Calculating accelerations in Induced Electric Fields

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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?

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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.