Solving Griffiths' Electrodynamics Ex. 10.2

[LuxAurum]

Homework Statement

I'm working through the 3rd edition of Griffiths' Electrodynamics book and have gotten stuck on some details in example 10.2, which describes an infinite straight wire carrying a current I₀ for t>0.

The figure included with the example illustrates a wire in the vertical z direction with element dz. A point P is located at distance s from the wire. The distance from dz to P is given by script_r, and forms a right triangle with base of z.

The part that is unclear to me is the following. Griffiths states:
"For t < s/c, the 'news' has not yet reached P and the potential is zero. For t > s/c, only the segment |z| [itex]\leq \sqrt{(ct)² - s²}[/itex]
contributes (outside this range t_r is negative..."

Homework Equations

t_r = t - (script_r/c)

The Attempt at a Solution

I understand that the expression t = s/c represents the time it takes for the wave to traverse the distance s to get to P. What I'm unclear about is how to properly setup and analyze this problem if it wasn't an example. Is the distance s used to establish a lower bound for the time? Griffiths then uses (ct)² in the expression for |z|.


Also, why does I(t_r) = 0 when t_r is negative?

Hopefully this doesn't come across as scattered. Sometimes I get stuck on minor details that should be obvious but aren't at the time.

 

[njoysci]

In this problem we are to derive the electric and magnetic fields at any point P away from an infinite straight wire carrying a constant current I0.
Generally, E = -∇V - ∂A/∂t and B = ∇ x A
Since the wire has equal positive and negative charges, V = 0 everywhere outside the wire. Once we determine A then the above eqn.s allow us to determine E and B.
I assume you agree with their eqn. for A(s,t) at point P, which was derived from eqn. 10.19 referring to Fig. 10.4 analysis.
As you noted the retarded time, tr = t - (script_r/c) ,where (script_r/c) is the time delay for the source (in this case the dz contribution of the current I0) to reach P traveling at speed c. The first potential contribution reaches P when t = s/c. Before that time A = 0. As t increases, the potential contributions are from +z to –z only, because beyond those the contributions haven’t had enough time to arrive (that is, their tr is negative, and hence their contribution to A is zero).
The integration for A(s,t) at P from + z to – z is determined as shown, and E and B are thus determined as a function of t. From the results you can see how E = 0 and B is as noted as t approached infinity.

 

[LuxAurum]

Thank you for your reply! This definitely helps clarify!

 

[almarpa]

Thank from my side as well