Bring a particle to rest within a field

In summary, a particle with a charge of 10^-24 C and a mass of 10^-24 kg has a linear velocity of 1 km/s. To bring the particle to a stop in 100 microseconds, an electric field E must be applied. This can be calculated using the equation F=ma, where a is the deceleration of the particle and m is its mass. Since the electric field is the only force acting on the particle, the force of the electric field must be equal to the force needed to decelerate the particle.
  • #1
Octavius1287
30
0

Homework Statement


A charged particle is moving in a magnetic field. what is the electric field E needed to bring the particle to a rest at 100 us, and v is 1 km/s

Homework Equations


I know the magnetic force is q*vxB
and Fc=(mv^2)/r

I have all the parts I think to solve this, I'm just not sure about what method to use
 
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  • #2
What is the relative orientation of the fields and the velocity?
The magnetic field will not influence the velocity directly, but it can lead to a curved path.
 
  • #3
v is perpendicular to the field. I know B, m, and q and r
 
  • #4
What about the electric field?
I know B, m, and q and r
Please post all values you know - ideally, copy the exact problem statement. It is pointless to try to help based on some fraction of the problem statement.
 
  • #6
A particle with a charge 10^-24 C and a mass 10^-24 kg has a linear velocity v of 1 km/s.
Part a of the problem was finding the radius of curvature if it was entering a magnetic field B perpendicular to v
part B was if v and b was parallel. But stated B=2T

Part c is what i posted.
Find the breaking electric field E needed to bring the particle to a rest at 100 us, and a linear v is 1 km/s
 
  • #7
So in part c are the B and E fields still in parallel?
 
  • #8
in part B it just says find R if v is parallel to B, which is 0 because sin(180) is 0.

C is worded exactly as i typed it. Part C is the first time the electric field is mentioned. I assume V is perpendicular to B, because if it was parallel everything would be 0
 
  • #9
Octavius1287 said:
in part B it just says find R if v is parallel to B, which is 0 because sin(180) is 0.
Actually, R = ∞ if B and v are parallel since there would be no bending of the beam path.
C is worded exactly as i typed it. Part C is the first time the electric field is mentioned. I assume V is perpendicular to B, because if it was parallel everything would be 0

No. v would be a straight line. If v were still parallel to B then the solution to c) would be easy. If v is perpendicular to B then I would revert to "get a cyclotron".

Hopefully someone brighter will give us a better idea ... you could set up an alternating E field with a square or sine or some other wave I suppose, but the math ... :eek:
 
  • #10
ugh this has been killing me for hours
 
  • #11
Octavius1287 said:
ugh this has been killing me for hours

EDITed:

It will keep killing you until you assume that B and v are still parallel (as in part b). :smile:
 
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  • #12
my brains so tired now I am not seeing it even if they are parallel
 
  • #13
Octavius1287 said:
my brains so tired now I am not seeing it even if they are parallel

If B and v are parallel what does B do to the path of the charged particle?
 
  • #14
i think nothing
 
  • #15
Right. I think you should assume that B is parallel to v, as in b, so you can ignore the magnetic field for part c. The electric field is aligned with the velocity as well (otherwise it cannot completely stop the particle).
Therefore, all motion is constrained to one direction, and the equations get easy.
 
  • #16
the force of the lectric field need to be equal to the force of the moving particle right
 
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  • #17
i guess i don't see the equation where time would play a factor
 
  • #18
Octavius1287 said:
the force of the lectric field need to be equal to the force of the moving particle right

The electric field applies a force to the moving particle to slow it down to a stop.

Just like throwing a ball up. It goes up a certain distance, then stops before returning.
 
  • #19
Octavius1287 said:
i guess i don't see the equation where time would play a factor

If you threw a ball up with a certain velocity v0 and the ball stopped going up after time T, could you compute g? It's exactly the same problem.
 
  • #20
i mean in equation form, i see how it would but reading my book i don't see how to set up the equation
 
  • #21
There is no "force of the moving particle". You want to decelerate the particle (can you calculate numbers for that? Using the velocity different and the stopping time...) with the force of the electric field (which formula can you use to calculate this?).
 
  • #22
F=ma, but with a - acceleration...damn it i didnt see that
 

What is meant by "bringing a particle to rest within a field"?

When we talk about bringing a particle to rest within a field, we are referring to the process of slowing down and eventually stopping the movement of a charged particle within an electric or magnetic field.

Why is it important to bring a particle to rest within a field?

Bringing a particle to rest within a field allows us to control and manipulate its movement and behavior. This is essential in many scientific experiments and technological applications.

What factors affect the ability to bring a particle to rest within a field?

The strength and direction of the field, as well as the charge and mass of the particle, are all factors that can affect the ability to bring a particle to rest within a field. The properties of the particle itself, such as its velocity and trajectory, also play a role.

Can any particle be brought to rest within a field?

No, not all particles can be brought to rest within a field. Only charged particles, such as electrons and protons, can be affected by electric and magnetic fields and brought to rest.

What techniques are used to bring a particle to rest within a field?

There are several techniques that can be used to bring a particle to rest within a field, including using electric or magnetic fields to exert a force on the particle, using collision processes, or using techniques such as laser cooling and trapping.

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