RE: The Poynting vector, wire length and inductance (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Wed, 26 Sep 2007 12:38:18 -0400
From: "Lau, Gary" <Gary.Lau@xxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: RE: The Poynting vector, wire length and inductance (fwd)

I must admit to having trouble with your math.  I knew integral calculus 30+ years ago, but it's a struggle now.  I do recognize your closing equation: 
> or simply:  w = ½ L (i)sqrd      (as expected!)
as the standard equation for the energy contained in an energized inductor, except that the dimensions of the result are energy (Joules), not power (Watts).  Is that what the "w" represents?

Is there a new idea or relationship that you are trying to highlight here?  I didn't follow your closing comment about the actual energy actually being 2X the ½ L (i)sqrd  value.  I'm quite sure it's just 1X, as COE requires equating that with bang energy ½ C (v)sqrd in a tank circuit.  

Gary Lau
MA, USA
... Drinking and deriving don't mix!


> From: Jared Dwarshuis <jdwarshuis@xxxxxxxxx>
> To: Pupman <tesla@xxxxxxxxxx>
> Subject: The Poynting vector, wire length and inductance
> 
>                      The Poynting vector, wire length and inductance
> 
> 
> 
> 
> 
> We will examine an ideal air cored solenoid:
> 
> 
> 
> Using Amperes law:  H = Ni / l  and B = u I N / l
> 
> 
> 
> Using Faraday's law:
> 
> 
> 
> Closed integral E  dot ds = 2 pi r = - double integral of the partial
> derivative of B with respect to time dot ds = - u  N/l  di/dt   ( pi
> Rsquared)
> 
> 
> Then the magnitude of the vector E at the surface of the wire equals:  ½ u
> di/dt R N/l
> 
>  Or by multiplying the numerator and denominator by 2pi we get:
> 
> 
> E = u (2 pi R N) / 4pi l  = u (wire length) / 4 pi l    di/dt
> 
> 
> 
> Now in this instance we can use the expression V = Ed :  (where the distance
> (d) equals the length of wire in the solenoid.)
> 
> 
> 
> Then:    V = - u (wire length)sqrd  / 4 pi l     di/dt
> 
> 
> 
> Which we recognize as: V = - L di/dt            (as expected!)
> 
> 
> 
> We will now  use the Poynting vector to show the total power flowing through
> the surface  as:
> 
> 
> 
> Power = E cross H = ½ u (N/l)sqrd  R  I di/dt
> 
> 
> 
> Then the total power flowing into a volume of length l is:
> 
> 
> 
> Power = dW/dt = ½  u (N/l)sgrd pi Rsgrd l i di/dt
> 
> 
> 
> Integrating both sides, then  multiplying the numerator and denominator by 4
> pi and regrouping  we get:
> 
> 
> 
> w = ½  u  (wire length)sqrd / 4 pi l    (i)sqrd
> 
> 
> 
> or simply:  w = ½ L (i)sqrd      (as expected!)
> 
> 
> 
> Comment:
> 
> 
> 
> The energy considered above is contained entirely in the  magnetic field
> outside the wire of  the inductor. An equal amount of energy is also
> contained within the wire of the inductor, it sustains this magnetic
> field.
>  ( This is not a double accounting it is consequence of conservation laws)
> 
> 
> 
> Jared Dwarshuis, Lawrence Morris
> 
> Sept. 07
>