Understanding Linear Motors: Right Hand Rule & Current Flow

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The discussion centers on the application of the right and left-hand rules in understanding linear motors. Participants clarify that the right-hand rule is the standard method for determining the direction of force in relation to current and magnetic fields, while the left-hand rule is less commonly used and typically associated with motors versus generators. Confusion arises regarding the direction of current flow and the application of the Lorentz force law, particularly with a magnetic field directed into the page. It is established that the current flows downward through the movable rail, and the right-hand rule can be effectively applied to determine the force direction. Overall, the right-hand rule is emphasized as the reliable method for analyzing linear motor dynamics.
TheWonderer1
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Hi there,

I attached an image of a basic linear motor. I'm a little confused if I use the right hand rule or left hand rule. I did it both ways and I think the left hand rule is the only rule that allows for a force being applied to the right. I assume the current is going down rather than upward like what it says in the left corner since the current moves clockwise.

Let me know if I'm correct on the rule and direction of the current. I'm assuming it moves through the coil in the middle.
 

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TheWonderer1 said:
I attached an image of a basic linear motor. I'm a little confused if I use the right hand rule or left hand rule. I did it both ways and I think the left hand rule is the only rule that allows for a force being applied to the right. I assume the current is going down rather than upward like what it says in the left corner since the current moves clockwise.
I am not aware of any "left hand rule" only the right hand rule for determining the direction of the cross product of two vectors. In this scenario, you should apply the Lorentz force law to the charges traveling through the movable wire.
 
So when is the left hand rule usable? I've seen that it's for motor vs. right hand for generators. How do I apply the law with a magnetic field into the page and the current going up? It's just an example that I haven't seen.
 
TheWonderer1 said:
So when is the left hand rule usable? I've seen that it's for motor vs. right hand for generators.
I never use it. There may be some instances were it works for a specific problem, but the strict application of the right hand rule for cross products always works.
TheWonderer1 said:
How do I apply the law with a magnetic field into the page and the current going up?
The Lorentz force law states
$$\mathbf{F}=q(\mathbf{E}+\mathbf{v\times B})$$
There is no electric field shown outside the wire so ##\mathbf{E}## is zero. The charges move in the direction of ##\mathbf{I}## and you know the direction of ##\mathbf{B}##, so what is the direction of ##\mathbf{F}##?
 
Is the current flowing upward? The magnet field is going into the page. I do the right hand rule and I find my palm facing away from me. My pointer finger is pointing down and my middle finger is away from me. The force is to the right.

However, do you use the thumb is the current direction, field are your fingers and your palm faces the force direction? It's all confusing because there are several ways to do it.
 
TheWonderer1 said:
Is the current flowing upward?
It appears that it is flowing downward through the movable rail. To use the right hand rule, extend your fingers on your right hand in the direction of the first vector ##\mathbf{v}##, now adjust your hand such that you can curl your finders in the direction of the second vector ##\mathbf{B}##. Your thumb should now point in the direction of the cross product ##\mathbf{F}##.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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