• Forces in a railgun

    From Paul.B.Andersen@21:1/5 to Aether Regained on Fri Aug 1 22:22:53 2025
    XPost: sci.physics.relativity

    Aether Regained wrote:

    https://www.feynmanlectures.caltech.edu/img/FLP_II/f26-06/f26-06_tc_big.svgz

    Compare Figure FLP-II-26-6 with the rail gun configuration.
    As in the figure, in the rail gun the currents in the rail
    and in the slider (cylinder) are orthogonal, so action and
    reaction of the electromagnetic forces will not balanced.

    Fig. 26–6. The forces between two moving charges are not always
    equal and opposite. It appears that "action" is not equal to
    "reaction."


    Let's see what the forces in a railgun are:

    ^
    | Fr
    ----|----------------------------------------|-------
    | Armature <- I | Power supply
    Fa <-| (cylinder) I -> | -> Fp
    ----|----------------------------------------|-------
    | Fr
    V

    If your editor clutters up the figure, see it here:

    http://paulba.no/temp/Railgun.pdf


    The with of the rail is W.
    The distance between the armature and power supply is L.

    The rail, armature and power supply make a rectangle.
    The current I is flowing around this rectangle.
    That will make a magnetic flux through the rectangle,
    perpendicular to the current.

    The flux density depend on the geometry of the rail,
    but it will be in the order B ≈ 4e-7⋅I/W T

    The force on the conductors in the rail will be:
    Fr = I⋅L⋅B
    This force is perpendicular to the current and the B field.
    Since the currents in the two parts of the rail are in opposite
    direction, these forces will act in opposite direction, and will
    be each other's reaction force. The rail must obviously be so solid
    that it can withstand these forces without breaking.

    NOTE: NO FORCES ON THE RAIL ARE ACTING PARALLEL TO THE RAIL!

    The force on the armature is Fa = I⋅W⋅B, acting perpendicular to
    the current and the B-field. To the left on the figure.
    This is the force that accelerates the armature.

    The force on the power supply, or rather on the conductors in
    the power supply carrying the current, is Fp = I⋅W⋅B.
    Since the current in the power supply and the armature
    are in opposite direction, so are the forces.
    Fa and Fp are equal in magnitude and in opposite direction.
    Fa and Fp are each other's reaction force.

    Since the power supply is fixed on the rail the reaction
    force are acting on the rail through the power supply.

    --
    Paul

    https://paulba.no/

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    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From J. J. Lodder@21:1/5 to Paul.B.Andersen on Sat Aug 2 13:03:05 2025
    XPost: sci.physics.relativity

    Paul.B.Andersen <[email protected]> wrote:

    Aether Regained wrote:

    https://www.feynmanlectures.caltech.edu/img/FLP_II/f26-06/f26-06_tc_big.svgz

    Compare Figure FLP-II-26-6 with the rail gun configuration.
    As in the figure, in the rail gun the currents in the rail
    and in the slider (cylinder) are orthogonal, so action and
    reaction of the electromagnetic forces will not balanced.

    Fig. 26–6. The forces between two moving charges are not always
    equal and opposite. It appears that "action" is not equal to
    "reaction."


    Let's see what the forces in a railgun are:

    ^
    | Fr
    ----|----------------------------------------|-------
    | Armature <- I | Power supply
    Fa <-| (cylinder) I -> | -> Fp
    ----|----------------------------------------|-------
    | Fr
    V

    If your editor clutters up the figure, see it here:

    http://paulba.no/temp/Railgun.pdf


    The with of the rail is W.
    The distance between the armature and power supply is L.

    The rail, armature and power supply make a rectangle.
    The current I is flowing around this rectangle.
    That will make a magnetic flux through the rectangle,
    perpendicular to the current.

    The flux density depend on the geometry of the rail,
    but it will be in the order B ≈ 4e-7?I/W T

    The force on the conductors in the rail will be:
    Fr = I?L?B
    This force is perpendicular to the current and the B field.
    Since the currents in the two parts of the rail are in opposite
    direction, these forces will act in opposite direction, and will
    be each other's reaction force. The rail must obviously be so solid
    that it can withstand these forces without breaking.

    NOTE: NO FORCES ON THE RAIL ARE ACTING PARALLEL TO THE RAIL!

    The force on the armature is Fa = I?W?B, acting perpendicular to
    the current and the B-field. To the left on the figure.
    This is the force that accelerates the armature.

    The force on the power supply, or rather on the conductors in
    the power supply carrying the current, is Fp = I?W?B.
    Since the current in the power supply and the armature
    are in opposite direction, so are the forces.
    Fa and Fp are equal in magnitude and in opposite direction.
    Fa and Fp are each other's reaction force.

    Since the power supply is fixed on the rail the reaction
    force are acting on the rail through the power supply.

    Yes, but if the current comes from some distance away,
    through flexible conductors,
    there are possibilities for muddling things up,

    Jan

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