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Maxwell's Equations

  1. Jul 28, 2009 #1
    I'm new to these equations but let's see if I have the talk down.

    1. The electric field surrounding a charge will go thru a surface and not come back into the surface. It's strength is proportional to the charge and inversely proportional to the perm of free space.

    2. The magnetic field surrounding a current or a magnet will go thru a surface and come back into the surface (therefore the divergence =0). It's strength is proportional to the current and the permeability of freespace and the permittivity of free space.

    3. A changing electric field produces a magnetic field perpendicular to the direction of the current by right hand rule. The magnitude of the magnetic field is proportional to the paralelliped formed by two vectors, one being the current vector whose magnitude is proportional to the number of coulombs along a certain defined length of the current vector determined by the defined length which produces a stronger electric field if the length of the wire investigated is longer, the other being the radius along B to the point of interest.

    4. A changing magnetic field produces an electric field. The resulting strength of the electric field is proportional to H and the length of wire that interacts with H.

    Do I have this about right?
     
  2. jcsd
  3. Aug 4, 2009 #2
  4. Aug 4, 2009 #3
    Thanks, I know the site. It's good.
    The eq Nabla cp E = part deriv B / part deriv t
    The right hand side is the rate of change of B wrt to t. Right?
    What does the left hand side mean in similar language?
    I would guess something like the rate of change of E in x hat, y hat and z hat but I'd certainly be wrong. So what's the real meaning of the left part of the equation?

    I would think that what would cause changes in B would be changes in E so one could incorrectly think that dE/dt=dB/dt. That's not the equation though. And the cross product throws me because I'm trying to think of two vectors, say, in the x y plane that lead to a vector in the z plane and how the correlates with E and how all that equates to the rate of change of B wrt t.

    Thx
     
  5. Aug 4, 2009 #4

    jtbell

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    Think of (e.g.) [itex]\vec \nabla \times \vec B[/itex] at a point as a measure of the tendency of a vector field to "circulate" or "run in loops" around that point. It's often called the "curl of [itex]\vec B[/itex]".

    This "circulation" is a spatial property of the field. Don't picture the field vectors as moving around in circles as time passes, like water swirling around as it goes into a drain.
     
  6. Aug 4, 2009 #5
    Woops. Don't you mean Nabla X E?
    That's the eq

    Nabla X E = pdB/pdt

    So is Nabla X E the measure of the tendency of a vector field at a specific point to run in loops around the point? It must be, based on your Nabla X B answer. :)

    And what in reality causes a point in an electric field to run in a loop around a point?

    Thx
     
  7. Aug 4, 2009 #6

    jtbell

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    The concept of "curl" ([itex]\vec \nabla \times[/itex]) can be applied to any vector field.

    It is the tendency of [itex]\vec E[/itex] to run in loops around the point. There are other vector fields besides [itex]\vec E[/itex]. :wink:
     
  8. Aug 4, 2009 #7
    So the equation states...

    the curl of E equals the rate of change of B?
     
  9. Aug 5, 2009 #8

    Born2bwire

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    Yeah, change in B in time though to be exact. You'll often here of a vector field that has zero curl being called irrotational, giving a little insight into the meaning of the curl operator.
     
  10. Aug 5, 2009 #9
    So here's a fundamental question that has been building up for me and only now do I have the chance to ask.

    If you take the electric field of a single point charge which radiates outward in ever grown spheres at the speed of light until stabilizing into a steady state...this is an irrotational field. Describe how B comes in and makes it otherwise. Say a magnet is introduced to the point charge. Now B moves slightly. Are you saying that E now experiences some of it's vectors having non-zero curl?
     
  11. Aug 5, 2009 #10

    Vanadium 50

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    Rocky,

    Have you read the book I suggested? (Schey's Div, Grad, Curl and All That) Last time, you gave as an excuse that you hadn't even ordered it, which is no excuse at all. If not, I can only conclude that you are deliberately wasting our time. You are too ignorant of vector calculus to understand the answers to your questions, and we have explained how to remove this ignorance. The ball is in your court.
     
  12. Aug 5, 2009 #11

    dx

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    Vector fields don't 'move', they change. If B is changing at some point, and the rate of change is R, then the curl of E at that point must be -R.
     
  13. Aug 5, 2009 #12
    So what you're saying is that B does not cause the vectors of E to curl?
    Or, if you think about it, maybe the intersection of two E fields which bends the lines of flux, is the curl you're talking about.
     
  14. Aug 5, 2009 #13

    Born2bwire

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    A bending of the vector field is not going to work I think. Take this vector:

    [tex]\mathbf{v} = (1-x)\hat{x}+y\hat{y}[/tex]

    We can see that this vector field will curve, curl as you have stated, however, since the x component is only dependent upon x and the y component only dependent upon y, the curl of this vector field will be zero. However take this vector field:

    [tex]\mathbf{v} = x^2\hat{y}[/tex]

    This field has no change of direction in the field but it has a non-zero curl.

    EDIT:

    The curl is a measure of the rotation of the vector field, but it is better to think of it as a rotation due to the change in flow. If the vector field represented the flow of a fluid, rotationality can be thought of as the movement of a paddle wheel in the fluid. If we place our paddle wheel in the second vector field, the vector field on the side of the paddle away from the x=0 axis i stronger than on the other side. Thus, the paddle wheel will rotate.
     
    Last edited: Aug 5, 2009
  15. Aug 5, 2009 #14
    Born2Bwired: You're equation above is a vector in the x y plane. I don't see why this equation, if plotted, and if the line of the graph were a flow of water, could not turn a paddlewheel whose axis were in the z plane?

    The only time a plotted equation in the x y plane whose graph is a flow of water, would NOT turn a paddlewheel whose axis is in the Z plane, is when the arrows are perfectly paralell in the x y plane and impinge upon the wheel plates equally on both sides of the axis.

    So I'm not understanding why your first eq has a curl of zero. I would think that the majority of eq would have curl of + or - something.

    Thx
     
  16. Aug 5, 2009 #15

    Born2bwire

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    Take a look at Stoke's Theorem. If we want to find whether or not the paddle wheel will turn, then we can think of the net force to be akin to:

    [tex]\oint \mathbf{F}\cdot d\mathbf{\ell}[/tex]

    Where, in the case of the curl operator, we take the line integral to be over a circle whose radius goes to the limit of zero. In essence, we are finding the contribution of the field normal to our paddle wheel at all points along the wheel. Using Stoke's Theorem,

    [tex]\oint \mathbf{F}\cdot d\mathbf{\ell} = \int_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}[/tex]
    where the integral is now over an area enclosed by the boundary of the line integral. We can see that if the curl is zero, there will be no net rotation on our paddle wheel. This is all elementary vector calculus.
     
  17. Aug 5, 2009 #16
    So would that not apply to everything leading to a zero curl for any equation? You didn't use your first equation to make your point.
     
  18. Aug 5, 2009 #17

    Born2bwire

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    I didn't use the first equation because it would be an annoying integral to work out. You are more than welcome to do it yourself. Let's cheat and use a square path:

    [tex] \oint \mathbf{F}\cdot d\mathbf{\ell} = \int_a^b ydy + \int_b^a ydy +\int_c^d (1-x)dx + \int_d^c (1-x) dx = 0 [/tex]

    It would not lead to a zero curl for every vector field. Take the vector field:

    [tex]\mathbf{F} = \hat{\phi} [/tex]

    This field is nothing more than vectors that point in circles. Obviously this will turn our paddle wheel and it does have a non-zero curl.
     
  19. Aug 5, 2009 #18
    I think where I'm getting confused is this. Take your square path and place the paddlewheel along any segment of it and it should turn if the path falls say on the left paddle and not the right. Since it's a line segment it can't fall on both.
    Or are talking about the field of vectors inside an area BORDEREd by the square?
     
  20. Aug 5, 2009 #19

    Born2bwire

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    The path integral is a measure of the net flow around the paddle. The paddle wheel does not follow along the path, the path is a trace of the line normal to the paddles on the wheel. We are trying to find the amount of fluid striking against the paddles. In any case, we will see that fluid striking one paddle will be exactly countered by the fluid striking the paddle on the opposite side for our given vector field. In the above case, if we look at the paddles that are parallel to the x-axis, the fluid striking it is along the y-axis. The component of the vector field along the y axis is "y". We can thus see that the fluid striking the paddle to the left of the hub is the same magnitude and direction as the fluid striking the paddle to the right of the hub because the vector field normal to these paddles is independent of x. Thus, there is no net rotation due to those two paddles. We can keep doing this all along the circumference of our paddle wheel and find the same result.
     
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