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Re: [Phys-l] induced electric field



I finally had the opportunity to down load the viewer and to look at both versions of your rendering of David's equation, Bob.
Most interesting!
I started wondering... anchored as I am in the concrete...what should I expect from a single turn of copper outside the perimeter where by definition the magnetic field is increasing? (I actually do expect a circulating current...)

Brian W


Bob Sciamanda wrote:
For those who don't have access to MATHCAD, the program MATHCAD 8 EXPLORER is a free MATHCAD viewer.
Download it at:
http://users.ecs.soton.ac.uk/msn/book/worksheets.html

Bob Sciamanda
--------------------------------------------------
From: "Bob Sciamanda" <treborsci@verizon.net>
I have written a MATHCAD program to generate a vector field plot of an example of David's solution. The program (to be run on MATHCAD is downloadable at:

http://h1.ripway.com/treborsci/Bowman.MCD

Bob Sciamanda
--------------------------------------------------
From: "David Bowman" <David_Bowman@georgetowncollege.edu>
/snip/ The sign-fixed version for B
increasing along the (+)z direction at a rate of B' is given here:

Let A == L/2 - x
Let B == L/2 + x
Let C == L/2 - y
Let D == L/2 + y

E_x = (-B'/(4*[pi]))*(A*ln((A^2 + C^2)/(A^2 + D^2)) +
B*ln((B^2 + C^2)/(B^2 + D^2))+ 2*C(arctan(A/C) + arctan(B/C))
- 2*D(arctan(A/D) + arctan(B/D)))



and

E_y = (B'/(4*[pi]))*(C*ln((A^2 + C^2)/(B^2 + C^2)) +
D*ln((A^2 + D^2)/(B^2 + D^2))+ 2*A(arctan(C/A) + arctan(D/A))
- 2*B(arctan(C/B) + arctan(D/B)))

David Bowman