Tag Archives: 电动力学和光学代写

电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

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The students are cultivated as high-quality innovative professionals. They are well developed in the
aspects of morality, intelligence, physique, and aesthetic. They possess the basic theoretical knowledge in
the fields of Materials Science and Physics，and are well trained in the applied research, technological
development, and engineering. They can research physical properties and laws of materials at the level of
molecule, atom and electron，and apply to develop new materials preparative technology，advanced
function materials and equipments. The graduates are expected to work in various industries, universities,
and research institutes in the fields of over function materials and the related (such as energy engineering,
electric power, etc.), engaging in product design, technological development, scientific research, and
management, and playing important and leading roles in the fields with international competitiveness and
innovation.

这是一份sydney悉尼大学PHYS3035/PHYS3935 的成功案例

电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写


问题 1.

$$
\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u-u_{\mathrm{e}}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}-2 u u_{\mathrm{e}}\right)\right),
$$
and for the lower section of windings, just above (1),
$$
\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u+u_{e}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}+2 u u_{e}\right)\right) \text {. }
$$


证明 .


For the upper section of the coil containing $N$ wires , combiningand yields an excess of positive charges:
$$
\Delta Q_{+}^{\prime}=N \Delta q_{+}^{\prime}=\frac{N q u_{\mathrm{e}} u}{c^{2}} .
$$
For the lower section of windings (1), the combination of yields an excess of negative charges:
$$
\Delta Q_{-}^{\prime}=N \Delta q_{-}^{\prime}=-\frac{N q u_{\mathrm{e}} u}{c^{2}}
$$







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PHYS3035/PHYS3935 COURSE NOTES ：

Derivation: Each end of the bar (or coil) produces a flux density $B_{t}=$ $\frac{\Phi}{4 \pi R^{2}}$ at the point of observation according. Only the difference of the two values is important, so that in the first principal orientation
$$
B=\frac{\phi}{4 \pi}\left(\frac{1}{(R-l / 2)^{2}}-\frac{1}{(R+1 / 2)^{2}}\right)
$$
When the distance $R$ is sufficiently large compared to the length $I$ of the bar or coil, we can neglect $P^{2}$ relative to $R^{2}$, and for the magnitude of $B$, we then obtain
$$
B=\frac{1}{2 \pi} \frac{\Phi l}{R^{3}}=\frac{\mu_{0}}{2 \pi} \frac{m}{R^{3}}
$$
Correspondingly, for the second principal orientation, we find
$$
B=\frac{\mu_{0}}{4 \pi} \frac{m^{3}}{R^{3}}
$$




















电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

0
Posted on by

The students are cultivated as high-quality innovative professionals. They are well developed in the
aspects of morality, intelligence, physique, and aesthetic. They possess the basic theoretical knowledge in
the fields of Materials Science and Physics，and are well trained in the applied research, technological
development, and engineering.

这是一份sydney悉尼大学PHYS3035/PHYS3935 的成功案例

电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写


问题 1.

$$
G(q, \omega)=\frac{1}{q^{2}-q_{0}^{2}}\left(U-\frac{q q}{q_{0}^{2}}\right)
$$
which sometimes is useful. Like the Huygens propagator, the dyadic Green function is also singular for $q=q_{0}$.

The plane-wave representation of the dyadic Green function for the magnetic field is obtained by inserting the Fourier integral transformation
$$
\boldsymbol{G}{\mathrm{M}}(\boldsymbol{R} ; \omega)=(2 \pi)^{-3} \int{-\infty}^{\infty} \boldsymbol{G}_{\mathrm{M}}(q, \omega) \mathrm{e}^{i \boldsymbol{q} \cdot \boldsymbol{R}} \mathrm{d}^{3} q
$$


证明 .


Calculations analogous to those carried out to determine $\boldsymbol{G}(\boldsymbol{R} ; \omega)$ result in
$$
\boldsymbol{G}{\mathrm{M}}(\boldsymbol{q}, \omega)=\frac{q}{q{0}} \frac{1}{q^{2}-q_{0}^{2}} \boldsymbol{U} \times \hat{\boldsymbol{q}}
$$
an expression which also is singular for $q=q_{0}$. The folding theorem in $\boldsymbol{r}$-space gives when applied to
$$
\boldsymbol{B}(q, \omega)=\frac{i \mu_{0} \omega}{c_{0}} \boldsymbol{G}{\mathrm{M}}(q, \omega) \cdot J(q, \omega) $$ A current density parallel to the $q$-direction does not give rise to a magnetic field, and the magnetic field generated by a current density perpendicular to $\hat{q}$ always lies in a plane perpendicular to $\hat{q}$. To prove the above-mentioned claims we expand the unit dyad after a triple set of orthogonal unit vectors $\hat{q}{\perp}^{(1)}, \hat{q}{\perp}^{(2)}$, and $\hat{q}$, with $\hat{\boldsymbol{q}}{1}^{(1)} \times \hat{\boldsymbol{q}}{2}^{(2)}=\hat{\boldsymbol{q}}$, i.e., $$ \boldsymbol{U}=\hat{\boldsymbol{q}}{\perp}^{(1)} \hat{q}{\perp}^{(1)}+\hat{\boldsymbol{q}}{\perp}^{(2)} \hat{\boldsymbol{q}}_{\perp}^{(2)}+\hat{q} \hat{q}
$$







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PHYS3035/PHYS3935 COURSE NOTES ：

In the mixed representation, the electric field, $\boldsymbol{E}(z ; q |, \omega)$, from a current density distribution, $\boldsymbol{J}(z: q |, \omega)$, is outside the distribution given by
$$
E\left(z ; q_{|}, \omega\right)=i \mu_{0} \omega \int_{-\infty}^{\infty} G\left(z-z^{\prime} ; q_{1}, \omega\right) \cdot J\left(z^{\prime} ; q_{|}, \omega\right) \mathrm{d} z^{\prime}
$$
where
$$
G\left(Z ; q_{|}, \omega\right)=\Gamma\left(s g n Z ; q_{1}, \omega\right) \mathrm{e}^{i x_{\perp}|Z|}
$$
with
$$
\Gamma\left(\operatorname{sgn} Z: q_{|}, \omega\right)=\frac{i}{2 q_{0}^{2} \kappa \perp}\left[q_{0}^{2} U-q_{|} q_{|}-\kappa_{\perp}^{2} \hat{z} \bar{z}-\left(q_{|}^{\hat{z}}+\hat{z} q_{|}\right) \kappa_{\perp} \operatorname{sgn} Z\right]
$$
Let us now assume that the source current density is nonvanishing only on a plane sheet located at $z^{\prime}=z_{0}$. Thus,
$$
J\left(z^{\prime} ; q_{|}, \omega\right)=J_{0}\left(q_{|}, \omega\right) \delta\left(z^{\prime}-z_{0}\right)
$$