Calculating accelerations in Induced Electric Fields

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SUMMARY

The discussion focuses on calculating the work done by an electric field on a proton moving in a time-varying magnetic field. The magnetic field, initially at 1.0 T, decreases to zero over 30 seconds, creating an induced electric field. The key equations involved include the Lorentz force (F=q(v×B)), which does not perform work on the proton, and the work equation (W=F·d). The solution requires calculating the electric field at two time points and using the charge of the proton to find the work done.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Faraday's law of induction.
  • Familiarity with the Lorentz force equation (F=q(v×B)).
  • Knowledge of work-energy principles in physics (W=F·d).
  • Basic calculus for evaluating integrals and derivatives.
NEXT STEPS
  • Study Faraday's law of electromagnetic induction in detail.
  • Learn about the relationship between changing magnetic fields and induced electric fields.
  • Explore the concept of line integrals in the context of electric fields.
  • Investigate the implications of the Lorentz force in various electromagnetic scenarios.
USEFUL FOR

Physics students, educators, and professionals in electromagnetism, particularly those interested in the dynamics of charged particles in varying magnetic fields.

Ignitia
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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?
 

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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.
 

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