91影视

Write the expression for the duality transformation equations.

$$\begin{gathered} \mathit{E}\mathbf{\text{'}}\mathbf{=}\mathit{E}\mathit{c}\mathit{o}\mathit{s}\mathit{伪}\mathbf{+}\mathit{c}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{伪} \\ \mathit{c}\mathit{B}\mathbf{\text{'}}\mathbf{=}\mathit{c}\mathit{B}\mathit{c}\mathit{o}\mathit{s}\mathit{伪}\mathbf{-}\mathit{E}\mathit{s}\mathit{i}\mathit{n}\mathit{伪} \\ \mathit{c}{\mathit{q}}_{\mathbf{e}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathit{c}\mathit{o}\mathit{s}\mathit{伪}\mathbf{+}{\mathit{q}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathit{伪} \\ {\mathit{q}}_{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{q}}_{\mathbf{m}}\mathit{c}\mathit{o}\mathit{s}\mathit{伪}\mathbf{-}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathit{s}\mathit{i}\mathit{n}\mathit{伪} \end{gathered}$$

Substitute $伪=90掳$in all the duality transformation equations.

$$\begin{gathered} E\text{'}=E\cos \left(90掳\right)+cB\sin \left(90掳\right) \\ E\text{'}=cB........\left(1\right) \end{gathered}$$

$$\begin{gathered} cB\text{'}=cB\cos \left(90掳\right)-E\sin \left(90掳\right) \\ cB\text{'}=-E.......\left(2\right) \end{gathered}$$

$$\begin{gathered} c{q}_e^{\text{'}}=c{q}_e\cos \left(90掳\right)+q_m\sin \left(90掳\right) \\ c{q}_e^{\text{'}}=q_m \end{gathered}$$

$$\begin{gathered} q_m^{\text{'}}=q_m\cos \left(90掳\right)-c{q}_e\sin \left(90掳\right) \\ {q}_n^{\text{'}}=-c{q}_e \end{gathered}$$

Write the expression for the magnetic field using the equation .

$B=-\frac{{渭}_0{p}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$

Here,$p_0=-\frac{m_0}{c}$ .

Hence, the above equation becomes,

$B=-\frac{{渭}_0{m}_0{蝇}^2}{4蟺c^2}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$

Substitute $B=-\frac{{渭}_0{m}_0{蝇}^2}{4蟺c^2}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$in equation (1).

$$\begin{gathered} E\text{'}=c\left[-\frac{{渭}_0{m}_0{蝇}^2}{4蟺c^2}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}\right] \\ E\text{'}=\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒} \end{gathered}$$

Write the expression for an electric field using the equation .

$E=\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$

Substitute $E=\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$in equation (2).

$$\begin{gathered} B\text{'}=\frac{1}{c}\left[-\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}\right] \\ {B}^{\text{'}}=\frac{-{渭}_0{m}_0{蝇}^2}{4蟺c^2}\left[\left(\frac{\sin 胃}{r}\right)\cos \left[蝇ct-\frac{r}{c}\right]\widehat{蠒}\right] \end{gathered}$$

Hence, these are identical to the fields of an ampere dipole.

Therefore, the fields that would be produced by an oscillating 鈥淕ilbert鈥� magnetic dipole is$E\text{'}=\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}\left(\frac{\sin 胃}{r}\right)\cos \left[蝇\left(t-\frac{r}{c}\right)\right]\widehat{蠒}$ and$B\text{'}=\frac{-{渭}_0{m}_0{蝇}^2}{4蟺c^2}\left[\left(\frac{\sin 胃}{r}\right)\cos \left[蝇ct-\frac{r}{c}\right]\widehat{蠒}\right]$ and the equations 11.36 and 11.37 are identical to the fields of an ampere dipole.