Question

The maximum allowed leakage of microwave radiation from a microwave oven is 5.0 mw/cm2. if microwave radiation outside an oven has the maximum value, what is the amplitude of the oscillating electric field?

Answer 1

Answer:

194 V/m

Explanation:

In order to find electric field, we can use the formula of power density

i.e Pd = E^2 / Z

where:

Pd = power density in W/m^2

E = electric field strength in V/m

Z = impedance of free space = 120 * π

E = sqrt(Pd * Z) -----> re-arranging it for E

before solving, convert Pd unit into W/m^2  

Pd= 5mW/cm^2 = 50 W/m^2    

Solving for E:

E= sqrt(50 * 120 * π)

E = 137.3 V/m  

the above value is RMS value

In order to find the peak amplitude of the oscillating field will therefore be 137.3 * sqrt(2) = 194 V/m

Answer 2

Answer:

E_peak = 194 N/C

Explanation:

Given:-

- The Power density Pd = 5.0 mW/cm^2

- The impedance of free space Z_o = 376.73 ohms

Find:-

If microwave radiation outside an oven has the maximum value, what is the amplitude of the oscillating electric field?

Solution:-

- The oscillating electric field strength E root mean squared can be determined by the Power Density formula applied for electromagnetic waves in free space. We have:

                                       Pd = E^2_rms / Z_o

                                       E_rms = sqrt ( Pd * Z_o )

- Solve for electric field strength E_rms:

                                       E_rms = sqrt ( 50 * 376.73 )

                                       E _rms= 137.24612 N / C

- The E obtained is the rms value. To obtained the peak value of E, we need to multiply the transformation factor √2.

                                      E_peak = E_rms * √2

                                      E_peak = 137.24612 * √2

                                      E_peak = 194 N/C