Calculating accelerations in Induced Electric Fields

AI Thread Summary
The discussion focuses on calculating the work done by an electric field on a proton moving in a decreasing magnetic field. The Lorentz force, which is perpendicular to the proton's velocity, does not perform work, leading to the need to consider the electric field generated by the changing magnetic field. Participants clarify that the electric field should be calculated at the initial and final times to find the difference, which is then multiplied by the proton's charge. The importance of using the definition of work as an integral of force over displacement is emphasized. The conversation highlights the necessity of understanding the relationship between the electric field and the work done on charged particles in a magnetic field.
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: 438
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.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top