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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. 26rCo6. The forces between two moving charges are not always
equal and opposite. It appears that "action" is not equal to
"reaction."
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. 26rCo6. 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 ree 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.
Aether Regained wrote:[...]
Let's see what the forces in a railgun are:
-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a ^
-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a | Fr
-a ----|----------------------------------------|-------
-a-a-a-a-a | Armature-a-a-a-a-a-a-a-a-a-a-a-a-a <- I-a-a-a-a-a-a-a-a-a-a-a-a | Power supply
-aFa <-| (cylinder)-a-a-a-a-a-a-a-a-a-a-a I ->-a-a-a-a-a-a-a-a-a-a-a-a | -> Fp
-a ----|----------------------------------------|-------
-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a | Fr
-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a 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 ree 4e-7riaI/W T
The force on the conductors in the rail will be:
-aFr = IriaLriaB
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 = IriaWriaB, 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 = IriaWriaB.
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.
Aether Regained wrote:
Hello Paul,
I'll have to agree with your straightforward analysis
on the slider end of the rail gun. In particular, even
if there are unbalanced action-reaction forces between
some pairs of orthogonal current elements, the action
and reaction between all current-element pairs on the
rails and the slider seems to balance out in the direction
of the rails.
On the breech-end or power supply side, things are not
so clear, at least to me. By the way, this turns out
to be quite a deep rabbit hole!
https://web.archive.org/web/20150924124500/http://www.dtic.mil/get-tr-doc/pdf?AD=ADA387401
https://apps.dtic.mil/sti/tr/pdf/ADA473387.pdf
https://web.archive.org/web/20150924130034/http://www.dtic.mil/get-tr-doc/pdf?AD=ADA514371https://apps.dtic.mil/sti/tr/pdf/ADA514371.pdf
Compare this by Matthew K. Schroeder:
regards,
Aether Regained
Paul.B.Andersen <relativity@paulba.no> 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. 26rCo6. 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 ree 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