The Poynting vector, wire length and inductance (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Tue, 25 Sep 2007 23:13:55 -0400
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