Transmission Line - Electrodynamic Calculations

L}{l} = \frac{N\phi}{Il} = \frac{N\mu_0 Jlw}{Il} = \mu_0 Jw(c) To find the capacitance per unit length, we need to use the formula C = \frac{Q}{V}. But in this case, we are dealing with a length of l, so we can write:\frac{C}{l} = \frac{Q}{Vl} = \frac{Q}{Edl} = \frac{Q}{Eld} = \frac{Q}{(V/d)ld} = \frac{Q
  • #1
malindenmoyer
31
0
Two very large parallel conducting plates of very large length [tex]l[/tex], and width [tex]w[/tex] are separated by a distance [tex]d[/tex]. A current [tex]I=Jw[/tex] flows to the right in the lower plate and to the left in the upper plate. Each of the two currents produces a magnetic field [tex]\frac{B}{2}[/tex] between the two plates.

(a) Show that the total field between the plates is [tex]B=\mu_0 J[/tex] via Ampere's Circuital Law.

(b) Find the flux [tex]\phi[/tex] and the self inductance per unit lenght, [tex]\frac{L}{l}[/tex] for this arrangement.

(c) Find the capacitance per unit length, [tex]\frac{C}{l}[/tex].

(d) Find [tex]\sqrt{\frac{L}{C}}[/tex] in terms of [tex]\mu_0, \epsilon_0[/tex] and geometrical factors

My Attempt at Solution
Part (a) is confusing as I have not used the circuital Law for rectangular geometry. I know that Ampere's Law is given by:

[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 I_{\mathrm{enc}}[/tex]

But am confused as to how to apply it as it is not circular geometry.

In part (b) I know that:

[tex]\phi=BA[/tex]

But am not sure as to what area to use, since we know B per part (a)

Solving for flux leads us one step closer to finding the self inductance which is:

[tex]L=\frac{N\phi}{I}[/tex]

But again, I do not know what value to substitute in for [tex]N[/tex].

I am pretty sure I can find the capacitance per unit length per (c), and then (d) is a matter of combing (b) and (c) so that would be self explanatory. Could somebody help me get this thing started? Please keep in mind that I have a very elementary understanding of this material.

Thanks.
 
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  • #2




Thank you for your post and for providing your attempt at the solution. Let's go through each part step by step to help you understand how to approach this problem:

(a) To apply Ampere's Circuital Law, we need to find a closed path that encloses the current. In this case, we can choose a rectangular path that goes around one of the plates, parallel to the direction of the current. The length of this path would be the same as the length of the plate (l) and the width would be the distance between the plates (d). Using Ampere's Law, we can write:

\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 I_{\mathrm{enc}}

But since the magnetic field is constant and parallel to the path, we can write:

B\oint_C \mathrm{d}\boldsymbol{\ell} = \mu_0 I_{\mathrm{enc}}

The integral of the length of the path is simply the perimeter of the rectangle, which is 2l + 2d. Therefore, we can write:

B(2l + 2d) = \mu_0 I

And solving for B, we get:

B = \frac{\mu_0 I}{2l + 2d}

But we know that the current in each plate is I=Jw, so we can substitute this in the equation and get:

B = \frac{\mu_0 Jw}{2l + 2d}

Since we are interested in the magnetic field between the plates, we can divide this equation by 2 to get:

\frac{B}{2} = \frac{\mu_0 Jw}{4l + 4d}

And this is the same result that we were given in the problem statement. Therefore, we have shown that the total magnetic field between the plates is B = \mu_0 J.

(b) To find the flux, we need to find the area enclosed by the closed path we chose in part (a). In this case, it would be the area of the rectangle, which is lw. Therefore, we can write:

\phi = B(lw) = \mu_0 Jlw

To find the self inductance per unit length, we can use the formula L = \frac{N\phi}{I}. But
 

1. What is a transmission line?

A transmission line is a specialized type of electrical circuit used to transmit high-frequency signals, such as radio or microwave signals, over long distances with minimal loss. It consists of a pair of conductors, typically in the form of wires or coaxial cables, separated by a dielectric material.

2. What are electrodynamic calculations?

Electrodynamic calculations involve the use of mathematical equations and formulas to analyze the behavior of electric and magnetic fields in a transmission line. These calculations are used to determine various parameters such as voltage, current, impedance, and power flow.

3. What factors affect the performance of a transmission line?

The performance of a transmission line can be affected by factors such as the type and length of the line, the materials used for the conductors and dielectric, the frequency of the signal being transmitted, and external environmental conditions such as temperature and humidity.

4. How are electrodynamic calculations used in the design of transmission lines?

Electrodynamic calculations play a crucial role in the design of transmission lines. They are used to determine the appropriate dimensions of the conductors and dielectric, as well as to evaluate the overall performance and efficiency of the line. These calculations also help in identifying potential issues and optimizing the design for maximum performance.

5. What are some common applications of transmission line electrodynamic calculations?

Transmission line electrodynamic calculations are used in a wide range of applications, including telecommunications, radio and television broadcasting, radar systems, and high-speed data transmission. They are also important in the design of power transmission and distribution systems, as well as in the development of electronic devices and equipment.

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