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

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

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:
$$\Delta Q_{-}^{\prime}=N \Delta q_{-}^{\prime}=-\frac{N q u_{\mathrm{e}} u}{c^{2}}$$

## 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
$$B=\frac{\mu_{0}}{4 \pi} \frac{m^{3}}{R^{3}}$$