# Finding the relative permittivity and permeability of the medium

Discussion in 'Homework Help' started by u-will-neva-no, Feb 29, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hey everyone!

I have been given an electric field and magnetic field equation which are:
$\vec{E} = \hat{x}10cos(6\pi10^7t - 0.8\pi.z)$

$\vec{H} = \hat{y}\frac{1}{6\pi}cos(6\pi10^7t - 0.8\pi.z)$

I have to find the relative permittivity an permeability of the medium. Does anyone have steps and or equations that I could use to solve this?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
Since you are trying to find two values, you need two equations. Permeability and permittivity will affect both wave speed and wave impedance. You need to study these two concepts and determine how to extract the information.

Hint: Field amplitude ratio, wave frequency and wavelength are key aspects to this problem.

We won't do your work for you, so you'll need to show some effort along these lines.

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3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
@steveb, thank you for the reply.

I understand the rules of a forum. I personally do not want the answer given to me. Hopefully my above question did not infer that I wanted the solution posted!

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Ah yes, I know you understand the rules. I answered so quickly that I didn't notice it was you asking. I just gave my standard warning when methods are asked for without work shown.

So, you should be able to take the hints and find the relevant formulas. Then I think the steps will be clear to you. In a nutshell, either field gives information about wavelength and wave frequency, which allow wave speed to be calculated. The ratio of the fields gives the wave impedance. Both wave impedance and wave speed are related to permeability and permittivity.

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5. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
No worries! I am still thinking of the solution and should be able to solve it based on the hints you have provided. Thanks for the help steveb!

6. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Look like I still need some help..

I first found the ratio of the fields to find the wave impedance (n) and so i calculated that to be $n = 60\pi$

I can extract $\omega = 6\pi.10^7$ so $f = \frac{6\pi.1x10^7}{2\pi} = 3x10^7 Hz$

Using the equation $v = f\lambda$ I can find $\lambda = \frac{3x10^8}{3x10^7} = 10 m^-^1$

I guess Im wrong as I assumed that $v = 3x10^8$ but I do not know how to calculate the wave speed given the proportional relationship between the frequency and wavelength.

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, yes you are wrong here. You have the correct formula for wave speed v=fλ. However you can't assume that v is the speed of light in vacuum because the problem explicitly says the wave is in a medium other than vacuum. The wavelength can be found from other information in the field. The sinusoidal variation of the wave is of the form cos(ωt-kz), where ω=2∏f and k=2∏/λ. There is also another formula for the speed which uses ω and k directly: v=ω/k.

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8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Most texts on the subject have equations similar to those shown in the attachment. Based on your E & H equations this appears to be a loss-less transmission situation.

β is the phase constant = 0.8*∏ in your example

The intrinsic wave impedance η would be defined as η=|E|/|H|

• ###### Lossless transmission equations.jpg
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41.5 KB
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9. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
I must still be doing something wrong so let me show you what I have done:

$w = 2\pi.f = 6\pi.10^7 => f = 3x10^7$

$k = \frac{2\pi}{k} => \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.8\pi} = 2.5$ so $v = 75 x10^6$

$v = \frac{c}{\sqrt{\mu_r.\varepsilon_r}} =>\mu_r.\varepsilon_r = (\frac{c}{v})^2 = 16$ (equation 1)

$n = \frac{|E|}{|H|} = \frac{|10|}{|\frac{1}{6\pi}|} = 60\pi$

So I then used the following two formulas:
$n = \sqrt{\frac{\mu_r}{\varepsilon_r}} => \frac{\mu_r}{\varepsilon_r}= (60\pi)^2 => \mu_r=(60\pi)^2.\varepsilon_r$ (equation 2)

Substituting equation 2 into equation 1, I get the values:

$\varepsilon_r = 0.02 \mu_r = 800$ which is wrong as my answers have $\varepsilon_r = 8, \mu_r = 2$

Last edited: Mar 3, 2012
10. ### steveb Senior Member

Jul 3, 2008
2,433
469
For starters, this equation in not correct. It should be

$n = \sqrt{\frac{\mu}{\varepsilon}}$

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11. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Oh I see. Thanks for all the help, I have the answer now!