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In freespace, the Electric Field component of a traveling electromagnetic wave is:

$$\vec{E}(t,z) = 10^3 sin(\omega t - \beta z) \hat{y}~~~\text{[V/m]}$$

Since the wave propagation is in freespace, the wave impedance is:

$$\eta = 120 \pi~~\Omega$$

Further, the wave impedance can be calculated as the ratio of the Electric field Amplitude to the Magnetic field Amplitude, that is:

$$\eta = \frac{E_y}{H_x}$$

Now for the part I don't get... when they calcuate wave impledence... they put a negative sign on the magnetic field amplitude as follows:

$$\eta = \frac{E_y}{-H_x}$$

And thus, the magnetic field can be calculated from the electric field as follows:

$$H_x = \frac{E_y}{-\eta}$$

$$\vec{H} = - \frac{10^3}{120\pi}\sin(\omega t - \beta z) \hat{x}~~~\text{[A/m]}$$

My question is this:

How did they know to put a negative on the $H_y$ term given that $\vec{E}(t,z) = 10^3 sin(\omega t - \beta z) \hat{z}$?

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  • $\begingroup$ This is confusing and I think you have typos. Is E along the z-axis or the y-axis? Whatever that turns out to be just use Maxwell's equations to get H from E and the propagation vector. Is the E field propagation in the z direction or pointing in the z direction? $\endgroup$
    – user196418
    Commented Jan 2, 2020 at 20:01
  • $\begingroup$ changed it to $\hat{y}$ $\endgroup$
    – pico
    Commented Jan 2, 2020 at 20:13

1 Answer 1

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First, it is necessary to make sure we are using a right-handed Cartesian coordinate system... to construct the x-y-z axis, we need to curl the fingers from x-axis to y axis with thumb pointing in direction of z axis.....

Case 1:

$$\vec{E}(t,z) = 10^3 sin(\omega t - \beta z) \hat{y}~~~\text{[V/m]}$$

Now we can plot the first cycle of E for t=0. Since, first half of sin phase is in positive y direction...if we point our thumb in the wave propagation direction z, and curl fingers from positive y axes of E field to 90 degrees... it turns out that the fingers will point in negative x axis direction...

$H_x$ will have a negative sign...

Case 2:

$$\vec{E}(t,z) = 10^3 sin(\omega t - \beta z) \hat{x}~~~\text{[V/m]}$$

plot first cycle of E for t=0. Since, first half of sin phase is in positive x direction...if we point our thumb in the wave propagation direction z, and curl fingers from positive x axes of E field to 90 degrees... it turns out that the fingers will point in postive y axis direction...

$H_y$ will have a postive sign...

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  • $\begingroup$ Looks like you answered your question. $\endgroup$
    – user196418
    Commented Jan 2, 2020 at 20:27
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    $\begingroup$ i just figured it out by hitting my head again the wall upside down... $\endgroup$
    – pico
    Commented Jan 2, 2020 at 20:30

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