Vector potential due to oscillating dipole

[_Andreas]

Homework Statement

Calculate the magnetic vector potential A at a point p located at a distance r from the axis of an oscillating dipole of length s.

It is assumed that r\gg s and that the current is the same throughout s.

Homework Equations

r=\sqrt{(x^2+(z-z')^2)}, where x,z is the horizontal and vertical coordinates of p, respectively, and z' is the vertical coordinate of the source point. The axis of the dipole lies on the z axis, and so x'=0. The problem is confined to the xz plane only.

A=c\int^{s/2}_{-s/2}\frac{\exp(ikr)}{r}dz' \hat{z},

where c is a constant and k is the wave number. The exponential comes from the fact that the current is a function of the retarded time, [t]=t-r/c.

The Attempt at a Solution

I really don't know how to calculate this integral. Without the exponential I would've been fine, but now... lol wut? Are there perhaps some approximations, expansions, or variable changes that I could do? Any tips?

If it is of any help, the answer is apparently the same answer as in the case of a current localized at the center of the dipole:

A=d*\frac{\exp(i\omega[t])}{r}s \hat{z}

(d is a constant.)

 

[gabbagabbahey]

r=\sqrt{(x^2+(z-z')^2)}, where x,z is the horizontal and vertical coordinates of p, respectively, and z' is the vertical coordinate of the source point.

The problem statement you provided defines r as the distance from the center of the dipole (z'=0 I presume) to the point p...you'll want to use a different letter, like say, (capital) R to represent the separation of the source point and field point:

r=\sqrt{x^2+z^2}

R=\sqrt{(x^2+(z-z')^2)}

A=c\int^{s/2}_{-s/2}\frac{\exp(ikr)}{r}dz' \hat{z},

where c is a constant and k is the wave number. The exponential comes from the fact that the current is a function of the retarded time, [t]=t-r/c.[/tex]

It seems like you are using c to represent both a constant with units of Tesla-meters (in SI) and the speed of light...that's pretty confusing notation to me.

Also, the current in an oscillating dipole varies with time, so why is there no t in this expression?

In SI units, the expression for the vector potential is something like

\textbf{A}(\textbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{I(t-\frac{R}{c})}{R}dz'\mathbf{\hat{z}}

 

[_Andreas]

Wow, that's a lot of annoying mistakes I did.

The problem statement you provided defines r as the distance from the center of the dipole (z'=0 I presume) to the point p...you'll want to use a different letter, like say, (capital) R to represent the separation of the source point and field point:

r=\sqrt{x^2+z^2}

R=\sqrt{(x^2+(z-z')^2)}

Yes, my bad.

It seems like you are using c to represent both a constant with units of Tesla-meters (in SI) and the speed of light...that's pretty confusing notation to me.

Oh boy. Yes, that is definitely confusing. The factor c in front of the integral is not supposed to be the speed of light.

Also, the current in an oscillating dipole varies with time, so why is there no t in this expression?

I shouldn't have called c a constant, since there's supposed to be a time-dependent exponential included.

In SI units, the expression for the vector potential is something like

\textbf{A}(\textbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{I(t-\frac{R}{c})}{R}dz'\mathbf{\hat{z}}

Yes. In my case it's (yes, I even forgot the minus sign in the exponential)

\frac{\mu_0I_0\exp(i\omega t)}{4\pi}\int_{-s/2}^{s/2}\frac{\exp(-ikR)}{R}dz'\hat{z},

where R is as you said, and I_0 is the amplitude of the current, which is independent of z'.

I still don't know how to calculate the integral, though.

 

[gabbagabbahey]

Yes. In my case it's (yes, I even forgot the minus sign in the exponential)

\frac{\mu_0I_0\exp(i\omega t)}{4\pi}\int_{-s/2}^{s/2}\frac{\exp(-ikR)}{R}dz'\hat{z},

where R is as you said, and I_0 is the amplitude of the current, which is independent of z'.

I still don't know how to calculate the integral, though.

You'll probably want to use \omega/c instead of k here; as you'll want to use the approximation that s\ll \frac{\omega}{c} at some point (basically that the dipole is very small compared to the wavelength of the radiation it produces)

You also are given that s\ll r, and since |z'|\leq s, you know z'\ll r...in order for this knowledge to be useful, you'll want to use the law of cosines to express R in terms of z', r and the angle between the source and field point vectors (As measured from the center of the dipole). You can then Taylor expand both 1/R and \text{exp}(-i\omega R/c) for small z'/r.

 

[_Andreas]

Thank you very much!

Doing as you say I get

R=\sqrt{z'^2+r^2-2z'r\cos(\theta)}=r\sqrt{(z'/r)^2+1-2z'/r\cos(\theta)}.

Expanding 1/R and e^{-ikR} around z'/r=0 yields 1/r and e^{-ikr}, respectively (the first order terms of both expansions contain factors z'/r^2 and therefore become insignificant).

The integral then simply becomes

\frac{\mu_0I_0e^{i\omega(t-kr)}}{4\pi r}\int dz' \hat{z},

since r is independent of z'. This gives me the correct answer.

 

[gabbagabbahey]

Expanding 1/R and e^{-ikR} around z'/r=0 yields 1/r and e^{-ikr}, respectively (the first order terms of both expansions contain factors z'/r^2 and therefore become insignificant).

Careful, you can definitely say terms with \left(\frac{z&#039;}{r}\right)^2[/tex] are insignificant (the square of a very small number is an even smaller number), but that doesn&#039;t necessarily mean that \frac{z&amp;#039;}{r^2} is negligible...As an example, if s=10^{-6} and r=10^{-4}, then |z&amp;#039;|\leq s\ll r as demanded by the problem statement, but \frac{z&amp;#039;}{r^2}\leq\frac{s}{r^2}=100 won&#039;t be negligible.<br /> <br /> You will need to keep the first order terms in your expansions, and use your second approximation (s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them.

 

[_Andreas]

I'm not exactly having my finest moment as an aspiring physicist here. I blame it on simply rushing through the calculations since I'd really like to be doing something else for a little while.

Careful, you can definitely say terms with \left(\frac{z&#039;}{r}\right)^2[/tex] are insignificant (the square of a very small number is an even smaller number), but that doesn&#039;t necessarily mean that \frac{z&amp;#039;}{r^2} is negligible...As an example, if s=10^{-6} and r=10^{-4}, then |z&amp;#039;|\leq s\ll r as demanded by the problem statement, but \frac{z&amp;#039;}{r^2}\leq\frac{s}{r^2}=100 won&#039;t be negligible.<br /> <br /> You will need to keep the first order terms in your expansions, and use your second approximation (s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them.

&lt;br /&gt; &lt;br /&gt; You&amp;#039;re right. I didn&amp;#039;t get any \frac{z&amp;amp;#039;}{r^2} terms, though; the expansions I did were incorrect. I did have to use the approximation s\ll \frac{\omega}{c}[/tex] in one of the expansions, though.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Thanks again!