1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Induced E by a solenoid with time-varying current

  1. May 29, 2007 #1
    Imagine a solenoid with n turns per length. Now, for an instant, in which everything looks static, the magnetic field inside the solenoid will be [tex]n \mu_0 I \mathbf e_z[/tex] (choosing solenoid alinged with z-axis), and zero field outside. Now, what would happen if we change the current in time?

    To keep the discussion simple, i consider a current varying linear with time, [tex]I=I_0 + ct[/tex], so magnetic field becomes [tex]\mathbf B=(B_0 + kt) \mathbf e_z[/tex] inside the solenoid.

    [tex]\mathbf \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t} = -k \mathbf e_z[/tex]

    So curl of E has only z component.

    [tex](\mathbf \nabla \times \mathbf E)_z = \left( \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right) = -k[/tex]

    valid solutions for [tex]E[/tex] are

    [tex]\mathbf E = -k/2 (-y,x,,0)[/tex]
    [tex]\mathbf E = -k(-y,0,0)[/tex]
    [tex]\mathbf E = -k(0,x,0)[/tex]

    but which one?? I thought about boundary conditions, like: since B is zero outside, so is time development and curl of E at the "wires", where [tex]x^2+y^2=R^2[/tex]... but i couldn't accommodate this with solutions, they seem to be incompatible...

    Any ideas about the field induced inside the solenoid?
    Last edited: May 29, 2007
  2. jcsd
  3. May 29, 2007 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]{\vec E}=({\vec r}\times{\vec k})/2.[/tex]
  4. May 30, 2007 #3
    What is the direction of the induced electric field? Can you work in cylindrical coordinates instead of cartesian first? (That will help you "see" whats happening and also, the form of the equations is simpler).

    Another hint: don't blindly apply curl and divergence all the time...try to think about the field first...this one is particularly easy (B is axial, E should be ...)

    The other solutions can be ruled out by symmetry considerations...
  5. May 30, 2007 #4
    Thanks for the replies!

    Meir Achuz, I'd ask "why?"

    In cylindirical coordinates, z-component of curl is

    [tex]\frac{1}{s} \left( \frac{\partial (s A_{\theta}) }{\partial s} -\frac{\partial A_s}{\partial \theta} \right) = -k[/tex]

    which has again 3 mathematically possible solutions

    [tex]-k/2(\frac{s}{2} \mathbf e_\theta - (s \theta) \mathbf e_s)[/tex]
    [tex]-k(\frac{s}{2} \mathbf e_\theta )[/tex]
    [tex]-k(- (s \theta) \mathbf e_s)[/tex]

    I also had the "intuitive" answer [tex]\mathbf E = -k/2 (-y,x,,0)[/tex], however, I want to know what is mathematically wrong with the other "solution"s. I can't simply say "i just didn't like the way other solutions looked", right?
    Last edited: May 30, 2007
  6. May 30, 2007 #5
    Cylindrical solution gave a better insight though. Since the system is symmetric around z-axis, the electric field should be independent of [tex]\theta[/tex], therefore, the solution is [tex]\mathbf E = -k \left(\frac{s}{2} \mathbf e_\theta \right)[/tex].

    Wish that I had a more rigorous way to show it, though.
  7. May 30, 2007 #6
    Okay so you want the math :biggrin: Fine.

    Let me first do some physics. Assume that the solenoid is infinite in length (as otherwise what you and I are saying are both wrong :rolleyes:). Now, A shouldn't have any [itex]\theta[/itex] dependence (everything is symmetric in [itex]\theta[/itex]). So that makes [itex]\partial A_{s}/\partial \theta = 0[/itex]. Which gives:

    [tex]\frac{1}{s}\frac{\partial(sA_{\theta})}{\partial s} = -k[/itex]

    which gives

    [tex]A_{\theta} = -\frac{k}{2}s[/tex]

    (upto within an additive constant)

    But this is what you did too :tongue2: and believe me, its sufficiently rigorous for Physics (someone correct me if I'm wrong, but read the paragraph below).

    The reasoning above by the way, rules out your first solution. You can state it formally as

    [tex]\frac{\partial \vec{A}}{\partial \theta} = 0[/tex]

    which is of course stronger than saying that [itex]\partial A_{s}/\partial \theta = 0[/itex].

    See you can't just write any general solution to a differential equation and expect it to behave like the field in question. What about boundary conditions/physical requirements?

    If you only want to be mathematical, then you will have to characterize your fields (solutions) into a category or class which asserts their properties...like they have to go to zero at infinity, they have such and such singularities and so on. Then you can explicitly enforce these conditions and work from there, without using "physical insight" or 'non-rigorous' arguments.

    We're not saying that we don't like the way the other solutions look, but it turns out that they're wrong anyway according to the physics :smile:
  8. May 30, 2007 #7

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's easier without coords:
    [tex]\nabla\times({\vec r}\times{\vec k})=({\vec k}\cdot\nabla}){\vec r}
    -{vec k}(\nabla\cdot{\vec r})={\vec k}-3{\vec k}=-2{\vec k}.[/tex]
    Last edited: May 30, 2007
  9. May 30, 2007 #8

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Pardon my LateX in the previous post, but it should be readable.
    Since only curl E is given, any gradient can be added to the solution.
    phi=k\pm xy leads to your other solutions.
    If you make the reasonable assumption that there is no scalar piotential here,
    you get only my solution.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook