Field at distance s from infinite wire .retarded potential.

[yungman]

Field at distance s from infinite wire...retarded potential.

This is an example from Griffiths p425.

An infinite straight wire carries the current $$I(t)=\left [ \begin{array}{c} 0, \;\;\; \; For \;t \leq 0\\ I_0,\;\;\; For \; t>0; \end{array}$$

That is, a constant current $I_0$ is turned on abruptly at t=0. Find the resulting electric and magnetic fields.



Book given solution:

The wire is presumably electrially neutral , so the scalar potential is zero. Let the wire lie along the z axis, the retarded vector potential at point P is

$$A_{(s,t)} =\hat z \frac {\mu_0}{4\pi}\int^{\infty}_{-\infty} \frac {I_{(t_r)}}{\eta} dz \;\;\hbox { where }\; \eta = +\sqrt{z^2+s^2}$$

For t < s/c, the "news" has not yet reached P, and the potential is zero. For t > s/c, only the segment

$$|z|\leq \sqrt{(ct)^2-s^2}$$

contributes ( outside this range $t_r$ is negative, so $I_{t_r}$ = 0)



1) From the above, my understand is the book consider the steady current $I_0$ appear on the WHOLE wire instantenously at t=0. The book did not take into account of the current must start at some point on the line and take time to propagate down the wire!

Am I correct from the solution of the book? This is not a good example because it is unrealistic! I just want to confirm this.

2) The book continue:

$$\vec A_{(s,t)}=\left ( \hat z \frac { \mu_0 I_0}{4\pi} \right ) 2 \int^{ \sqrt{(ct)^2-s^2} } _0 \frac {dz}{\sqrt{ s^2+z^2}} = \left ( \hat z \frac { \mu_0 I_0}{4\pi} \right ) ln( \sqrt{s^2+z^2} +z) \right | \;^{\sqrt{(ct)^2-s^2}}_0$$

I don't get this answer, this is my work:

Let $$tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s$$

$$\int \frac {dz}{\sqrt{ s^2+z^2 }} = \int sec \;\theta \;d\theta = ln| sec \;\theta + tan \;\theta| = ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right |$$

But from the book:

$$\int \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |}$$

Can anyone help?

 

[mark726]

Thank you for bringing up these questions. I am always happy to clarify any confusion and help deepen understanding.

To answer your first question, yes, you are correct in pointing out that the book's solution does not take into account the time it takes for the current to propagate down the wire. In this example, the book is assuming an ideal scenario where the current is turned on instantaneously along the entire length of the wire. This is a simplification often used in physics problems to make calculations easier. However, in reality, the current would indeed take some time to propagate down the wire, and the resulting fields would be different.

As for your second question, the book's solution is correct. It may seem different from your work because of the way the integration is approached. In the book's solution, the substitution used is \eta = \sqrt{z^2 + s^2}, and the limit of integration is from 0 to \sqrt{(ct)^2 - s^2}. In your work, you used the substitution \tan \theta = \frac{z}{s}, and the limit of integration is from 0 to \arctan \left(\frac{\sqrt{(ct)^2 - s^2}}{s}\right). Both approaches should yield the same result.

I hope this helps clarify any confusion. Please feel free to ask any further questions or seek clarification on any other aspects of this problem. As scientists, it is important to have a thorough understanding of the concepts and methods we use in our work. Happy studying!