自感和互感问题
电磁感应,通常称为感应,是一个过程,其中将导体放置在特定位置并且磁场随着导体移动而变化或保持静止。因此,在电导体上产生电压或 EMF(电动势)。
迈克尔法拉第发现了这个电磁感应规则。他建立了一条类似于下图的导线,并将其连接到一个测量电路两端电压的设备上。当条形磁铁穿过设备时测量电路中的电压。这样做的意义在于,它是一种通过使用磁场而不是电池在电路中产生电能的方法。发电机、变压器和电动机等设备使用电磁感应原理。
电感器和电感
A resistor is an electrical component that opposes and regulates the flow of electricity. Resistance is the ability of a conductor to resist current flow. While a capacitor is a device that briefly stores charge and energy, capacitance refers to a capacitor’s ability to store energy.
电感器只不过是一个线圈。为了制作线圈,将导线紧紧缠绕在一起。当直流电通过它时,它将产生恒定的磁场。相反,如果我们使用交流电或不断变化的电源,流过线圈的电流就会发生变化。由于磁通量的变化,它将在与电源相反的方向上感应其电动势。这被称为电感。因此,电感是指导体抵抗电流变化的能力,但电感器是一种电气设备。
自感
如果变阻器电阻保持恒定,电池将在线圈中产生恒定电流。由于提供给它的恒定电流,线圈内部将感应出连续的磁场。
随着变阻器电阻的变化,流过线圈的电流也会发生变化。由于电流在变化,它们将是线圈内不断变化的磁通量。由于磁通量变化的影响,在这个线圈内部会产生一个电动势,试图对抗磁通量。因此,由于感应电动势,感应电流的方向将与提供的电流方向相反。
The current flowing in the coil determines the flux induced in the coil:
ϕ ∝ I
As a result, the flux-to-current ratio must be constant, as this will define the coil’s capacity to create magnetic flux in relation to the current provided. Self-inductance is the name for this constant (L).
L=ϕ/I
An inductor is the result of this process. If there are N turns on the coil, then
L=Nϕ/I
Nϕ=LI
Differentiating this equation in terms of time on both sides
d/dt (Nϕ)=d/dt (LI)
N dϕ/dt=L dI/dt
But, according to Faraday’s law of EMI, emf induced in a coil is,
ε=–N dϕ/dt
The negative sign indicates that the induced emf is in the opposite direction of the present rate of change.
ε=–L dI/dt
L=–ε /dI/dt
As a result, self-inductance may alternatively be defined as the emf induced per unit rate of current change within the coil minus the emf induced per unit rate of current change.
The SI unit of inductance is: Wb/A=Vs/A=Henry (H)
Magnetic Energy Stored in an Inductor
An inductor stores energy in the form of a magnetic field because it may induce its emf. Let us calculate the magnetic energy stored in an inductor using the following equation:
The rate of work done for a current I in the circuit may be represented as:
dW/dt=ε I
Substituting the following equation for the coil’s induced emf:
dW/dt=–L dI/dt I
dW=–LI dI
Integrating both sides of this equation:
∫dW=∫–LI dI
W=–1/2 LI2
Work done is the inverse of energy stored. So, if the inductor’s initial magnetic energy is zero, the energy stored in the inductor is:
U = 1/2 LI2
互感
自感类似于互感。相反,它仅由两个线圈组成。其中一个被赋予电流,而另一个被赋予电动势。
线圈 1 连接到图中的电源和变阻器,而线圈 2 连接到电流计。线圈 1 将开始以波动电流流动。因此,它将被移动的磁通量感应。因为线圈非常靠近,所以变化的磁通量将连接到另一个。由于通过线圈 2 的磁通量随时间不断变化,因此在线圈 2 中将感应出与该磁通量相反的电动势,从而导致电流以与提供的电流相反的方向流动。因此,我们可以声称线圈 1 中传递的电流通过线圈 2 引起磁通量。
φ 2 ∝ I 1
因此,通过线圈 2 的通量和线圈 1 中传递的电流将具有恒定比率。由于线圈 1 导致的线圈 2 中的互感是该常数的名称。
M 21 =φ 2 / I 1
我们可以使用以下等式将两个线圈之间的互感定义为与一个线圈相关的每单位电流在另一个线圈中传递的磁通量的比率。
φ 2 =M 21 I 1
取前面的等式并根据时间对其进行微分
d/dt(φ 2 )=d/dt(M 21 I 1 )
使用法拉第 EMI 定律
ε 2 =–M 21 dI 1 /dt
M 21 =–ε 2 /dI 1 /dt
结果,互感可替代地定义为感应线圈中每单位电流变化率感应的电动势减去感应线圈中每单位电流变化率感应的电动势。
电感器的应用
- 扼流电路主要采用电感器(扼流圈)。
- 在变压器中,应用了互感概念。
- 正如公式所示,它们被用来储存能量。
- 为了产生交流电,它被用于 LC 振荡器电路。
- 也用于电源转换器(ac-ac 或 dc-dc)。
示例问题
问题 1:当 2mA 的电流提供给 100 匝的线圈时,会产生 0.2Wb 大小的磁通量。求这个线圈的自感。
回答:
Current supplied to the coil I=2mA
Number of turns in the coil (N)=100
Magnetic flux linked with the coil (ϕ)=0.2Wb
Self-inductance of a coil is given by the equation
L=Nϕ/I
Substituting the values
L=100×0.2/2
∴ L=10H
This coil has a self-inductance of 10H.
问题 2:确定电感为 100mH 的电感在通过 0.2A 电流时存储的能量。
回答:
The inductance of the inductor (L)=100mH
Current passed through it (I)=0.2A
Energy stored in an inductor is given by the equation
U=1/2LI2
Substituting the values
U=1/2×100×10–3×0.2×0.2
∴U=2mJ
The energy stored in this coil is 2mJ.
问题3:互感原理是什么?
回答:
Mutual induction is based on the electromagnetic induction concept. When one coil’s magnetic field connects to the other, the second coil generates its own emf. Mutual induction is the term for this procedure.
问题 4: 两个长度为 L 且横截面积为 A 的长螺线管放置成其轴线重合。求这个系统的互感。
回答:
A long solenoid’s magnetic field is provided by:
B=μ0NI/L
As a result, if electricity is delivered through one of the solenoids, the magnetic flux within it will be connected to the other. As a result, this system’s mutual induction will be:
M=B1A/L
∴ M=μ0N1N2A/L
问题5:电感的应用有哪些?
回答:
- Inductors are commonly used in choking circuits (choke coil).
- The mutual inductance notion is used in transformers.
- They are used to store energy, as the formula says.
- It’s utilised in an LC oscillator circuit to generate alternating current.
- It’s also used in power converters (ac-ac or dc-dc).
问题6:互感有什么用?
回答:
Mutual induction is used in transformers, generators, motors, etc.
问题7:电感有什么用?
回答:
Inductance is required to induce back emf and keep current flowing even after the switch is turned off.
问题 8:长螺线管有 500 转。当通过 2 A 的电流时,与螺线管每匝相连的磁通量为 4×10 -3 Wb。找到自感。
回答:
Φ=4×10−3 Wb/turn
N= Number of turns =500
I= Current flowing =2A
N×Φ=L×I
∴500×4×10−3
=L×2
∴L= 500×4×10−3/2
∴L=1H
∴ Self-inductance of solenoid is 1.0H
问题 9:螺线管(空心)有 400 圈,长 20 厘米,横截面为 4 厘米2 。然后求自感应系数。
回答:
n=400turns ,l=20cm ,A=4cm2
L= 1⋅25×10−6Hm−1×400×400×4×100 / 100×100×20
=4×10−4H
问题 10:当两个线圈之间的 4A 电流在 0.5 秒内在初级变为 12A 并在次级感应出 50mV 的电动势。计算两个线圈之间的互感。
回答:
M= e2/ di2/dt
=(50 x 10-3) / (8/0.5)
=3.125 x 10-3 H
=3.125 mH