Question

Suppose$\mathbf{v}\mathbf{=}\mathbf{0}$ andlocalid="1654682194645" $\mathit{A}\mathbf{=}{\mathit{A}}_{\mathbf{0}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{‐}\mathbf{\omega }\mathbf{t}\mathbf{)}\hat{\mathbf{y}}$, wherelocalid="1654682226085" ${\mathit{A}}_{\mathbf{0}\mathbf{}\mathbf{,}}\mathit{\omega }$, and kare constants. Find E and B, and check that they satisfy Maxwell’s equations in a vacuum. What condition must you impose localid="1654682236104" $\mathit{\omega }$on andk?

Short answer

The electric field is $\mathbf{E}=A_0\omega \cos \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{\mathrm{y}}$, the magnetic field is$\mathbf{B}=A_{0}k\cos \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{z}$, and the imposing condition on $\omega$and k is$\omega =ck$.

Step-by-step solution

Expression for Maxwell’s equation of electromagnetism in a vacuum:

Write the expression for Maxwell’s equation of electromagnetism in a vacuum.$$\begin{gathered} \mathbf{\nabla }\mathbf{}\mathbf{.}\mathbf{}\mathit{E}\mathbf{=}\mathbf{0} \\ \mathbf{\nabla }\mathbf{\times }\mathit{E}\mathbf{=}\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}} \\ \mathbf{\nabla }\mathbf{}\mathbf{.}\mathbf{}\mathit{B}\mathbf{=}\mathbf{0} \\ \mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{t}} \end{gathered}$$

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

Determine the electric and magnetic field:

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

$\mathbf{E}=-\nabla \mathrm{V}\frac{\partial \mathbf{A}}{\partial \mathrm{t}}$

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

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

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

$\mathbf{B}=\nabla \times \mathbf{A}$

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

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

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

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

Satisfy Maxwell’s equation ∇ .E =0and ∇×E=-∂B∂t :

Calculate $\nabla \mathbf{.}\mathbf{E}\mathbf{.}$

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

Hence, Maxwell’s first equation is satisfied.

Calculate$\nabla \times \mathbf{E}$:

$\nabla \times \mathbf{E}\mathbf{\left|}\begin{array}{ccc}\hat{\mathbf{X}}& \hat{\mathbf{Y}}& \hat{\mathbf{Z}}\\ \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{X}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{Y}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{X}}\\ \mathbf{0}& {\mathbf{A}}_{\mathbf{0}}\mathbf{\omega }\mathbf{}\mathbf{cos}\mathbf{(}\mathbf{kx}\mathbf{-}\mathbf{\omega t}\mathbf{)}& \mathbf{0}\end{array}\mathbf{\right|}$

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

$$\begin{gathered} \nabla \times \mathbf{E}\mathbf{=}\frac{\mathbf{\partial }}{\mathbf{\partial }}\left[{\mathrm{A}}_0\mathrm{\omega cos}\left(\mathrm{Kx}-\mathrm{\omega t}\right)\right]\hat{\mathbf{Z}} \\ \nabla \times \mathbf{E}=-{\mathrm{A}}_0\mathrm{\omega sin}\left(\mathrm{Kx}-\mathrm{\omega t}\right)\hat{\mathrm{Z}} \end{gathered}$$

Calculate $\frac{\partial \mathbf{B}}{\partial t}$.

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

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

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

$\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial \mathrm{t}}$

Hence, Maxwell’s second equation is satisfied.

Satisfy Maxwell’s equation ∇ .B=0 and ∇×B= μ0ε0 ∂E∂t :

Calculate$\nabla .\mathbf{B}\mathbf{}\mathbf{.}$

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

Hence, Maxwell’s third equation is satisfied.

Calculate $\nabla \times \mathbf{B}={\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\frac{\partial \mathbf{E}}{\partial \mathrm{t}}$

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

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

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

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

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

Now, if localid="1654685007918" $k^2={\mu }_0{\epsilon }_0{\omega }^2$ then, the value of localid="1654685017837" $\nabla \times \mathbf{B}$becomes,

localid="1654685026709" $\nabla \times \mathbf{B}=-{\mathrm{A}}_0{\mathrm{\mu }}_0{\mathrm{\epsilon }}^2\sin \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{\mathrm{y}}......\left(3\right)$

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

localid="1654685058720" $\left({\mu }_0{\epsilon }_0\right)\frac{\partial \mathbf{E}}{\partial t}=A_0{\mu }_0{\epsilon }_0{\omega }^2\sin \left(kx-\omega t\right)\hat{y}$

Substitute in equation (3).

localid="1654685076197" $\left({\mu }_0{\epsilon }_0\right)\frac{\partial \mathbf{E}}{\partial t}=A_0{\mu }_0{\epsilon }_0{\omega }^2\sin \left(kx-\omega t\right)\hat{y}$

Substitute localid="1654685099465" $\left({\mu }_0{\epsilon }_0\right)\frac{\partial \mathbf{E}}{\partial t}=A_0{\mu }_0{\epsilon }_0{\omega }^2\sin \left(kx-\omega t\right)\hat{y}$in equation (3).

localid="1654685109547" $\nabla \times \mathbf{B}={\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\frac{\partial \mathbf{E}}{\partial \mathrm{t}}$

Hence, Maxwell’s fourth equation is satisfied.

Determine the imposing condition on ω and k:

Write the relation between k and $\omega$.

$k^2={\mu }_0{\epsilon }_0{\omega }^2......\left(4\right)$

It is known that:

$\frac{1}{C^2}={\mu }_0{\epsilon }_0$

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

$k^2=\frac{1}{C^2}{\omega }^2$

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

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