# Vector potential due to sudden burst of current in a wire

1. Oct 26, 2013

### Aziza

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 dont understand why we cannot just integrate wrt z. Help is much appreciated!

2. Oct 26, 2013

### TSny

Note that the argument of the delta function is some function of $z$. So, if you are going to use $z$ as the integration variable, you will need to use property (7) given at this link: http://mathworld.wolfram.com/DeltaFunction.html

Last edited: Oct 26, 2013