Question

A uniform plane wave in air with H=4sin(ωt−5x)A1​,A/m is normally incident on a plastic region (x=0) Jith the parameters μ−μ0,​=δ=4εok​ and σ=0. (a) Obtain the total electric field in air. (b) Calculate the time-average power density in the plastic region. (c) Find the standing wave ratio.

Answer 1

Total Electric Field in Air is
`\[ E = Z_0 H \]`.where H is the magnetic field amplitude of the incident wave and Z0 is the impedance of free space.

Given a uniform plane wave in air with
`\( H = 4 \sin(\omega t - 5x) \, \text{A/m} \)`, the electric field in air
`(\( E \))` is related to the magnetic field (\( H \)) by the impedance of free space
`(\( Z_0 \))`:

`\[ E = Z_0 H \]`

The impedance of free space
`(\( Z_0 \))` is given by:

`\[ Z_0 = \sqrt{(\mu_0)/(\varepsilon_0)} \]`

where
`\( \mu_0 \)` is the permeability of free space, and
`\( \varepsilon_0 \)` is the permittivity of free space.

(a) Total Electric Field in Air:

`\[ E = Z_0 H \]`

(b) Time-Average Power Density in the Plastic Region:

The time-average power density
`(\( P_{\text{avg}} \))`in the plastic region is given by:

`\[ P_{\text{avg}} = (1)/(2) \text{Re} \left\{ E * H^* \right\} \]`

where
`\( H^* \)` is the complex conjugate of
`\( H \)`.

(c) Standing Wave Ratio (SWR):

The standing wave ratio (SWR) is given by:

`\[ \text{SWR} = \frac{E_{\text{max}}}{E_{\text{min}}} \]`

where
`\( E_{\text{max}} \)`is the maximum amplitude of the electric field, and
`\( E_{\text{min}} \)` is the minimum amplitude of the electric field.

Let's calculate each of these:

(a) Total Electric Field in Air:

`\[ E = Z_0 H \]`

(b) Time-Average Power Density in the Plastic Region:

`\[ P_{\text{avg}} = (1)/(2) \text{Re} \left\{ E * H^* \right\} \]`

(c) Standing Wave Ratio (SWR):

`\[ \text{SWR} = \frac{E_{\text{max}}}{E_{\text{min}}} \]`

If you have specific values for
`\( \mu_0 \)`,
`\( \varepsilon_0 \)`, and
`\( \omega \)`, please provide them so that we can proceed with the calculations.