Re: [Phys-l] induced electric field

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding Carl's request for details of my analytic result:

> David Bowman:
>
> Fantastic, I thought this thread was dying and then you guys came
> along and resurrected it with some incredible insights.
>
> Okay, can you please provide some details on how you obtained your
> analytic result?

OK.

> I think this is how one could proceed, but please correct me: We
> can obtain E-vector as 1/4*pi multiplied by a "Helmholtz" (in this
> case a "Biot-Savart") type integral of R-vector cross-producted
> with (c k-hat) divided by the cube of the magnitude of R-vector.
> Here R-vector is defined as the difference between the field
> position (r-vector) and the source position (r'-vector), and the
> integral is a volume integral over all source elements. Also c is
> the (constant) value of dB/dt where k-hat is the unit vector along
> the z-axis.  Cylindrical coordinates are used, with axes aligned
> along the symmetry axes of the square solenoid.
>
> Am I on the right track?

Mostly, but I did the problem in rectangular coordinates which are much easier to deal with when one has a square geometry to work with.  But you noted that in a subsequent post.

Here's the scoop.  First let me say a few things in general for solving a generic pair of linear PDEs where one has specified (mostly localized) sources for both the curl and the divergence of a vector field where the field is taken to approach zero in the limit far from the sources.  In particular suppose we are given:

div(F) = s  and   curl(F) = v

such that s(r) is a specified field of scalar sources (s is the scalar density of these sources), and v(r) is a specified vector field of other sources (v is a vector current density of the other sources), and the unknown vector field F(r) is taken to approach zero in regions where location r is asymptotically far from locations where the sources have nonzero values.  Because v is a curl it is required to be transverse (i.e. be divergence free and thus obey a Kirchoff current-type law).  We can break up any differentiable vector field (including field F) into a sum of an irrotational field term and a transverse field term (F = F_i + F_t).  The irrotational part F_i of F can be found by solving the pair of equations:

div(F_i) = s and curl(F_i) = 0.

Likewise, the transverse part F_t of F can be found by solving the pair of equations:

curl(F_t) = v and div(F_t) = 0.

By adding F_i to F_t giving F, the field that results obeys the original inhomogeneous pair of equations for F.

Now F_i can be expressed as a Coulomb-type of integral over the sources in s and F_t can be expressed as a Biot-Savart-type of integral over the sources in v.  Adding the separate pieces gives the resulting equation for F(r) in terms of a composite volume integral:

F(r) =(1/(4*[pi]))*
      Integral[d^3r'*(s(r')*(r - r') + v(r') x (r - r'))/|r - r'|^3]

The first s(r')-containing term contribution to F is the irrotational part, F_i, and the second v(r')-containing term contribution to F is the transverse part, F_t.

For the particular problem of our square cross-section solenoid our unknown field F is the electric field E.  The scalar source term s is zero since for this problem there is no net density of electric charges anywhere to be found--even the surface currents in the walls of the solenoid are electrically neutral having no net charge.  The specified vector source field v is -dB/dt which is taken to be spatially uniform inside the solenoid and be zero outside it.  This means that our integrated expression for the E-field only has the second transverse contribution, so that the mathematics involved is, essentially, just doing a Biot-Savart-style of integral.

If we take dB/dt = B'*(z-hat) and make the z-axis of the coordinate system coincide with the central longitudinal symmetry axis of the solenoid our E-field integral expression becomes:

E(r) =(-B'/(4*[pi]))*Integral[d^3r'*((z-hat) x (r - r'))/|r - r'|^3]


The region of integration is the entire interior of the solenoid:

- L/2 < x' < L/2,  -L/2 < y' < L/2, -[infinity] < z' < [infinity]

The numerator cross product in the integrand is independent of the z' integration variable.  This means that we can integrate out the z' coordinate using an integrand involving just the denominator factor 1/|r - r'|^3.  Doing this initial 1-dimensional integral over z' is fairly straightforward and becomes easy if one makes a substitution of u = (z' - z)/|r' - r|.  The result of integrating 1/|r - r'|^3 over all z' values is 2/((x'-x)^2 + (y'-y)^2).  After doing this longitudinal integration we still have to do the x' and y' integrations over the interior cross-section of the solenoid.  After working out the cross product in the numerator the remaining 2-dimensional integral has the form:

E(r) = (-B'/(4*[pi]))*Integral[dx'*dy'*(2*(y'-y)*x-hat
        - 2*(x'-x)*y-hat)/((x'-x)^2 + (y'-y)^2)]  .

The x-component of the E-field is easiest to integrate if we integrate it over y' first and integrate the result over x' last.  But y-component is easiest to integrate if we integrate it over x' first and integrate that result over y' last.  Doing it this way is easy since the initial integrand in both cases can be written as the differential

d ln((x'-x)^2 + (y'-y)^2)  which is trivial to integrate in each case using the fundamental theorem of calculus.

We introduce the auxiliary quantities A = L/2 - x, B = L/2 + x, C = L/2 - y, and D = L/2 + y.  Each component of the E-field has one integral left to do.  The x-component is

E_x = (-B'/(4*[pi]))*Integral[dx'*(ln(C^2 + (x'-x)^2)
      - ln(D^2 + (x'-x)^2))] .

Likewise the y-component is

E_y = (B'/(4*[pi]))*Integral[dy'*(ln(A^2 + (y'-y)^2)
      - ln(B^2 + (y'-y)^2))] .

In both cases the integration variables range between -L/2 and +L/2.

Now the remaining integrals can each be integrated using the integration formula:

Integral[du*ln(a^2 + u^2)] = u*ln(a^2 + u^2) - 2*u + 2*a*arctan(u/a).

The result after all the remaining integrals are done are just my previously given formulas for E_x and E_y.

Thus, the exact solution is simply found from a straightforward integration of the Biot-Savart-style integral representation of the E-field, where the integrals are just done one Cartesian direction at a time.  No fancy tricks are needed by transforming to exotic coordinate systems, no conformal transformations, nothing glamourous or sexy at all--just straightforward sequential Cartesian integrations do the job.

> I checked the above procedure for a circular solenoid and it
> gives the familiar solution.

It better.

> I also noticed (yes, I'm a bit slow compared to you guys) that if I
> take the curl of the familiar solution (namely r*c/2 inside and
> R^2*c/2r outside, both in the azimuthal direction) that I correctly
> get c inside and zero outside. (That would make a good homework
> problem.) Great, I assume that I could similarly take the curl of
> your solution for the square solenoid and get exactly the same
> answers.

Yep, but doing so is more of a pain.

> David, could I STRONGLY urge you to write up your solution and
> send it to AJP? I'm confident that John M would agree with me that
> this is a very interesting and instructive result that MANY people
> will be interested in.

I'll consider it.  Thanks for the vote of confidence about the possible general interest in the result.  However I would be more enthusiastic about writing it up if the solution procedure involved some sort of sexy transformation.  But it is just simply doing straightforward Cartesian integrals one dimension at a time.

> Much thanks also to Bob S for the fantastic MathCad rendering and
> Viewer program. It really helped to see the field vectors.

Amen.  The diagram *is* slick-looking.

> *This* is the kind of thread that makes PHYS-L so great. I wish
> sometimes we had more such threads.... -Carl

Do you want to start another one?  Have at it.

> --
> Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
> Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
> mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/

David Bowman