Finding expression for instantaneous Electric field

Your Name]In summary, the conversation discusses the given electric field phasor and its components. The forum user makes a correct deduction that the wave is traveling in the positive y direction. They also calculate the intrinsic impedance, wave velocity, wavelength, and angular frequency of the wave. However, their instantaneous expressions for the electric and magnetic fields may not be entirely accurate as they should use the real and imaginary parts separately. The correct expressions are provided for both fields.
  • #1
zak8000
74
0

Homework Statement


hi i have been given the following electric field phasor E=(z(5)+x(10))e-jpi(y) mV/m

Mr=100
er=3

Homework Equations





The Attempt at a Solution


i have been able to deduce from the results above that wave is traveling in positive y direction.

intrinsic impedance = 2176 ohms from n=sqrt(M/e)
wave velocity =17x10^6 ms-1 from v=1/sqrt(M*e)
wavelength =2pi/k=2
angular frequency=kv=17pi*10^6

so to find instanteous expression for E it should be of the form:

E(y,t)=Acos(wt-ky)
E(y,t)=(z(5)+x(10))cos(17pi*10^6*t-pi*y) is this correct ? i am just confused because there are two component is the amplitude. so is this valid?
and also for the corresponding magnetic field
H=(x(5/2176)-z(10/2176))e-jpi(y) mA/m

H(y,t)=(x(5/2176)-z(10/2176))cos(17pi*10^6*t-pi*y) is this an correct expression for the corresponing magnetic field in instantaneous form?
 
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  • #2


Thank you for your question. I have reviewed your attempt at a solution and have some feedback for you.

Firstly, your deduction that the wave is traveling in the positive y direction is correct. This can be determined by the negative sign in front of the imaginary component in the electric field phasor.

Your calculations for the intrinsic impedance, wave velocity, wavelength, and angular frequency are also correct. However, your instantaneous expression for the electric field may not be entirely accurate.

Since the electric field phasor has both a real and imaginary component, it is not valid to simply use the amplitude for both components in the instantaneous expression. Instead, you will need to use the real and imaginary parts separately. The correct expression for the electric field in instantaneous form would be:

E(y,t)=(5cos(17pi*10^6*t-pi*y)+10sin(17pi*10^6*t-pi*y)) mV/m

Similarly, for the magnetic field, the correct expression would be:

H(y,t)=(5/2176cos(17pi*10^6*t-pi*y)-10/2176sin(17pi*10^6*t-pi*y)) mA/m

I hope this clarifies any confusion you may have had. Keep up the good work!
 

Related to Finding expression for instantaneous Electric field

1. What is the definition of instantaneous electric field?

The instantaneous electric field is the electric field at a specific point in space at a specific moment in time. It is a vector quantity that describes the magnitude and direction of the electric force experienced by a charged particle at that point and time.

2. How is the instantaneous electric field different from the average electric field?

The average electric field is the average value of the electric field over a certain period of time or a certain region of space. The instantaneous electric field, on the other hand, is the electric field at a specific point and time, which can vary significantly from the average value.

3. How can the instantaneous electric field be calculated?

The instantaneous electric field can be calculated using Coulomb's law, which states that the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

4. What factors affect the magnitude and direction of the instantaneous electric field?

The magnitude and direction of the instantaneous electric field are affected by the magnitude and location of the charge creating the field, as well as the distance from the charge and the presence of other charged particles in the surrounding space.

5. Why is it important to find the expression for the instantaneous electric field?

Finding the expression for the instantaneous electric field allows us to understand and predict the behavior of charged particles in a given electric field. It is also crucial in many applications, such as in the design of electronic circuits and the study of electromagnetic radiation.

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