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# 物理代写|电磁学代写Electromagnetism代考|Eddy Currents in Circular Cores

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## 物理代写|电磁学代写Electromagnetism代考|Eddy Currents in Circular Cores

Consider a long conducting core with a circular cross-section of radius $R$, as shown in Figure 5.9. The excitation winding carrying ac currents is simulated by a current sheet with surface current density $\boldsymbol{K}$, placed on the core surface, where
$$\boldsymbol{K}=K_o \cdot e^{j \omega t} \cdot \boldsymbol{a}{\varphi}$$ At power frequencies, the magnetic field outside the core being negligible, we have $$\left.H_z\right|{\rho=R}=K_o \cdot e^{j \omega t}$$
Let
$$H_z=F(\rho) \cdot e^{j \omega t}$$
Neglecting displacement currents,
$$\nabla \times H \cong J=\sigma E$$
Therefore,
$$\nabla \times \nabla \times H=\sigma \nabla \times E=-\mu \sigma \frac{d \boldsymbol{H}}{d t}=-j \omega \mu \sigma \boldsymbol{H}=-j \omega \mu \sigma F(\rho) \cdot e^{j \omega t} \cdot \boldsymbol{a}{\varphi}$$ Now, since $$\nabla \times \boldsymbol{H}=\left[\frac{1}{\rho} \cdot \frac{\partial H_z}{\partial \varphi}-\frac{\partial H{\varphi}}{\partial z}\right] \boldsymbol{a}\rho+\left[\frac{\partial H\rho}{\partial z}-\frac{\partial H_z}{\partial \rho}\right] \boldsymbol{a}{\varphi}+\frac{1}{\rho} \cdot\left[\frac{\partial\left(\rho H{\varphi}\right)}{\partial \rho}-\frac{\partial H_z}{\partial \varphi}\right] \boldsymbol{a}z$$ and $$H\rho=H_{\varphi}=0$$
$$\nabla \times H=-\frac{\partial H_z}{\partial \rho} a_{\varphi}=\left[-\frac{d F(\rho)}{d \rho} \cdot e^{j \omega t}\right] a_{\varphi}$$
Thus, $\nabla \times \nabla \times H=\nabla \times\left[-\frac{d F(\rho)}{d \rho} \cdot e^{j \omega t}\right] \boldsymbol{a}{\varphi}=-\frac{1}{\rho} \cdot\left[\frac{d}{d \rho}\left{\rho \frac{d F(\rho)}{d \rho}\right}\right] \cdot e^{j \omega t} a{\varphi}$

## 物理代写|电磁学代写Electromagnetism代考|Distribution of Current Density in Circular Conductors

Skin effects in slot-embedded conductors with trapezoidal, ${ }^{10}$ rectangular $^{11}$ and circular ${ }^{12}$ cross-sections have been obtained. For long isolated conductors with a rectangular cross-section, skin effect has been obtained ${ }^{13}$ by assuming constant current density over the conductor surface. These assumptions, however, are not needed for conductors with a circular cross-section located in free space. Consider a long solid conductor with a circular cross-section of radius $R$, shown in Figure 5.10. This conductor carries an ac current $I$, at power frequency $\omega$. The current density $J$ in the conductor section is a function of the radial distance, $\rho$. There is only axial component of this current $J_{z^{\prime}}$ satisfying eddy current equation in the cylindrical system of space coordinates as below:
$$\frac{d^2 J_z}{d \rho^2}+\frac{1}{\rho} \cdot \frac{d J_z}{d \rho}+k^2 \cdot J_z=0$$
where
$$k=j^{3 / 2} \sqrt{\omega \mu \sigma} \stackrel{d e f}{=} j^{3 / 2} \cdot \ell$$
Ignoring the factor $e^{\text {jot, }}$ in the expression of the current density in the circular conductor, the solution of this equation is given $\mathrm{as}^9$
$$J_z=c \cdot J_o(k \rho)$$
where $c$ indicates an arbitrary constant.
Let the total current in the conductor be $I$, then
$$I=\int_0^R c \cdot J_o(k \rho) \cdot 2 \pi \rho d \rho=c \cdot 2 \pi \cdot \int_0^R J_o(k \rho) \cdot \rho d \rho=c \cdot \frac{2 \pi R}{k} \cdot J_1(k R)$$

## 物理代写|电磁学代写Electromagnetism代考|Eddy Currents in Circular Cores

$$\boldsymbol{K}=K_o \cdot e^{j \omega t} \cdot \boldsymbol{a}{\varphi}$$在功率频率下，磁芯外的磁场可以忽略不计，我们得到$$\left.H_z\right|{\rho=R}=K_o \cdot e^{j \omega t}$$

$$H_z=F(\rho) \cdot e^{j \omega t}$$

$$\nabla \times H \cong J=\sigma E$$

$$\nabla \times \nabla \times H=\sigma \nabla \times E=-\mu \sigma \frac{d \boldsymbol{H}}{d t}=-j \omega \mu \sigma \boldsymbol{H}=-j \omega \mu \sigma F(\rho) \cdot e^{j \omega t} \cdot \boldsymbol{a}{\varphi}$$现在，既然$$\nabla \times \boldsymbol{H}=\left[\frac{1}{\rho} \cdot \frac{\partial H_z}{\partial \varphi}-\frac{\partial H{\varphi}}{\partial z}\right] \boldsymbol{a}\rho+\left[\frac{\partial H\rho}{\partial z}-\frac{\partial H_z}{\partial \rho}\right] \boldsymbol{a}{\varphi}+\frac{1}{\rho} \cdot\left[\frac{\partial\left(\rho H{\varphi}\right)}{\partial \rho}-\frac{\partial H_z}{\partial \varphi}\right] \boldsymbol{a}z$$和$$H\rho=H_{\varphi}=0$$
$$\nabla \times H=-\frac{\partial H_z}{\partial \rho} a_{\varphi}=\left[-\frac{d F(\rho)}{d \rho} \cdot e^{j \omega t}\right] a_{\varphi}$$

## 物理代写|电磁学代写Electromagnetism代考|Distribution of Current Density in Circular Conductors

$$\frac{d^2 J_z}{d \rho^2}+\frac{1}{\rho} \cdot \frac{d J_z}{d \rho}+k^2 \cdot J_z=0$$

$$k=j^{3 / 2} \sqrt{\omega \mu \sigma} \stackrel{d e f}{=} j^{3 / 2} \cdot \ell$$

$$J_z=c \cdot J_o(k \rho)$$

$$I=\int_0^R c \cdot J_o(k \rho) \cdot 2 \pi \rho d \rho=c \cdot 2 \pi \cdot \int_0^R J_o(k \rho) \cdot \rho d \rho=c \cdot \frac{2 \pi R}{k} \cdot J_1(k R)$$

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