Physics 12: Acceleration of Proton in B-Field

In summary: The acceleration is v^2/r.In summary, a proton is accelerated from rest at the positive plate of two charged parallel plates with a potential difference of 2000 v. After leaving the plates, it enters a uniform magnetic field of 0.50 T in a direction perpendicular to the magnetic field directed out the page. The acceleration of the proton can be found by using the equations F=qvB and a=v^2/2d, where F is the force, q is the charge of the proton, v is the final velocity, B is the magnetic field, and d is the distance between the plates. Additionally, the centripetal acceleration in the magnetic field can be found using F=qvB sin(theta) and a
  • #1
marysaf
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Homework Statement


[/B]
A proton is accelerated from rest at the positive plate of two charged parallel plates with a potential difference of 2000 v. After leaving the plates through a small hole in the negative plate, it enters a uniform magnetic field of 0.50 T in a direction perpendicular to the magnetic field directed out the page as shown in the diagram. Find the acceleration of the proton.
upload_2017-6-4_14-1-59.png

https://drive.google.com/file/d/0B_FmHNxGrZAcN21ic0kwQ191alpram5qc0loclFhRFZTN3Yw/view?usp=sharing

Homework Equations



a=v^2/2d

The Attempt at a Solution



I tried to use the equation above but I don't have d and don't know how to find it. should I use f=ma?
 
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  • #2
marysaf said:
should I use f=ma?
Yes.
 
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  • #3
marysaf said:

Homework Statement


[/B]
A proton is accelerated from rest at the positive plate of two charged parallel plates with a potential difference of 2000 v. After leaving the plates through a small hole in the negative plate, it enters a uniform magnetic field of 0.50 T in a direction perpendicular to the magnetic field directed out the page as shown in the diagram. Find the acceleration of the proton.

https://drive.google.com/file/d/0B_FmHNxGrZAcN21ic0kwQ191alpram5qc0loclFhRFZTN3Yw/view?usp=sharing

https://lh6.googleusercontent.com/SWII9lhrczWbmznSniycfEcDFtVfg1orKhlr4ua9AmkZbKIVm5eUmqZWPcBqWt-2xArx66svycn4BCHaQm8nXaSDU10Jd3dVW85c=w878-h1043-rw

Homework Equations



a=v^2/2d

The Attempt at a Solution



I tried to use the equation above but I don't have d and don't know how to find it. should I use f=ma?
F = qvB also helps.
 
  • #4
There are two accelerations in this problem.
The linear one, between the plates.
To find this one you must find first the final velocity. Use energy to do that.

The centripetal one. In the magnetic field. Do find this use F=qvB sin(theta).
 

1. What is the formula for calculating the acceleration of a proton in a B-field?

The formula for calculating the acceleration of a proton in a B-field is a = qvB/m, where q is the charge of the proton, v is its velocity, B is the strength of the magnetic field, and m is the mass of the proton.

2. How does the acceleration of a proton in a B-field affect its path?

The acceleration of a proton in a B-field causes it to move in a circular path perpendicular to both its velocity and the direction of the magnetic field. This is known as circular motion.

3. What factors affect the acceleration of a proton in a B-field?

The acceleration of a proton in a B-field is affected by the strength of the magnetic field, the velocity of the proton, and the mass and charge of the proton.

4. Can the direction of the acceleration of a proton in a B-field change?

Yes, the direction of the acceleration of a proton in a B-field can change if the strength or direction of the magnetic field changes. It can also change if the velocity of the proton changes.

5. How is the acceleration of a proton in a B-field related to its kinetic energy?

The kinetic energy of a proton is directly proportional to its acceleration in a B-field. This means that as the proton's acceleration increases, so does its kinetic energy.

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