Calculating accelerations in Induced Electric Fields

Click For Summary

Homework Help Overview

The problem involves calculating the work done by an electric field on a proton moving in a circular path within a region of a spatially uniform magnetic field that decreases over time. The context includes concepts from electromagnetism, specifically the interaction between electric and magnetic fields.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the Lorentz force and the work done by the electric field, questioning how to calculate the acceleration and the work done without explicit values for velocity and field strength. There is also an exploration of the electric field's behavior over time and its impact on the proton's motion.

Discussion Status

Some participants have provided insights regarding the nature of the Lorentz force and its implications for work done on the proton. There is an ongoing exploration of how to calculate the electric field at different times and its relevance to the work done, with no clear consensus yet on the approach to take.

Contextual Notes

Participants note the importance of considering the electric field generated by the changing magnetic field and the need to evaluate it at specific time intervals. There is an emphasis on understanding the definitions of work and force in this context.

Ignitia
Messages
21
Reaction score
5

Homework Statement


Over a region of radius R, there is a spatially uniform magnetic field B →. (See below.) At t =0, B=1.0T, after which it decreases at a constant rate to zero in 30 s.

(b) Assume that R=10.0cm. How much work is done by the electric field on a proton that is carried once clock wise around a circular path of radius 5.0 cm?

cff5f780a549fca32486e67d6d45f6095be630d0.jpg

Homework Equations


W = F*d
F=q(vxB)
F=m*a

d=2πr r=5cm

The Attempt at a Solution


Since velocity and field are not given, have to go with F=m*a, but I don't see how [a] can be calculated.

with W = F*d, then W=(m*a)*(2πr)

Is there something I'm missing?
 

Attachments

  • cff5f780a549fca32486e67d6d45f6095be630d0.jpg
    cff5f780a549fca32486e67d6d45f6095be630d0.jpg
    11.3 KB · Views: 463
Physics news on Phys.org
Ignitia said:
Is there something I'm missing?
You are missing the fact that the Lorentz force (##q\vec v\times \vec B##) does no work on the proton because it is always perpendicular to ##\vec v##. Besides the problem asks you to find the work done by the electric field. Where do you think that comes from?
 
kuruman said:
You are missing the fact that the Lorentz force (##q\vec v\times \vec B##) does no work on the proton because it is always perpendicular to ##\vec v##. Besides the problem asks you to find the work done by the electric field. Where do you think that comes from?

Okay, in part (a) I had calculated the Electric Field both inside and outside the magnetic field. (for reference, inside was was dB/dt * r/2 < answer checks out) If I'm reading this right, I have to take that equation, calculate the field at t = 0 and t = 30, find the difference, and multiply that by the charge of the proton?
 
Ignitia said:
Okay, in part (a) I had calculated the Electric Field both inside and outside the magnetic field.
Good. The proton is in the region inside the magnetic field.
Ignitia said:
... I have to take that equation, calculate the field at t = 0 and t = 30, find the difference, and multiply that by the charge of the proton?
Is that consistent with the definition of work ##W = \int \vec F \cdot d\vec s## ? I think you should find the force on the proton (remember there is an electric field at its location) and then do the line integral.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K
Replies
8
Views
2K