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Write the expression for Maxwell鈥檚 equation of electromagnetism in a vacuum.$$\begin{gathered} \mathbf{鈭�}\mathbf{}\mathbf{.}\mathbf{}\mathit{E}\mathbf{=}\mathbf{0} \\ \mathbf{鈭�}\mathbf{脳}\mathit{E}\mathbf{=}\frac{\mathbf{鈭�}\mathbf{B}}{\mathbf{鈭�}\mathbf{t}} \\ \mathbf{鈭�}\mathbf{}\mathbf{.}\mathbf{}\mathit{B}\mathbf{=}\mathbf{0} \\ \mathbf{鈭�}\mathbf{脳}\mathit{B}\mathbf{=}{\mathit{渭}}_{\mathbf{0}}{\mathit{蔚}}_{\mathbf{0}}\frac{\mathbf{鈭�}\mathbf{E}}{\mathbf{鈭�}\mathbf{t}} \end{gathered}$$

Here, E is the electric field, Bis the magnetic field,${渭}_0$is the permeability of free space, and${蔚}_0$is the permittivity of free space.

Write the expression for the electric field for a given vector and scalar potential.

$\mathbf{E}=-鈭�\mathrm{V}\frac{鈭�\mathbf{A}}{鈭�\mathrm{t}}$

Substitute$V=0$and$\mathbf{A}={\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}$in the above expression.

$$\begin{gathered} \mathbf{E}=-鈭�\left(0\right)\frac{鈭�}{鈭�\mathrm{t}}\left({\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}\right) \\ \mathbf{E}=-鈭�{\mathrm{A}}_0\left(-\mathrm{蝇}\right)\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}} \\ \mathbf{E}={\mathbf{A}}_0\mathrm{蝇}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}} \end{gathered}$$

Write the expression for the magnetic field for a given vector and scalar potential.

$\mathbf{B}=鈭�脳\mathbf{A}$

Substitute$\mathbf{A}={\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}$the above expression.

$\mathbf{B}=\left|\begin{array}{ccc}\hat{x}& \hat{y}& \hat{z}\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ 0& A_0\sin \left(kx-蝇t\right)& 0\end{array}\right|脳\left[A_0\sin \left(kx-蝇t\right)\right]$

$\mathbf{B}=\left[\begin{array}{ccccc}\hat{\mathrm{x}}\left(\frac{鈭�}{鈭�\mathrm{y}}\left(0\right)-\frac{鈭�}{鈭�\mathrm{z}}\left({\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\right)\right)-& \hat{\mathrm{y}}& \left(\frac{鈭�}{鈭�\mathrm{x}}\left(0\right)-\frac{鈭�}{鈭�\mathrm{z}}\left(0\right)\right)+& & \\ & & & & \\ \hat{\mathrm{z}}\left(\frac{鈭�}{鈭�\mathrm{x}}\left({\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)-\frac{鈭�}{鈭�\mathrm{y}}\left(0\right)\right)\right)& & & & \end{array}\right]$

$$\begin{gathered} \mathbf{B}=\hat{\mathrm{y}}\left(0-0\right)-\hat{\mathrm{y}}\left(0\right)+\left[\frac{鈭�}{鈭�\mathrm{X}}{\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\right]\hat{\mathrm{Z}} \\ \mathbf{B}={\mathrm{A}}_0\mathrm{k}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{Z}} \end{gathered}$$

Calculate $鈭�\mathbf{.}\mathbf{E}\mathbf{.}$

$$\begin{gathered} 鈭�.\mathbf{E}=\left(\hat{\mathrm{X}}\frac{鈭�}{鈭�\mathrm{X}}+\hat{\mathrm{Y}}\frac{鈭�}{鈭�\mathrm{Y}}+\hat{\mathrm{Z}}\frac{鈭�}{鈭�\mathrm{Z}}\right).\left[{\mathrm{A}}_0\mathrm{蝇}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}\right] \\ 鈭�.\mathbf{E}=\frac{鈭�}{鈭�\mathrm{Y}}\left[{\mathrm{A}}_0\mathrm{蝇}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\right] \\ 鈭�\mathbf{}\mathbf{.}\mathbf{E}=0 \end{gathered}$$

Hence, Maxwell鈥檚 first equation is satisfied.

Calculate$鈭�脳\mathbf{E}$:

$鈭�脳\mathbf{E}\mathbf{\left|}\begin{array}{ccc}\hat{\mathbf{X}}& \hat{\mathbf{Y}}& \hat{\mathbf{Z}}\\ \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{X}}& \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{Y}}& \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{X}}\\ \mathbf{0}& {\mathbf{A}}_{\mathbf{0}}\mathbf{蝇}\mathbf{}\mathbf{cos}\mathbf{(}\mathbf{kx}\mathbf{-}\mathbf{蝇t}\mathbf{)}& \mathbf{0}\end{array}\mathbf{\right|}$

$鈭�脳\mathbf{E}=\left[\begin{array}{cc}\hat{\mathrm{x}}\left[\frac{鈭�}{鈭�\mathrm{y}}\left(0\right)-\frac{鈭�}{鈭�\mathrm{Z}}\left({\mathrm{A}}_0\mathrm{蝇cos}\left(\mathrm{Kx}-\mathrm{蝇t}\right)\right)\right]-\hat{\mathrm{y}}\left[\frac{鈭�}{鈭�\mathrm{X}}\left(0\right)-\frac{鈭�}{鈭�\mathrm{Z}}\left(0\right)\right]& +\\ \hat{\mathrm{Z}}\left[\frac{鈭�}{鈭�\mathrm{X}}\left({\mathrm{A}}_0\mathrm{蝇cos}\left(\mathrm{Kx}-\mathrm{蝇t}\right)\right)-\frac{鈭�}{鈭�\mathrm{y}}\left(0\right)\right]& \end{array}\right]$

$$\begin{gathered} 鈭�脳\mathbf{E}\mathbf{=}\frac{\mathbf{鈭�}}{\mathbf{鈭�}}\left[{\mathrm{A}}_0\mathrm{蝇cos}\left(\mathrm{Kx}-\mathrm{蝇t}\right)\right]\hat{\mathbf{Z}} \\ 鈭�脳\mathbf{E}=-{\mathrm{A}}_0\mathrm{蝇sin}\left(\mathrm{Kx}-\mathrm{蝇t}\right)\hat{\mathrm{Z}} \end{gathered}$$

Calculate $\frac{鈭�\mathbf{B}}{鈭�t}$.

Substitute $\mathbf{B}={\mathrm{A}}_0\mathrm{k}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{z}}$in the above value.

$$\begin{gathered} \frac{鈭�\mathbf{B}}{鈭�t}=\frac{鈭�}{鈭�t}\left[A_{0}k\cos \left(kx-蝇t\right)\right]\hat{z} \\ \frac{鈭�\mathbf{B}}{鈭�t}=-A_{0}k\sin \left(kx-蝇t\right)\left(-蝇\right)\hat{z} \\ \frac{鈭�\mathbf{B}}{鈭�t}=A_{0}k蝇\sin \left(kx-蝇t\right)\hat{z} \end{gathered}$$

Substitute$\frac{鈭�\mathbf{B}}{鈭�t}=A_{0}k蝇\sin \left(kx-蝇t\right)\hat{z}$in equation (1).

$鈭�脳\mathbf{E}=-\frac{鈭�\mathbf{B}}{鈭�\mathrm{t}}$

Hence, Maxwell鈥檚 second equation is satisfied.

Calculate$鈭�.\mathbf{B}\mathbf{}\mathbf{.}$

$$\begin{gathered} 鈭�.\mathbf{B}=\left(\hat{\mathrm{x}}\frac{鈭�}{鈭�\mathrm{x}}+\hat{\mathrm{y}}\frac{鈭�}{鈭�\mathrm{y}}+\hat{\mathrm{z}}\frac{鈭�}{鈭�\mathrm{z}}\right).\left[{\mathrm{A}}_0\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{z}}\right] \\ 鈭�.\mathbf{B}=\frac{鈭�}{鈭�\mathrm{z}}\left[{\mathrm{A}}_0\mathrm{k}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\right] \\ 鈭�.\mathbf{B}=0 \end{gathered}$$

Hence, Maxwell鈥檚 third equation is satisfied.

Calculate $鈭�脳\mathbf{B}={\mathrm{渭}}_0{\mathrm{蔚}}_0\frac{鈭�\mathbf{E}}{鈭�\mathrm{t}}$

$\mathbf{B}=\left|\begin{array}{ccc}\hat{\mathrm{x}}& \hat{\mathrm{y}}& \hat{\mathrm{z}}\\ \frac{鈭�}{鈭�\mathrm{x}}& \frac{鈭�}{鈭�\mathrm{y}}& \frac{鈭�}{鈭�\mathrm{z}}\\ 0& {\mathrm{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)& 0\end{array}\right|脳\left[{\mathbf{A}}_0\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\right]$

localid="1654684353464" $$\begin{gathered} 鈭�脳\mathbf{B}=-\frac{鈭�}{鈭�\mathrm{X}}\left[{\mathrm{A}}_0\mathrm{k}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\right]\hat{\mathrm{y}} \\ 鈭�脳\mathbf{B}=-{\mathrm{A}}_0{\mathrm{k}}^2\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}} \end{gathered}$$

Calculate localid="1654684988366" $\frac{鈭�\mathbf{B}}{鈭�t}$

Substitute localid="1654684993827" $\mathbf{E}={\mathrm{A}}_0\mathrm{蝇}\cos \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}$in the above value.

localid="1654685000985" $$\begin{gathered} \frac{鈭�\mathbf{E}}{鈭�t}=\frac{鈭�}{鈭�t}\left[A_0蝇\cos \left(kx-蝇t\right)\hat{y}\right] \\ \frac{鈭�\mathbf{E}}{鈭�t}=-A_0蝇\sin \left(kx-蝇t\right)\left(-蝇\right)\hat{y} \\ \frac{鈭�\mathbf{E}}{鈭�t}=A_0{蝇}^2\sin \left(kx-蝇t\right)\left(-蝇\right)\hat{y}......\left(2\right) \end{gathered}$$

Now, if localid="1654685007918" $k^2={渭}_0{蔚}_0{蝇}^2$ then, the value of localid="1654685017837" $鈭�脳\mathbf{B}$becomes,

localid="1654685026709" $鈭�脳\mathbf{B}=-{\mathrm{A}}_0{\mathrm{渭}}_0{\mathrm{蔚}}^2\sin \left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}......\left(3\right)$

Multiply by localid="1654685051493" ${渭}_0{蔚}_0$on both the sides in equation (2).

localid="1654685058720" $\left({渭}_0{蔚}_0\right)\frac{鈭�\mathbf{E}}{鈭�t}=A_0{渭}_0{蔚}_0{蝇}^2\sin \left(kx-蝇t\right)\hat{y}$

Substitute in equation (3).

localid="1654685076197" $\left({渭}_0{蔚}_0\right)\frac{鈭�\mathbf{E}}{鈭�t}=A_0{渭}_0{蔚}_0{蝇}^2\sin \left(kx-蝇t\right)\hat{y}$

Substitute localid="1654685099465" $\left({渭}_0{蔚}_0\right)\frac{鈭�\mathbf{E}}{鈭�t}=A_0{渭}_0{蔚}_0{蝇}^2\sin \left(kx-蝇t\right)\hat{y}$in equation (3).

localid="1654685109547" $鈭�脳\mathbf{B}={\mathrm{渭}}_0{\mathrm{蔚}}_0\frac{鈭�\mathbf{E}}{鈭�\mathrm{t}}$

Hence, Maxwell鈥檚 fourth equation is satisfied.

Write the relation between k and $蝇$.

It is known that:

$\frac{1}{C^2}={渭}_0{蔚}_0$

Substitute $\frac{1}{C^2}={渭}_0{蔚}_0$in equation (4).

$k^2=\frac{1}{C^2}{蝇}^2$

$$\begin{gathered} k=\frac{1}{C}蝇 \\ 蝇=ck \end{gathered}$$

Therefore, the electric field is $\mathbf{E}={\mathrm{A}}_0\mathrm{蝇cos}\left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{y}}$, the magnetic field is $\mathbf{A}={\mathrm{A}}_0\mathrm{kcos}\left(\mathrm{kx}-\mathrm{蝇t}\right)\hat{\mathrm{z}}$, and the imposing condition on$蝇$and k is $蝇=ck$.