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物理代写|电磁学代写Electromagnetism代考|Arbitrary Orientation of Two Teeth

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物理代写|电磁学代写Electromagnetism代考|Arbitrary Orientation of Two Teeth

In Section 4.4.1, we have considered the potential distribution between two identical coaxial teeth wherein each tooth is located between deep and wide slots. The modelling of potential distributions discussed in this section assumes that the axes of these two teeth are displaced by $(2 \delta)$ from each other as shown in Figure 4.7. As indicated in this figure, the air space is divided into an air gap and four slot regions referred to as regions 1 to 4 . This figure shows deep and wide slots around two identical teeth each of width $t$, separated by the air gap of length $g$. Let the tooth above the air gap $(z \geq g / 2)$ be at a magnetic potential of $-0.5 \mathrm{~A}$ and that below the air gap $(z \leq-g / 2)$ be at $+0.5 \mathrm{~A}$. The distribution of scalar magnetic potential $\mathcal{V}$, in each air region, satisfies the Laplace equation.
In region $1,(z>g / 2)$ since the value of the magnetic potential at $y=(\delta+t / 2)$ is taken as $-0.5 \mathrm{~A}$, we may tentatively express the potential distribution in this region as
$$\mathcal{V}_1^{\prime}=-\frac{1}{\pi} \cdot \tan ^{-1}\left(\frac{z-g / 2}{y-\delta-t / 2}\right)$$
This expression gives the required potential at $y=(\delta+t / 2)$ and also at $z=\infty$. Further, it gives zero potential at $y=\infty$. To this expression, a supplementary solution $\mathcal{V}_1^{\prime \prime}$ could be added that provides an arbitrary potential distribution at $z=g / 2$, and vanishes at $y=(\delta+t / 2)$ as well as at $y=\infty$. This supplementary solution must also vanish as $z \rightarrow \infty$. Therefore, using Fourier integral representation for an arbitrary function describing the potential distribution at $z=g / 2$, we have
$$\mathcal{V}_1^{\prime \prime}=-\frac{2}{\pi} \int_0^{\infty} f_1(w) \cdot \sin \left(w \cdot y_1\right) \cdot e^{-w(z-g / 2)} \cdot d w$$
Since $\mathcal{V}_1=\mathcal{V}_1^{\prime}+\mathcal{V}^{\prime \prime}$, we get
$$\mathcal{V}=-\frac{1}{\pi} \cdot \tan ^{-1}\left(\frac{z-g / 2}{y-\delta-t / 2}\right)-\frac{2}{\pi} \int_0^{\infty} f_1(w) \cdot \sin \left(w \cdot y_1\right) \cdot e^{-w(z-g / 2)} \cdot d w$$
for, $z>g / 2$ and $(y-\delta-t / 2 \geq 0$.

物理代写|电磁学代写Electromagnetism代考|Evaluation of Unknown Functions

To find values of potentials $\mathcal{V}1, \mathcal{V}_2, \mathcal{V}_3, \mathcal{V}_4$ and $\mathcal{V}_0$, the arbitrary functions $f_1(w), f_2(w), F_1^{\prime}(u), F_1^{\prime \prime}(u), F_2^{\prime}(u)$ and $F_2^{\prime \prime}(u)$ involved in Equations 4.67 through 4.70 and 4.72 are to be evaluated. For evaluation of these functions, first consider the values of $\mathcal{V}_0$ given by Equation 4.72 at $z=g / 2$ and $z=-g / 2$ These are \begin{aligned} & \left.\mathcal{V}_o\right|{z=g / 2}=\int_0^{\infty}\left{F_1^{\prime}(u) \cdot \cos (u \cdot y)+F_1^{\prime \prime}(u) \cdot \sin (u \cdot y)\right} \cdot d u \ & \left.\mathcal{V}o\right|{z=-g / 2}=\int_0^{\infty}\left{F_2^{\prime}(u) \cdot \cos (u \cdot y)+F_2^{\prime \prime}(u) \cdot \sin (u \cdot y)\right} \cdot d u \end{aligned}
In view of the anti-symmetry:
$$\left.\mathcal{V}o(y, z)\right|{z=-g / 2}=-\left.\mathcal{V}o(-y, z)\right|{z=g / 2}$$
Thus,
$$F_1^{\prime}(u)=-F_2^{\prime}(u) \stackrel{\operatorname{def}}{=} F^{\prime}(u)$$
and
$$F_1^{\prime \prime}(u)=-F_2^{\prime \prime}(u) \stackrel{\text { def }}{=} F^{\prime \prime}(u)$$
Here, it is to be noted that in view of Equations $4.74 \mathrm{a}$ and $4.74 \mathrm{~b}$ the number of arbitrary functions reduces from six to four viz. $f_1(w), f_2(w), F^{\prime}(u)$ and $F^{\prime \prime}(u)$. Thus, Equation 4.72 can be modified to
\begin{aligned} \mathcal{V}_0= & \int_0^{\infty}\left[F^{\prime}(u) \cdot \cos (u \cdot y) \cdot\left{\frac{\sinh (u \cdot z)}{\sinh (u \cdot g / 2)}\right}+F^{\prime \prime}(u) \cdot \sin (u \cdot y)\right. \ & \left.\times\left{\frac{\cosh (u \cdot z)}{\cosh (u \cdot g / 2)}\right}\right] \cdot d u \end{aligned}
Equation 4.73 a can also be rewritten as
$$\left.\mathcal{V}o\right|{z=g / 2}=\int_0^{\infty}\left{F^{\prime}(u) \cdot \cos (u \cdot y)+F^{\prime \prime}(u) \cdot \sin (u \cdot y)\right} \cdot d u$$
Continuity of potentials: In view of the continuity of potentials at $z=g / 2$, we can write
\begin{aligned} \left.\mathcal{V}o\right|{z=g / 2} & =\left.\mathcal{V}2\right|{z=g / 2} \quad \text { over }-\infty \leq y \leq-\delta \ & =1 / 2 \quad \text { over }-\delta \leq y \leq \delta \ & =\left.\mathcal{K}\right|_{z=g / 2} \quad \text { over } \delta \leq y \leq \infty \end{aligned}

物理代写|电磁学代写Electromagnetism代考|Arbitrary Orientation of Two Teeth

$$\mathcal{V}_1^{\prime}=-\frac{1}{\pi} \cdot \tan ^{-1}\left(\frac{z-g / 2}{y-\delta-t / 2}\right)$$

$$\mathcal{V}_1^{\prime \prime}=-\frac{2}{\pi} \int_0^{\infty} f_1(w) \cdot \sin \left(w \cdot y_1\right) \cdot e^{-w(z-g / 2)} \cdot d w$$

$$\mathcal{V}=-\frac{1}{\pi} \cdot \tan ^{-1}\left(\frac{z-g / 2}{y-\delta-t / 2}\right)-\frac{2}{\pi} \int_0^{\infty} f_1(w) \cdot \sin \left(w \cdot y_1\right) \cdot e^{-w(z-g / 2)} \cdot d w$$

MATLAB代写

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