Charged particles moving through a velocity selector

[enTropeeeeeeeee]

It seems that Work is one of those concepts that I have an issue understanding. So in an attempt to address that, I decided to apply it to a velocity selector.

If we have a particle accelerating thru a velocity selector and both electric force and magnetic force are equal, the particle will go right through. In this situation, no work is being done on the particle! Correct?

However, if the particle bends either up or down, then I can assume that work is being done on the particle. Correct?

Thanks for the help!

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[ilovemcat]

It seems that Work is one of those concepts that I have an issue understanding. So in an attempt to address that, I decided to apply it to a velocity selector.

If we have a particle accelerating thru a velocity selector and both electric force and magnetic force are equal, the particle will go right through. In this situation, no work is being done on the particle! Correct?

However, if the particle bends either up or down, then I can assume that work is being done on the particle. Correct?

Thanks for the help!

I'm assuming you got this from TPRH example So let's say you have a parallel plate capacitor in some magnetic field and it just so happens that the Magnetic Force and Electric for are equal but opposite directions. Then a particle is shot directly through the middle of the plate. What would happen then is that (according to Newton's first Law), the charged particle will pass through without bending in either direction. For this scenario no work is done because in either case - both forces are perpendicular to the movement of the particle. If however the particle curved towards one of the plates, then yes it would be doing work. Though it would only be the component of the velocity that's parallel to Electric Field that does work.

Also, Magnetic Force NEVER does any work because it's always perpendicular to the charged particle (just think of the weird hand rules). Magnetic Force is essentially THE centripetal force for a particle. It doesn't speed up the particle or slow it down. It just changes the direction.

 

[ilovemcat]

Also I just want to clarify - Electric Force and Magnetic Force are both vectors. For the scenario I provided above, they were pointing in exactly opposite directions. This is what allowed the particle to pass through undeflected.

On the other hand, the Electric Force and Magnetic Force could be equal but that doesn't always mean the particle will do no work. If the Magnetic Force and Electric Force were perpendicular to each other, the particle would just spiral (clockwise or counterclockwise) in the direction of the electric force.

 

[enTropeeeeeeeee]

I'm assuming you got this from TPRH example So let's say you have a parallel plate capacitor in some magnetic field and it just so happens that the Magnetic Force and Electric for are equal but opposite directions. Then a particle is shot directly through the middle of the plate. What would happen then is that (according to Newton's first Law), the charged particle will pass through without bending in either direction. For this scenario no work is done because in either case - both forces are perpendicular to the movement of the particle. If however the particle curved towards one of the plates, then yes it would be doing work. Though it would only be the component of the velocity that's parallel to Electric Field that does work.

Also, Magnetic Force NEVER does any work because it's always perpendicular to the charged particle (just think of the weird hand rules). Magnetic Force is essentially THE centripetal force for a particle. It doesn't speed up the particle or slow it down. It just changes the direction.

Magnetic force never does work because it is always perpendicular to the vectored velocity of the particle. Work = (F) (D)cos. Anything that is cos90 = 0.

So if this is true, then the bending of the particle will be work done on the particle by electric force. Is this correct?

 

[ilovemcat]

Magnetic force never does work because it is always perpendicular to the vectored velocity of the particle. Work = (F) (D)cos. Anything that is cos90 = 0.

So if this is true, then the bending of the particle will be work done on the particle by electric force. Is this correct?

Yes and Yes.

Take a tip from me though and ditch the whole cos(theta) part of the work equation. It just complicates things and slows you down. Once you realize that both force and the distance traveled must be parallel (or antiparallel) to each other in order to do work - that part of the equation becomes unnecessary. With this in mind, you'll quickly realize that any force perpendicular to the movement of an object does no work (as you stated correctly). If a force is at an angle such that a portion of the force lies in the same/opposite direction as the displacement of the object - work is done (but only for that component parallel/anti-parallel to the displacement).

For the second statement - yes. The fact that a particle is "bending" means that it's being acted against/by an Electric Field (either attracted towards or repelled away).

 

[enTropeeeeeeeee]

Yes and Yes.

Take a tip from me though and ditch the whole cos(theta) part of the work equation. It just complicates things and slows you down. Once you realize that both force and the distance traveled must be parallel (or antiparallel) to each other in order to do work - that part of the equation becomes unnecessary. With this in mind, you'll quickly realize that any force perpendicular to the movement of an object does no work (as you stated correctly). If a force is at an angle such that a portion of the force lies in the same/opposite direction as the displacement of the object - work is done (but only for that component parallel/anti-parallel to the displacement).

For the second statement - yes.

Thanks!

 

[Rabolisk]

Take a tip from me though and ditch the whole cos(theta) part of the work equation. It just complicates things and slows you down. Once you realize that both force and the distance traveled must be parallel (or antiparallel) to each other in order to do work - that part of the equation becomes unnecessary. With this in mind, you'll quickly realize that any force perpendicular to the movement of an object does no work (as you stated correctly). If a force is at an angle such that a portion of the force lies in the same/opposite direction as the displacement of the object - work is done (but only for that component parallel/anti-parallel to the displacement).

Your understanding is correct, but that's no reason to ditch the cos(&#952 part of the equation, which can be useful for calculations. After all W = Fdcos(&#952 comes directly from W = F·d (· being dot product), which is a specific case of the more general formula W = path integral (F · dr). I think it's better for one to memorize the cos(&#952 than having to realize that the parallel component is cos(&#952 and the perpendicular component is sin(&#952. Of course, as a math major, I think this is incredibly simple, but that's beside the point.