# Maxwell's equations

1. Nov 1, 2004

### AXIS

Having some problems with my Electromag Unit again.

Heres the questions I'm having problems with, any help would be appreciated

1

For plane electromagnetic waves in a homogeneous, linear, uncharged non conductor

Formulate the Maxwell equations for the electromagnetic fields E and B.

2

Given are the

i) linearly polarised, plane wave E= Eo sin (kz-wt)
ii) circularly polarised, plane wave E = Eo(cos(kz-wt)ex + sin(kz-wt)ey)

Compute for both waves the magnetic flux density B and the Poynting vector P = E x H

Last edited: Nov 1, 2004
2. Nov 1, 2004

### Tide

You're not going to formulate the Maxwell equations. You're going go formulate wave equations and their solutions from Maxwell's equations.

Your basic approach will be to work with Ampere's and Faraday's laws (the curl equations). Curl both sides of either one and use the other one to remove the curl in the partial derivatives with respect to time. You have just derived the wave equations! Now apply them to your specific problem. (HINT: You'll have use a vector identity on the curl of a curl in deriving the wave equations which is almost certainly contained in your textbook - if not then demand a refund!)

3. Nov 1, 2004

### AXIS

Are you talking about Question 1 or 2

4. Nov 1, 2004

### AXIS

Okay forget about Question 1, I managed to work it out for myself,

So can anyone do Question 2?

5. Nov 1, 2004

### Tide

You should be able to do Question 2 after I told you how to do it?

6. Nov 1, 2004

### marlon

No Tide, the equations describing the waves in this particular medium are the Maxwell-equations. Ofcourse the resulting E and B equations have to respect these specific Maxwell-equations. Besides a problem in electrostatics and magnetostatics has the property that the general Maxwell-equations loose their time derivatives and become perfectly separated from each other. Yet this is NOT the case : E is dependent of time so you cannot just loose the time derivative.

This is the Maxwell-equation you should use : rotor E = -dB/dt. Since E is given in the two situations, just calculate the rotor (you know, with a determinant). Then B can be found (or H = B/mu) by integrating the corresponding components with respect to the time. So the x component of B equals the integral with respect to t of the x-component of the rotor. This gives you the B or H vector and once this one is know it is real easy to calculate the Poynting vector. Just calculate the vectorproduct and your problem is solved.

Just to be clear : the real part of the Poynting vector corresponds to the outward dissipated energy through the surface of an encapsulated volume. This volume arises when working with Gauss's law...The energy arises from currents in the volume, that generate the EM-fields. This energy is delivered by these currents in order to maintain the generated EM-fields E and B. Ofcourse conservation of energy states that this energy needs to dissipate out of the surface so that the net amount of energy in the encapsulated volume is always CONSTANT.

regards
marlon

Last edited: Nov 1, 2004
7. Nov 1, 2004

### Tide

Marion,

Two points: First, Maxwell's equations have already been formulated - by Maxwell, incidentally! Axis is not reformulating Maxwell's equations but applying them. It sounds picky but it's an important distinction.

Second, perhaps I didn't make myself clear but I am not suggesting the time derivatives go away. I am saying that when you take the curl of the time derivative terms you can interchange the order of spatial derivatives and temporal derivatives so the curl of the time derivative becomes the time derivative of the curl allowing simplification of the wave equation.

8. Nov 1, 2004

### marlon

Tide,
in that case i stand corrected and i apologize for misreading your words...

regards
marlon

9. Nov 1, 2004

### AXIS

I need to Compute for both waves the magnetic flux density B and the Poynting vector P = E x H.

Now I figure I need to use faradays law to find B, but how do I do Del cross E where E= Eo sin (kz-wt) and again for E = Eo(cos(kz-wt)ex + sin(kz-wt)ey)

10. Nov 1, 2004

### AXIS

Basically I don't know how to find the curl of a plane wave.

11. Nov 2, 2004

### Tide

Well, yes, you can but that's not the point.

Here's a simplified version:

If

$$\nabla \times \vec E = -\frac {1}{c} \frac {\partial \vec B}{\partial t}$$

and

$$\nabla \times \vec B = \frac {1}{c} \frac {\partial \vec E}{\partial t}$$

then

$$\nabla \times \nabla \times \vec E = - \frac {1}{c} \frac {\partial}{\partial t} \nabla \times \vec B$$

from which you obtain the wave equation

$$\nabla^2 \vec E = \frac {1}{c^2} \frac {\partial^2 E}{\partial t^2}$$

using the identity $\nabla \times \nabla \times \vec E = \nabla \nabla \cdot \vec E - \nabla^2 \vec E$ and using the fact that $\nabla \cdot \vec E = 0$ in free space. $\vec B$ is governed by a similar equation.

You can easily verify the expressions you provided in your original post are solutions this wave equation. Of course, you will use the form of Maxwell's equations written in the system of units to which you are accustomed and include additional physics as needed such as electrical currents and dielectric effects.

12. Nov 2, 2004

### ehild

You can find the curl of a vector given with its cartesian components as

$$curl\vec A = (\frac {\partial A_y}{\partial z}- \frac {\partial A_z}{\partial y})\vec e_x + (\frac {\partial A_x}{\partial z}- \frac {\partial A_z}{\partial x})\vec e_y+(\frac {\partial A_y}{\partial x}- \frac {\partial A_x}{\partial y})\vec e_z$$

The electric field in the planar wave in question 2 propagates in the z direction, it depends only on z (and the time, t). Both the electric and magnetic fields are perpendicular to the direction of propagation in an isotopic medium, (and you have such simple wave) so you can suppose that the electric field is parallel to the x axis.

$$\vec E = E_x\vec e_x = E_{x0} \sin (kz-\omega t) \vec e_x$$.

As this wave has only x component and depends only on the coordinate z, its curl is very simple.

$$curl \vec E = \frac {\partial E_x}{\partial z}\vec j = E_{x0}*k*\cos (kz-\omega t)\vec e_y$$

According to Maxwell's second equation :

$$curl \vec E = - \frac{\partial\vec B}{\partial t}$$

so

$$\frac{\partial \vec B}{\partial t}=-E_{x0}*k*\cos (kz-\omega t)\vec e_y$$

Now you have to integrate with respect to time and you get the magnetic flux density.

ehild

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