generator effect produces pd. Current requires a conducting loop.
发电机效应产生电压，电流还需要闭合导电回路。
Do not use “motion exists” as the sole induction condition.
不要只根据存在运动就断定一定发生电磁感应。
A wire moving upwards through a magnetic field gave a reading of plus nought point four milliamperes.
一根导线在磁场中向上运动，电表读数为正零点四毫安。
Then the field is reversed and the motion repeated.
然后把磁场反向，再做同样的运动。
Explain the first current in three links: the wire cuts the field lines, a potential difference is induced, and the circuit is complete.
用三个环节解释第一次的电流：导线切割磁感线，产生感应电势差，电路是闭合的。
Then predict the new signed reading.
再预测新的读数，包括正负号。
Pause and try it.
暂停并试着回答。
With unchanged speed and field strength, reversing polarity changes direction rather than magnitude.
在速度和磁场强度不变时，反转磁极改变方向而不改变大小。
Opposition is to the change, not always to the original field direction.
感应磁场阻碍的是变化，并不总与原磁场方向相反。
A stronger magnet or more coil turns can also increase induced pd for the stated motion and geometry.
在运动和几何条件相同的情况下，更强磁铁或更多匝数也可增大感应电压。
For the simple rotating-coil model, an alternator reverses output each half-turn; a commutator dynamo gives one-directional pulses.
在简单转动线圈模型中，交流发电机每半周反转输出；带换向器的直流发电机输出单向脉动。
Real bicycle devices called dynamos can generate ac.
现实中叫作“自行车发电机”的装置可能输出交流。
Use the construction supplied, not the nickname.
根据给定结构判断，不要只看名称。
DC does not mean a perfectly flat graph, and its sign depends on terminal labels.
直流不意味着图像一定是水平直线，其正负也取决于端子标注。
The diaphragm and coil vibrate.
振膜和线圈振动。
Use the generator effect to link motion to the electrical output.
利用发电机效应把运动与电信号输出联系起来。
Do not say the incoming sound directly pushes electrical current through the wire.
不要说入射声波直接把电流推过导线。
The primary is connected to ac.
原线圈接交流电。
The changing magnetic field links the secondary and induces pd across it.
变化磁场通过铁芯连接副线圈，在副线圈两端感应电压。
Give the full primary-current, changing-core-field, induced-secondary-pd chain.
完整说明原线圈电流、变化的铁芯磁场及副线圈感应电压之间的因果链。
The coils need not move.
线圈不必移动。
A changing field can induce pd in a stationary coil.
变化磁场也能在静止线圈中感应电压。
These are the paper’s values.
这些是原试题数据。
Do not replace 400 kV with the separate handout example’s 4000 V.
不要把400 kV换成讲义中另一个例题的4000 V。
Turns are counts.
匝数是计数。
State which coil is input before naming step-up or step-down.
判断升压还是降压前，先说明哪个线圈是输入端。
Voltage ratio alone does not give input current.
只用电压比不能直接求输入电流。
Use the ideal power balance explicitly.
应明确使用理想功率平衡。
The power equality is conditional on 100% efficiency, not a property of every real transformer.
输入与输出功率相等只在效率100%的假设下成立，不是所有真实变压器的性质。
Hold transferred power and cable resistance fixed.
保持传输功率与电缆电阻相同。
A higher pd alone does not establish reduced loss.
仅凭电压更高不能断定损耗更小。
Use step-down transformers for suitable local supply pd; lower supply pd is still hazardous.
降压变压器提供合适的本地供电电压；较低的供电电压仍然有危险。
A strong answer names the condition, causal chain and appropriate model.
好的答案应说明条件、因果链和适用模型。
Return to sheets 7.1–7.3 for independent practice and to the handout for the reference knowledge.
回到练习7.1至7.3独立训练，基础知识可查阅讲义。