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# 物理代写|电磁学代写Electromagnetism代考|PHYS457 Self-inductance

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## 物理代写|电磁学代写Electromagnetism代考|Self-inductance

There is a difference between the emfs and the current produced by a battery or other sources and those induced by changing the magnetic field flux.

In general, the source emf and source current describe parameters associated with a physical source. In contrast, the induced emf and induced current describe parameters associated with changing magnetic field flux.

Consider the simple electric circuit shown in Fig. 10.5. It consists of a source emf, $\epsilon$, a resistance, $R$, and a switch, $S$. When the switch is closed the source current does not instantly increase to its maximum value $I_{\max }$ :
$$I_{\max }=\frac{\epsilon}{R}$$
At some instance of time $t$ the current in the circuit is $I(t)$. Besides, the current passing through straight wire line produces a magnetic field $B=\mu_0 I / 2 \pi r$ (where $r$ is the distance from the wire), and thus $B \sim I$. Since the current increases to reach its maximum value $I_{\max }$, so does the magnetic field. Furthermore, the magnetic flux passing through the surface area enclosed by the circuit is
$$\Phi_B=\int_{\mathcal{S}} \mathbf{B} \cdot d \mathbf{S}$$
where $\mathcal{S}$ is the surface area enclosed by the circuit. Therefore, since $B$ increases, $\Phi_B$ increases with time, that is, $d \Phi_B / d t \neq 0$, which in turn creates an induced emf in the circuit:
$$\epsilon_L=-\frac{d \Phi_B}{d t}$$

## 物理代写|电磁学代写Electromagnetism代考|Mutual Inductance

The magnetic flux passing through the surface area of a loop is (see also Fig. 10.8):
$$\Phi_B=\int_{\mathcal{S}} \mathbf{B} \cdot d \mathbf{S}$$
where $|\mathbf{B}| \sim I$. Therefore, $\Phi_B$ varies with time because $I$ varies with time. Thus, an induced emf occurs through the process of mutual inductance. This is related with the fact that it depends on the interaction between two circuits.

Consider two parallel coils of $N_1$ and $N_2$ turns, respectively, as shown in Fig. $10.9$. Through the coil I is passing the current $I_1$ and coil II the current $I_2$. Suppose the current $I_1$ is creating a magnetic field with magnetic field lines as depicted in Fig. 10.9. Some of these lines pass through the coil II. We denote by $\Phi_{12}$ the magnetic flux of the magnetic field created by coil I through the coil II. The mutual inductance, namely, $M_{12}$, of coil II with respect to coil I is

$$M_{12}=N_2 \frac{\Phi_{12}}{I_1}$$
Assuming that the current $I_1$ is varying with time, then an induced emf is created at coil II, given as
$$\epsilon_{\text {ind }, 2}=-N_2 \frac{d \Phi_{12}}{d t}$$
where $\Phi_{12}$ is calculated from Eq. (10.34) as
$$\Phi_{12}=I_1 \frac{M_{12}}{N_2}$$

## 物理代写|电磁学代写Electromagnetism代考|Self-inductance

$$I_{\max }=\frac{\epsilon}{R}$$

$$\Phi_B=\int_{\mathcal{S}} \mathbf{B} \cdot d \mathbf{S}$$

$$\epsilon_L=-\frac{d \Phi_B}{d t}$$

## 物理代写|电磁学代写Electromagnetism代考|Mutual Inductance

$$\Phi_B=\int_{\mathcal{S}} \mathbf{B} \cdot d \mathbf{S}$$

$$M_{12}=N_2 \frac{\Phi_{12}}{I_1}$$

$$\epsilon_{\text {ind }, 2}=-N_2 \frac{d \Phi_{12}}{d t}$$

$$\Phi_{12}=I_1 \frac{M_{12}}{N_2}$$

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