Question

Problem 3: Recall that the general expression for the vector potential is: A(. ,t).n f įe.war where tR is "retarded time", t.-t A very long (effectively infinite) neutral wire on the z-axis has zero current before t-to, anda steady current lo is suddenly turned on at t-to in the direction of +2. At a point a distance "s" from the wire, the B-field first becomes nen-zero at t,to+s/c At any time t>, the magnitude of the vector potential can be written in this form: 2t dz ż What are the appropriate limits of integration? (Note: I ONLY want the limits of integration, I'm not asking you to compute A!) Clearly explain and show all your work. (Unlike the homework we did, note that our current here turns on at t-to not t-0) lo +x (10 pts)

Answer 1

For t < s/c the news has not reached at observation point p. so, potential is zero for t > s/cVertical-bar z Vertical-bar = squareroot (c t)^2 - s^2 [therefore r = c t] so, this will contribute. so the limit of integration will be O to squareroot (c t)^2 - s^2 so, [A^vector (s, t) = mu_0/4 pi integral_0^squareroot (c t)^2 - s^2 I_0/squareroot s^2 + z^2 dz]