- #1
yungman
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Field at distance s from infinite wire...retarded potential.
This is an example from Griffiths p425.
An infinite straight wire carries the current [tex]I(t)=\left [ \begin{array}{c} 0, \;\;\; \; For \;t \leq 0\\ I_0,\;\;\; For \; t>0; \end{array} [/tex]
That is, a constant current [itex]I_0[/itex] 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
[tex]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}[/tex]
For t < s/c, the "news" has not yet reached P, and the potential is zero. For t > s/c, only the segment
[tex]|z|\leq \sqrt{(ct)^2-s^2}[/tex]
contributes ( outside this range [itex]t_r[/itex] is negative, so [itex]I_{t_r}[/itex] = 0)
1) From the above, my understand is the book consider the steady current [itex]I_0[/itex] 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:
[tex]\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[/tex]
I don't get this answer, this is my work:
Let [tex] tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s [/tex]
[tex]\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 | [/tex]
But from the book:
[tex]\int \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |} [/tex]
Can anyone help?
This is an example from Griffiths p425.
An infinite straight wire carries the current [tex]I(t)=\left [ \begin{array}{c} 0, \;\;\; \; For \;t \leq 0\\ I_0,\;\;\; For \; t>0; \end{array} [/tex]
That is, a constant current [itex]I_0[/itex] 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
[tex]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}[/tex]
For t < s/c, the "news" has not yet reached P, and the potential is zero. For t > s/c, only the segment
[tex]|z|\leq \sqrt{(ct)^2-s^2}[/tex]
contributes ( outside this range [itex]t_r[/itex] is negative, so [itex]I_{t_r}[/itex] = 0)
1) From the above, my understand is the book consider the steady current [itex]I_0[/itex] 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:
[tex]\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[/tex]
I don't get this answer, this is my work:
Let [tex] tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s [/tex]
[tex]\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 | [/tex]
But from the book:
[tex]\int \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |} [/tex]
Can anyone help?
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