- #1
Aziza
- 189
- 1
There is a wire oriented in the z direction with no current in it for t<0. At t=0, there is a burst of current: I(t) = qδ(t). What is the vector potential \vec{A}? (\vec{B}=∇×\vec{A})
My solution:
let point P be located distance s from the wire, z be length along wire, and Ω be distance from point on wire to point P. Then:
\vec{A}(s,t) = \frac{μq\hat{z}}{4\pi}\int^{∞}_{-∞}\frac{δ(t-Ω/c)}{Ω}dz = \frac{μq\hat{z}}{4\pi} \frac{1}{ct}
t-Ω/c is the retarded time, thus δ(t) becomes δ(t-Ω/c), and I substituded Ω = √(s2+z2) to get the integral in terms of z. The limits are from -Inf to +Inf because for a given time t, the only contribution to the potential is from z=±√( (ct)2-s2 ). Thus I integrated the expression by plugging in this z value.
The answer is wrong however, and despite perusing the solution I cannot figure it out. The solution rewrites the integral in terms of Ω instead of z, but i don't understand why we cannot just integrate wrt z. Help is much appreciated!
My solution:
let point P be located distance s from the wire, z be length along wire, and Ω be distance from point on wire to point P. Then:
\vec{A}(s,t) = \frac{μq\hat{z}}{4\pi}\int^{∞}_{-∞}\frac{δ(t-Ω/c)}{Ω}dz = \frac{μq\hat{z}}{4\pi} \frac{1}{ct}
t-Ω/c is the retarded time, thus δ(t) becomes δ(t-Ω/c), and I substituded Ω = √(s2+z2) to get the integral in terms of z. The limits are from -Inf to +Inf because for a given time t, the only contribution to the potential is from z=±√( (ct)2-s2 ). Thus I integrated the expression by plugging in this z value.
The answer is wrong however, and despite perusing the solution I cannot figure it out. The solution rewrites the integral in terms of Ω instead of z, but i don't understand why we cannot just integrate wrt z. Help is much appreciated!