Magnetic force Particle trajectory

In summary: Just to make sure. i can not find the distance doing the calculation of the first post. I have to use the radius (post 3).
  • #1
arierreF
79
0

Homework Statement



A particle of charge −q is moving with a velocity [itex]\vec{v}[/itex]. It then enters midway between two plates where there exists a uniform magnetic field pointing into the page, as shown in attached image.

1.png
(b) Compute the distance between the left end of the plate and where the particle strikes

Homework Equations



[itex]\vec{F_{B}} = q\vec{v}\times\vec{B}[/itex]

The Attempt at a Solution



I discovered that the charged particle is going to be deflected downward. Then it is going to strike the bottom plate before leaving the region between two plates. So the distance between the end of the bottom plate and where the particle strikes is what this problem is asking for.

See the following attached image.
2.png
[itex]y = yo -\frac{1}{2}at^{2}[/itex]

[itex]y = 0[/itex] because it strikes the end of the plate.
[itex]y_{0} = \frac{d}{2}[/itex]

so [itex]t = \sqrt{da}[/itex]

[itex]\vec{F_{B}} = q\vec{v}\times\vec{B}[/itex]

to find the magnitude of acceleration:[itex]ma = qvB[/itex]

[itex]a = \frac{qvB}{m}[/itex]

then [itex]t = \sqrt{d\frac{qvB}{m}}[/itex]

[itex]x = x_{0} + v_{0}t[/itex]

[itex]x_{0} = 0[/itex]

[itex]x= v_{0}t[/itex]

because [itex]v_{0} = v[/itex]

[itex]x = v\sqrt{d\frac{qvB}{m}}[/itex]Then the distance is [itex]l- v\sqrt{d\frac{qvB}{m}}[/itex]Is that correct? cause i have feeling that it is wrong.
 
Last edited:
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  • #2
I only skimmed, but I think you have assumed that the deflecting force always acts in the -y direction. Recall - a charge moving in a uniform/constant magnetic-field follows a circular path. So ... what kind of acceleration would that be and what is it's relationship to the radius of the circle?
 
  • #3
i could do it by the radius too.

2.png


[itex] m \frac{v^{2}}{R} = qvB[/itex]

so [itex] R = \frac{mv^{2}}{qvB}[/itex]


[itex] R^2 = (R- \frac{d}{2} )^{2} + x^{2}[/itex]


[itex] x^{2} = Rd - (\frac{d}{2})^{2}[/itex]

[itex] x^{2} = d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2} [/itex]

[itex] x = \sqrt{d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2}} [/itex]

Then the distance is [itex] l - x = \sqrt{d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2}} [/itex]
 
Last edited:
  • #4
Compute the distance between the left end of the plate and where the particle strikes
 
  • #5
Simon Bridge said:
Compute the distance between the left end of the plate and where the particle strikes

That's the problem. I am not understating what the problem is asking for.
 
  • #6
The bottom plate is the one that is going to be hit.
The bottom plate has two ends.
One end is on the left of the picture, the other is on the right.
Which end did you measure distance from (per your diagram)?
What is the length ##x## on your diagram?

Also - the last line in your calculations (post #3) does not follow from the previous line.
 
  • #7
Simon Bridge said:
The bottom plate is the one that is going to be hit.
The bottom plate has two ends.
One end is on the left of the picture, the other is on the right.
Which end did you measure distance from (per your diagram)?
What is the length ##x## on your diagram?

Also - the last line in your calculations (post #3) does not follow from the previous line.
oh right. I was calculating for the right end, not the left end.

so my [itex]x[/itex] is the desired distance (distance between the end of the left [itex]x_{0} = 0[/itex] and where the particle hits the plate [itex]x[/itex].
 
  • #8
Well done.
I used to get caught out by that sort of thing all the time :)
 
  • #9
Simon Bridge said:
Well done.
I used to get caught out by that sort of thing all the time :)

Thanks for your time. :)

just to make sure. i can not find the distance doing the calculation of the first post. I have to use the radius (post 3).
 

1. What is magnetic force?

Magnetic force is the force that is exerted on a charged particle moving through a magnetic field. This force is perpendicular to both the direction of the particle's motion and the direction of the magnetic field.

2. How does magnetic force affect particle trajectory?

Magnetic force causes charged particles to follow a curved path when moving through a magnetic field. The exact trajectory of the particle depends on the strength and direction of the magnetic field, as well as the velocity and charge of the particle.

3. What is the relationship between magnetic force and particle velocity?

The force exerted on a particle by a magnetic field is directly proportional to the particle's velocity. As the velocity of the particle increases, so does the strength of the magnetic force.

4. Can magnetic force be used to control particle motion?

Yes, magnetic force can be used to manipulate the trajectory of charged particles. This is the basis for technologies such as particle accelerators and MRI machines.

5. How is magnetic force calculated?

The magnitude of magnetic force on a charged particle can be calculated using the equation F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and magnetic field.

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