# Homework Help: Vector potential due to oscillating dipole

1. Jan 12, 2010

### _Andreas

1. The problem statement, all variables and given/known data

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.

2. Relevant 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, $$[t]=t-r/c$$.

3. 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.)

Last edited: Jan 12, 2010
2. Jan 13, 2010

### gabbagabbahey

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:

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

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

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

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

3. Jan 13, 2010

### _Andreas

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

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

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

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.

4. Jan 13, 2010

### gabbagabbahey

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 $z'/r$.

5. Jan 14, 2010

### _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.

Last edited: Jan 14, 2010
6. Jan 14, 2010

### gabbagabbahey

Careful, you can definitely say terms with $\left(\frac{z'}{r}\right)^2[/tex] are insignificant (the square of a very small number is an even smaller number), but that doesn't necessarily mean that [itex]\frac{z'}{r^2}$ is negligible.....As an example, if $s=10^{-6}$ and $r=10^{-4}$, then $|z'|\leq s\ll r$ as demanded by the problem statement, but $\frac{z'}{r^2}\leq\frac{s}{r^2}=100$ won't be negligible.

You will need to keep the first order terms in your expansions, and use your second approximation ($s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them. 7. Jan 14, 2010 ### _Andreas I'm not exactly having my finest moment as an aspiring physicist here. I blame it on simply rushing through the calculations since I'd really like to be doing something else for a little while. You're right. I didn't get any [itex]\frac{z'}{r^2}$ terms, though; the expansions I did were incorrect. I did have to use the approximation [itex]s\ll \frac{\omega}{c}[/tex] in one of the expansions, though.

Thanks again!