Work Done by an Induced Electric Field

[KC374]

Homework Statement

Over a region of radius R, there is a spatially uniform magnetic field 𝐁⃗ . (See below.) At 𝑡=0, 𝐵=1.0T, after which it decreases at a constant rate to zero in 30 s. (a) What is the electric field in the regions where 𝑟≤𝑅 and 𝑟≥𝑅 during that 30-s interval? (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? (c) How much work is done by the electric field on a proton that is carried once counterclockwise around a circular path of any radius 𝑟≥𝑅?

Relevant Equations

$W = \oint q \vec E \cdot d \vec l$

$\oint \vec E \cdot d \vec l = \frac {d\Phi_m} {dt}$

I have drawn a picture of what the induced electric field will look like, and I have determined its magnitude both within and outside of the magnetic field. I was able to get the right answer for part (b) with this information, but I don't understand why the answer for part (c) is 0 J. It implies that one of the components in the work equation need to be zero in order to yeild W = 0 J. I know that the work done by an electrostatic field over a closed path is zero because it is a conservative vector field, but I can't use that fact here since the induced electric field is non-conservative and does net work. What then is the reason why the induced electric field does no work in moving a proton outside the magnetic field counterclockwise one revolution?

 

[TSny]

I agree with you that the work done by E will be nonzero for both r < R and r > R. I think the answer of 0 for part (c) is wrong.

 

[KC374]

I was thinking that too. The work done by the induced electric field is positive when the proton is moved counterclockwise because the displacement is in the same direction as the electric field line. If the proton was moved clockwise, then the work would be negative, not zero, since the displacement is in the opposite direction. In other words, the induced electric field opposes the movement of the proton, reducing its kinetic energy. That makes sense as opposed to the answer given by the book.