# Electric field vector.

Discussion in 'Homework Help' started by lam58, May 19, 2014.

1. ### lam58 Thread Starter Member

Jan 3, 2014
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Q) A harmonic time-dependent electromagnetic plane wave, of angular frequency ω, propagates along the positive z-direction in a source-free medium with σ = 0, ε = 1 and µ = 3. The magnetic field vector for this wave is: H = Hy uy. Use Maxwell’s equations to determine the corresponding electric field vector.

Ans) I've pretty much forgotten all this stuff from 1st year, so I'm not sure if my answer is correct.

$\bigtriangledown \times H = \varepsilon_0 \frac{\partial E}{\partial t}$, $H = (0, Hy, 0)$

$\bigtriangledown \times H = \begin{vmatrix} \hat{u}x & \hat{u}y & \hat{u}z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & Hy & 0 \end{vmatrix}$

$= \hat{u}x (\frac{-\partial Hy}{\partial z}) - \hat{u}y(0) + \hat{u}z (\frac{\partial Hy}{\partial z}) = (\frac{-\partial Hy}{\partial z}, 0, 0)$

and

$\varepsilon_0 \frac{\partial E}{\partial t}= (\varepsilon_0 \frac{\partial E}{\partial t}, 0, 0)$

$\Rightarrow \varepsilon_0 \frac{\partial E}{\partial t}= \frac{-\partial Hy}{\partial z}$

2. ### studiot AAC Fanatic!

Nov 9, 2007
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was the uy at the end a typo?

and did you solve the partial equation in the last line since you are asked for E (x,y,z), not partial dE/dt?

3. ### lam58 Thread Starter Member

Jan 3, 2014
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Not sure about the uy, that's just what it said in the question.

I think I need to put it in the form Hy to get E(x,y,z)???

4. ### studiot AAC Fanatic!

Nov 9, 2007
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Perhaps the uy means u sub y the unit vector in the y direction?

Do you not think the question supplied these for a reason?

5. ### lam58 Thread Starter Member

Jan 3, 2014
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Would I be right in say the E is then E = $(\frac{-\partial Hy}{\partial z} \frac{j}{\omega \varepsilon_0}, 0, 0)$

6. ### lam58 Thread Starter Member

Jan 3, 2014
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Oh that's a good point. Lol.

7. ### studiot AAC Fanatic!

Nov 9, 2007
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Last edited: May 19, 2014
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8. ### lam58 Thread Starter Member

Jan 3, 2014
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But would't that mean J = 0?

Tbh I'm not entirely sure how you get from curl(H) to E. I'm trying to base this on what I remember and from maths and eng physics from 2 years ago, but I'm really confused now.

Last edited: May 19, 2014
9. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes since the conductivity is zero there is no conduction current.

Therefore the current is all displacement current.

It's terribly slow tonight

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10. ### studiot AAC Fanatic!

Nov 9, 2007
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You need the following Maxwell relations

$\nabla \times H = J + \frac{{\partial D}}{{\partial t}}$

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

$D = \varepsilon E$

$J = \sigma E$

$B = \mu H$

Since the wave travels in the z direction, the electric field is in the x direction and the magnetic field in the y direction. Thus every component of E is zero except Ex

So nabla cross E reduces to

$\frac{{\partial {E_x}}}{{\partial z}}{u_y} = - \frac{{\partial B}}{{\partial t}}$

Since J=0 the magnetic equation is reduced to

$- \frac{{\partial {H_y}}}{{\partial z}} = \frac{{\partial {D_x}}}{{\partial t}}$

The electric terms may be eliminated by differentiating the first derived partial equation with respect to z and the second with respect to t and sunstitutions made from the remaining Maxwell relations to arrive at a wave equation in Hy.

This may be solved using the numbers given for ε and μ to find the required equation in Hy

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11. ### lam58 Thread Starter Member

Jan 3, 2014
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Going by what you say, I think then because $\frac{{\partial {E_x}}}{{\partial z}}{u_y} = - \frac{{\partial B}}{{\partial t}} = -\mu \mu_0 \frac{ \partial Hy}{\partial t}$,

if I take the partial derivatives of both (one with respect to t and the other to z), I get $\varepsilon_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 H_y}{\partial t \partial z}$

and

$\mu \mu_0 \frac{\partial^2 H_y}{\partial t \partial z} = -\frac{\partial^2 E_x}{\partial z^2}$.

This then give $\mu \mu_0 \varepsilon_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 E_x}{\partial z^2} \Rightarrow \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}$

Subbing in the numbers for $\mu \mu_0 \varepsilon_0$

$\frac{\partial^2 E_x}{\partial t^2} = {3{\times 10^{16}} \frac{\partial^2 E_x}{\partial z^2}$

12. ### studiot AAC Fanatic!

Nov 9, 2007
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You have eliminated H to find a wave equation in E, but you are asked to eliminate E to find a wave equation in H.

This is not so bad since they are essentially the same except that E is a function of x only and H is a function of y only.

So how about the last line - the solution of the equation?
Don't forget this is an equation showing Hy as a function of y only (since Hx=Hx=0)

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13. ### studiot AAC Fanatic!

Nov 9, 2007
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I know I was saying you have to find the mag field, but I've just re-read the question and, of course, I was wrong you need to find the E field so

This is the homework section so I won't tell you the answer but since you have done all that work on the E filed equation here is the answer for the magnetic field, (without putting in the constants)

${H_y} = {H_0}\exp \left[ {j\left( {\omega t - kz} \right)} \right]$

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14. ### lam58 Thread Starter Member

Jan 3, 2014
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Hey thanks, you've really helped a lot. I was so stuck yesterday with this question and was getting really frustrated but I'm starting to understand it all now.

15. ### studiot AAC Fanatic!

Nov 9, 2007
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Did you note that every piece of the information in the question was needed?

For instance the constitutive relations only apply for homogeneous, isotropic media.
For non isotropic/homogeneous media mu, sigma and epsilon are higher order tensors and the realtions are transormations between vecotr fields, which can lead to some hairy maths.

The wave was stated to be time dependent and plane (and therefore travelling) which picks out certain solutions of the wave equation from others.

Glad to have helped, I'm sure you covered a few sheets of paper filling in the many gaps I left. Well done.