Question on finding inductance of parallel plate transmission line.

[yungman]

Find the inductance per unit length of parallel plate tx line width = w and space =d between the two plates. w>>d.

My question is whether the total inductance is twice the calculate inductance because total inductance is the sum of inductance of the top and the bottom plate. But it seems all books only use one side for calculation. This is my calculation of the bottom plate only.



For w>>d, B is consider uniform and parallel to the plates.

I use the boundary condition of lower plate:

[tex] \int_s \nabla \times \vec B \cdot d \vec s = \int_c \vec B\cdot d\vec l = \mu_0 I [/tex]

I take medium 1 is space between the plates and medium 2 be the copper of bottom plate. [itex]\hat T_1[/itex] is same direction as B1.

[tex]\Rightarrow \int_c \hat T_1 \cdot ( \vec B_1 - \vec B_2) dl = \mu_0 I [/tex]

[tex] \vec B_2 = 0 \;\Rightarrow B_1= \frac {\mu_0 I}{w} [/tex]

[tex] \Phi = \int_s \vec B \cdot d\vec s = \mu_0 \frac {d I }{w} \;\hbox { for unit length =1 } [/tex]

[tex] L = \frac {\Phi}{I} = \mu_0 \frac d w [/tex]

I thought this is inductance of only one plate. The total inductance of the parallel plate is twice of the calculation. Why don't I have to multiply by two?

Thanks

 

[yungman]

Anyone?

I understand that in order to make the assumption of uniform field, you need both plates and w>>d. So it is not as if you can calculate the inductance totally separate. But still what ever your inductance calculation is, it apply on the other side.

This question is true when calculating capacitance of the parallel plate tx line also using electric boundary condition also( normal boundary ).