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Radial component of B field of solenoid

  1. Mar 27, 2014 #1
    Hi everyone.

    I'm having a relatively hard time in understanding Griffith's explanation of why the radial component of a magnetic field of a solenoid should be zero. He says, "If B(r) were positive, if we reversed the direction of the current, B(r) would be negative. But switching I is physically equivalent to turning the current upside down, and that certainly should not alter the radial field".

    I understand the first half of the explanation but I'm just not able to get why the radial field should not be altered if I turn the current upside down. I'll be thankful if anyone explains why.

    - Madhav
     
  2. jcsd
  3. Mar 28, 2014 #2

    TSny

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    Hello, madhavpr. Welcome to PF!

    Griffiths is considering an infinitely long, ideal solenoid. Reversing the current is equivalent to turning the solenoid upside down.

    For the infinitely long solenoid, the B-field must be invariant under translations of the solenoid along its axis and invariant under rotations about the axis. Now, consider a point, p, inside the solenoid a distance d from the axis. And suppose that the field at point p has a radial component.

    What can you say about the the radial component of the field at all points inside the solenoid that are a distance d from the axis?

    Imagine turning the solenoid upside down. What can you say now about the radial component of the field at all points d from the axis?
     
  4. Mar 28, 2014 #3
    Hello TSny.

    I assumed a surface current pointing in the positive [itex]\phi[/itex] (rotating counterclockwise) and calculated, by the Biot Savart law, a component of the magnetic field in the +s (Griffith's notation for the radial distance in cylindrical coordinates).
    I reversed the direction of surface current, now flowing clockwise and found out that the component of the B field in the radial direction reversed its sign.

    I know that the magnetic field should mimic the symmetry of the source. If I turn my solenoid upside down, doesn't that change the direction of the surface current K, and hence the symmetry? So the direction of radial field definitely reverses. Griffiths says it doesn't.

    Secondly, what's the difference between turning the solenoid upside down and rotating my head by 180 degrees? Are they equivalent or different?
     
  5. Mar 28, 2014 #4
    " What can you say about the the radial component of the field at all points inside the solenoid that are a distance d from the axis? "

    - Radial components at a distance d will point away from the axis. (If I assume a positive radial component of the magnetic field)

    Imagine turning the solenoid upside down. What can you say now about the radial component of the field at all points d from the axis?

    - B field along the radius reverses its direction. It now points towards the axis.

    But I don't understand why or how is that a contradiction? Clearly, the source has changed its symmetry (current's direction has been reversed or the solenoid has turned upside down). So the field should change its direction as well.

    I guess something's wrong with my reasoning. I'm not sure what.

    Thanks,
    Madhav
     
  6. Mar 28, 2014 #5
    I think I got it. But it looks more or less like a guesswork to me.

    Reversing the current is equivalent to turning my head 180 degrees and thus, I cannot have two radial components (in and out) of the B field. So the radial field has to be zero.
     
  7. Mar 28, 2014 #6

    TSny

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    Right. Turning the solenoid upside down (or letting the observer stand on his head) does not change the direction of the radial component. Switching the direction of the current changes the direction of the radial component.

    But these two actions must lead to the same result.

    Note how the argument does not go through for a finite solenoid.
     
  8. Mar 17, 2017 #7

    TSny

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    -------------------------------------------------------------------------------------------------------------------------
    EDIT: The following post was meant to be a response to a question that I received in my INBOX. It was meant to go directly to the questioner. Somehow I inadvertently posted it here. The questioner asked:
    ---------------------------------------------------------------------------------------------------------------------------

    Consider the finite solenoid shown:
    upload_2017-3-17_11-7-59.png
    Note how the radial components of B (red arrows) are outward near the top and inward near the bottom. Thus, if you rotated the solenoid 180o about an axis perpendicular to the page and through the center of the solenoid, the radial components near the top will now be inward and the radial components near the bottom will now be outward. But this is in agreement with the fact that rotating the solenoid is equivalent to reversing the direction of the current (which reverses the direction of B everywhere). So, a finite solenoid can have a nonzero radial component of B without violating the fact that rotating the solenoid is equivalent to reversing the current.

    But for an infinitely long solenoid, if the radial component of B is outward at some point that is a distance d from the axis, then the radial component of B would have to be outward for all points at a distance d from the axis. But then, rotating the solenoid would lead to the radial component still being outward at all points a distance d from the axis. But this would not agree with what would happen if you reversed the current. Since rotating the solenoid and reversing the current must lead to the same result, we can conclude that an infinitely long solenoid cannot have a radial component of B at any point.
     
    Last edited: Mar 17, 2017
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