The Motor Effect & Principles of the Electric Motor | 马达效应与电动机原理

📚 The Motor Effect & Principles of the Electric Motor | 马达效应与电动机原理

The motor effect is one of the most fundamental concepts in electromagnetism. It describes the force experienced by a current-carrying conductor placed in a magnetic field. This principle forms the working basis of electric motors, which convert electrical energy into mechanical energy. In this revision guide, we will break down the physics, the mathematics, and the exam-style applications of this topic step by step.

马达效应是电磁学中最基本的概念之一。它描述了载流导体置于磁场中所受到的力。这一原理构成了电动机的工作基础，电动机将电能转化为机械能。在本复习指南中，我们将逐步拆解这一主题的物理原理、数学表达以及考试型应用。

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1. What Is the Motor Effect? | 什么是马达效应？

The motor effect occurs when a wire carrying an electric current is placed inside a magnetic field. The magnetic field produced by the current interacts with the external magnetic field, resulting in a physical force acting on the wire. This is not the same as the force between two magnets; rather, it is the force between a magnetic field and moving charges within a conductor.

马达效应发生在载流导线置于磁场中时。电流产生的磁场与外磁场相互作用，导致导线受到一个物理力。这不同于两块磁铁之间的力；确切地说，这是磁场与导体内部移动电荷之间的力。

• Key requirement: The current must not be parallel to the magnetic field. Maximum force occurs when they are perpendicular.
• 关键条件：电流不能与磁场平行。当两者垂直时，力最大。

For example, if a straight copper wire carrying current is placed between the poles of a horseshoe magnet, the wire will jump upward or downward depending on the direction of the current and the direction of the magnetic field.

例如，若一根通有电流的直铜线置于马蹄形磁铁两极之间，导线会向上或向下跳动，具体方向取决于电流方向和磁场方向。

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2. Fleming’s Left-Hand Rule | 弗莱明左手定则

To determine the direction of the force on a current-carrying conductor in a magnetic field, we use Fleming’s Left-Hand Rule. Hold your left hand so that the thumb, index finger, and middle finger are mutually perpendicular.

为了确定磁场中载流导体所受力的方向，我们使用弗莱明左手定则。将左手摊开，使大拇指、食指和中指两两相互垂直。

• First finger (index): Points in the direction of the magnetic field, from North to South.
• Middle finger: Points in the direction of conventional current (from positive to negative).
• Thumb: Points in the direction of the force (motion) on the conductor.

• 食指：指向磁场方向，从N极到S极。
• 中指：指向电流方向（从正极到负极）。
• 大拇指：指向导体所受力的方向（运动方向）。

It is crucial to remember that this rule applies to conventional current, not electron flow. In exam questions, always check the direction of conventional current first before applying the rule.

必须记住，该定则适用于传统电流方向，而非电子流动方向。在考试题目中，务必先确认传统电流方向，再应用此定则。

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3. Calculating the Force: F = BIL | 力的计算：F = BIL

When a conductor of length L carries a current I perpendicular to a uniform magnetic field of flux density B, the magnitude of the force is given by:

当长度为 L 的导体在磁感应强度为 B 的均匀磁场中，通以垂直于磁场的电流 I 时，力的大小为：

F = BIL

Where F is the force in newtons (N), B is the magnetic flux density in teslas (T), I is the current in amperes (A), and L is the length of the conductor in the magnetic field in metres (m).

其中 F 为力（单位：牛顿 N），B 为磁感应强度（单位：特斯拉 T），I 为电流（单位：安培 A），L 为导体在磁场中的长度（单位：米 m）。

If the conductor is not perpendicular to the magnetic field, the component of the length perpendicular to the field must be used:

若导体与磁场不垂直，则必须使用垂直于磁场的长度分量：

F = BIL sin θ

Here, θ is the angle between the conductor and the magnetic field direction.

其中 θ 是导体与磁场方向之间的夹角。

Quantity | 物理量	Symbol | 符号	Unit | 单位
Force | 力	F	N
Magnetic flux density | 磁感应强度	B	T
Current | 电流	I	A
Length | 长度	L	m

One tesla is defined as the magnetic flux density that produces a force of one newton per metre of conductor carrying a current of one ampere perpendicular to the field.

1特斯拉的定义是：在垂直于磁场的方向，长度为1米的导体通以1安培电流时所受到的力为1牛顿，此时的磁感应强度即为1特斯拉。

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4. Why Does the Force Occur? | 力为何会产生？

The force arises because the current in the wire generates its own magnetic field around the conductor. This circular magnetic field interacts with the uniform external magnetic field. On one side of the wire, the two magnetic fields point in the same direction and reinforce each other, creating a region of stronger field. On the opposite side, the fields oppose each other, creating a region of weaker field. This imbalance in field strength produces a net force pushing the wire from the stronger field region toward the weaker field region.

力的产生是因为导线中的电流在导体周围形成了自身的环形磁场。这个环形磁场与均匀的外磁场相互作用。在导线的一侧，两个磁场方向相同而相互加强，形成较强的磁场区域。在另一侧，两个磁场方向相反，形成较弱的磁场区域。这种磁场强度的不平衡产生了净力，将导线从强磁场区域推向弱磁场区域。

This explanation is often tested in conceptual questions, so it is important to be able to describe the “field crowding” and “field rarefaction” arguments clearly.

这一解释经常在概念题中考查，因此能够清晰地描述”磁场密集”和”磁场稀疏”的论证非常重要。

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5. Construction of a Simple DC Motor | 简易直流电动机的构造

A simple direct current (DC) motor consists of the following components:

一个简易直流（DC）电动机由以下部件组成：

• Armature (coil): A rectangular coil of wire that rotates in the magnetic field.
• Magnetic field: Provided by permanent magnets or electromagnets (stator).
• Split-ring commutator: A pair of half-rings that reverses the direction of current in the coil every half turn.
• Brushes: Carbon contacts that press against the commutator, allowing current to flow into the rotating coil.
• Axle: The central rod on which the coil rotates, delivering mechanical output.

• 电枢（线圈）：在磁场中转动的矩形线圈。
• 磁场：由永磁体或电磁铁提供（定子）。
• 换向器（半环）：每半圈反转线圈中电流方向的一对半圆环。
• 电刷：压在换向器上的碳质触点，使电流流入旋转的线圈。
• 转轴：线圈绕其转动的中心杆，输出机械能。

When a current flows through the coil, the side where current runs in one direction experiences a force upward, while the opposite side experiences a force downward. These two forces create a couple (torque) that rotates the coil.

当电流通过线圈时，电流沿一个方向流动的一侧受到向上的力，而另一侧受到向下的力。这两个力形成一个力偶（转矩），使线圈转动。

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6. The Role of the Split-Ring Commutator | 换向器（半环）的作用

Without a commutator, the coil would rotate to the vertical position and then oscillate back and forth, never completing a full rotation. This is because when the coil passes through the vertical position, the torque becomes zero, and the previously upward force on one side would become downward if the current direction remained unchanged.

如果没有换向器，线圈会转到竖直位置后反复振荡，永远无法完成整圈旋转。原因是当线圈经过竖直位置时，力矩变为零，如果电流方向不变，原来一侧向上的力会变为向下。

The split-ring commutator solves this problem. Every half turn, the two halves of the commutator swap which brush they are in contact with, reversing the direction of the current in the coil. This reversal ensures that the force on each side of the coil always points in the same rotational direction, allowing continuous rotation.

换向器解决了这个问题。每转半圈，换向器的两个半环互换所接触的电刷，从而反转线圈中的电流方向。这种反转确保线圈两侧所受的力始终指向相同的旋转方向，从而实现连续转动。

Position of Coil | 线圈位置	Torque | 力矩	Commutator Action | 换向器作用
Horizontal | 水平	Maximum | 最大	Current unchanged | 电流不变
Vertical | 竖直	Zero | 零	Commutator switches contacts | 换向器切换接触
Horizontal (opposite side up) | 水平（另一侧在上）	Maximum | 最大	Current reversed | 电流已反转

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7. Torque on the Coil | 线圈所受的力矩

The torque acting on the coil is the product of the force on one side and the perpendicular distance between the two parallel forces. For a coil of width w and N turns, the torque is:

线圈所受的力矩等于一侧的力与两个平行力之间垂直距离的乘积。对于宽度为 w、匝数为 N 的线圈，力矩为：

T = B I A N cos θ

Where A is the area of the coil (A = L × w), and θ is the angle between the normal to the plane of the coil and the magnetic field direction. The torque is maximum when the plane of the coil is parallel to the magnetic field (θ = 0°), and zero when the plane is perpendicular to the field (θ = 90°).

其中 A 是线圈面积（A = L × w），θ 是线圈平面法线与磁场方向之间的夹角。当线圈平面平行于磁场（θ = 0°）时力矩最大；当线圈平面垂直于磁场（θ = 90°）时力矩为零。

In practical motors, the torque is not constant throughout the rotation. To produce smoother rotation, multiple coils arranged at different angles are used, ensuring that one coil always experiences a significant torque.

在实际电动机中，力矩在整个旋转过程中并不是恒定的。为了获得更平稳的转动，通常使用多个不同角度排列的线圈，以确保始终至少有一个线圈受到较大的力矩。

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8. Back EMF in a Motor | 电动机中的反电动势

As the coil rotates in the magnetic field, it cuts magnetic field lines, inducing an electromotive force (EMF) in the coil. According to Lenz’s Law, this induced EMF opposes the change that produced it. In a motor, the induced EMF opposes the applied voltage, and is therefore called the back EMF.

当线圈在磁场中转动时，它会切割磁感线，在线圈中感应出电动势（EMF）。根据楞次定律，这个感应电动势会阻碍产生它的变化。在电动机中，感应电动势与外加电压方向相反，因此被称为反电动势。

The back EMF affects the current flowing through the motor. The net voltage driving the current is the difference between the applied voltage and the back EMF:

反电动势会影响流过电动机的电流。驱动电流的净电压是外加电压与反电动势之差：

I = (V − ε_back) / R

Where V is the applied voltage, ε_back is the back EMF, and R is the resistance of the coil.

其中 V 是外加电压，ε_back 是反电动势，R 是线圈电阻。

When the motor is first switched on, the coil is stationary and the back EMF is zero. This causes a large surge of current, which is why motors can draw a high starting current. As the motor speeds up, the back EMF increases, reducing the current to its normal operating value.

电动机刚启动时，线圈静止不动，反电动势为零，这会导致电流突增，这就是为什么电动机在启动时会吸入很大的启动电流。随着电动机转速加快，反电动势增大，电流降至正常工作值。

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9. Power and Efficiency of an Electric Motor | 电动机的功率与效率

For a motor, the total electrical power supplied is:

对于电动机，输入的总电功率为：

P_in = V I

The useful mechanical power output is the product of the torque and the angular velocity:

有用的机械功率输出等于力矩与角速度的乘积：

P_out = T ω

Where T is the torque in N·m and ω is the angular velocity in rad/s. The power lost as heat in the coil is:

其中 T 是力矩（单位：N·m），ω 是角速度（单位：rad/s）。线圈中以热量形式损耗的功率为：

P_loss = I² R

The efficiency of the motor is the ratio of useful output power to input power:

电动机的效率是有用输出功率与输入功率之比：

η = P_out / P_in × 100%

In exam questions, you may be asked to calculate the efficiency, or to explain why a motor gets warm during operation. The answer is always the same: current flowing through a resistance generates heat by the Joule heating effect (I²R), and additionally, magnetic losses in the iron core and friction in the bearings contribute to energy losses.

考试中可能要求计算效率，或者解释为什么电动机运行时会发热。答案始终是：电流流过电阻产生焦耳热效应（I²R），此外铁芯中的磁损耗和轴承中的摩擦也会造成能量损失。

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10. Motor vs. Generator: Two Sides of the Same Coin | 电动机与发电机：一枚硬币的两面

An electric motor and a generator are essentially the same physical device used in reverse. When electrical energy is supplied, the device acts as a motor and produces mechanical rotation. When mechanical energy is supplied to rotate the coil, the device acts as a generator and produces electrical energy.

电动机和发电机本质上是同一个物理装置的反向使用。当输入电能时，装置作为电动机运转，产生机械转动。当用机械能转动线圈时，装置作为发电机运行，产生电能。

Feature | 特征	Motor | 电动机	Generator | 发电机
Energy conversion | 能量转换	Electrical → Mechanical | 电能 → 机械能	Mechanical → Electrical | 机械能 → 电能
Commutator | 换向器	Split-ring (reverses current) | 半环（反转电流）	Slip rings (AC) or split-ring (DC) | 滑环（交流）或半环（直流）
Physical effect | 物理效应	Motor effect | 马达效应	Electromagnetic induction | 电磁感应

In a DC generator, a split-ring commutator is used to produce direct current output. In an AC generator (alternator), slip rings are used so that the output current alternates direction as the coil rotates.

在直流发电机中，使用换向器（半环）来产生直流输出。在交流发电机中，使用滑环，使输出电流随线圈转动而改变方向。

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11. Common Exam Pitfalls and Tips | 常见考试陷阱与解题技巧

The following are common mistakes students make when answering questions on the motor effect and electric motors:

以下是学生在回答马达效应和电动机题目时常见的错误：

• Using the wrong hand rule: Use Fleming’s Left-Hand Rule for motors (force) and Fleming’s Right-Hand Rule for generators (induced current). Mixing these up is the single most common error.
• Forgetting conventional current direction: Remember that current flows from positive to negative in the external circuit.
• Confusing back EMF with applied EMF: Back EMF opposes the applied voltage. When the motor is stalled (not turning), back EMF is zero and current is at its maximum.
• Not using the perpendicular component: In F = BIL sin θ, always identify the angle between the conductor and the magnetic field correctly.

• 用错手定则：电动机（求力）用弗莱明左手定则，发电机（求感应电流）用弗莱明右手定则。混淆这两个定则是最常见的错误。
• 忘记传统电流方向：记住，在外电路中，电流从正极流向负极。
• 混淆反电动势与外加电动势：反电动势与外加电压方向相反。当电动机堵转（不转）时，反电动势为零，电流最大。
• 未使用垂直分量：在 F = BIL sin θ 中，务必正确识别导体与磁场之间的夹角。

Additionally, be careful with units in numerical problems. Magnetic flux density is often given in milliteslas (mT), and length may be given in centimetres. Always convert to SI units before calculation.

此外，在数值计算中要注意单位。磁感应强度常以毫特斯拉（mT）给出，长度可能以厘米给出。计算前务必将其转换为国际单位制（SI）单位。

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12. Worked Example | 例题讲解

Problem: A straight wire of length 5.0 cm carries a current of 2.0 A. It is placed in a uniform magnetic field of flux density 0.40 T, perpendicular to the field. (a) Calculate the force on the wire. (b) If the wire is rotated so that it makes an angle of 30° with the field, calculate the new force.

题目：一根长度为5.0厘米的直导线通以2.0安的电流，置于磁感应强度为0.40特斯拉的均匀磁场中，导线与磁场垂直。(a) 求导线所受的力。(b) 若将导线转动，使其与磁场成30°角，求新的力。

Solution (a):

解 (a)：

F = BIL = 0.40 × 2.0 × 0.050 = 0.040 N

Solution (b): When the wire makes an angle of 30° with the field, the angle between the wire and the field is 30°, so:

解 (b)：当导线与磁场成30°角时，导线与磁场的夹角为30°，因此：

F = BIL sin θ = 0.40 × 2.0 × 0.050 × sin 30° = 0.020 N

Note how the force is halved because sin 30° = 0.5. This demonstrates why motors are designed to keep the conductor as close to perpendicular to the field as possible for maximum torque.

注意，因为 sin 30° = 0.5，力减半。这表明电动机设计时为何要尽量使导体保持与磁场垂直以获得最大力矩。

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