Vector potential due to oscillating dipole

You will need to keep the first order terms in your expansions, and use your second approximation ([itex]s\ll \frac{\omega}{c}[/tex] )to later rid yourself of [itex]\frac{z'}{r^2}[/tex] (this is, after all, what allows you to ignore the second order terms in the expansion of 1/R).In summary, the problem involves calculating the magnetic vector potential A at a point p located at a distance r from the axis of an oscillating dipole of length s. The equation for A involves an integral with an exponential term and a constant, where the
  • #1
_Andreas
144
1

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 [tex]r\gg s[/tex] and that the current is the same throughout s.

Homework Equations



[tex]r=\sqrt{(x^2+(z-z')^2)},[/tex] 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.

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

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

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:

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

(d is a constant.)
 
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  • #2
_Andreas said:
[tex]r=\sqrt{(x^2+(z-z')^2)},[/tex] 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 [itex]r[/itex] as the distance from the center of the dipole ([itex]z'=0[/itex] I presume) to the point [itex]p[/itex]...you'll want to use a different letter, like say, (capital) [itex]R[/itex] to represent the separation of the source point and field point:

[tex]r=\sqrt{x^2+z^2}[/tex]

[tex]R=\sqrt{(x^2+(z-z')^2)}[/tex]

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

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

It seems like you are using [itex]c[/itex] 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 [itex]t[/itex] in this expression?

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

[tex]\textbf{A}(\textbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{I(t-\frac{R}{c})}{R}dz'\mathbf{\hat{z}}[/tex]
 
  • #3
Wow, that's a lot of annoying mistakes I did.

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

[tex]r=\sqrt{x^2+z^2}[/tex]

[tex]R=\sqrt{(x^2+(z-z')^2)}[/tex]

Yes, my bad.

gabbagabbahey said:
It seems like you are using [itex]c[/itex] 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.

gabbagabbahey said:
Also, the current in an oscillating dipole varies with time, so why is there no [itex]t[/itex] in this expression?

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

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

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

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

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

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

I still don't know how to calculate the integral, though.
 
  • #4
_Andreas said:
Yes. In my case it's (yes, I even forgot the minus sign in the exponential)

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

where R is as you said, and [tex]I_0[/tex] 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 [itex]\omega/c[/itex] instead of [itex]k[/itex] here; as you'll want to use the approximation that [itex]s\ll \frac{\omega}{c}[/itex] at some point (basically that the dipole is very small compared to the wavelength of the radiation it produces)

You also are given that [itex]s\ll r[/itex], and since [itex]|z'|\leq s[/itex], you know [itex]z'\ll r[/itex]...in order for this knowledge to be useful, you'll want to use the law of cosines to express [itex]R[/itex] in terms of [itex]z'[/itex], [itex]r[/itex] and the angle between the source and field point vectors (As measured from the center of the dipole). You can then Taylor expand both [itex]1/R[/itex] and [itex]\text{exp}(-i\omega R/c)[/itex] for small [itex]z'/r[/itex].
 
  • #5
Thank you very much!

Doing as you say I get

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

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

The integral then simply becomes

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

since [tex]r[/tex] is independent of [tex]z'[/tex]. This gives me the correct answer.
 
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  • #6
_Andreas said:
Expanding [tex]1/R[/tex] and [tex]e^{-ikR}[/tex] around [tex]z'/r=0[/tex] yields [tex]1/r[/tex] and [tex]e^{-ikr}[/tex], respectively (the first order terms of both expansions contain factors [tex]z'/r^2[/tex] and therefore become insignificant).

Careful, you can definitely say terms with [itex]\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}[/itex] is negligible...As an example, if [itex]s=10^{-6}[/itex] and [itex]r=10^{-4}[/itex], then [itex]|z'|\leq s\ll r[/itex] as demanded by the problem statement, but [itex]\frac{z'}{r^2}\leq\frac{s}{r^2}=100[/itex] won't be negligible.

You will need to keep the first order terms in your expansions, and use your second approximation ([itex]s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them.
 
  • #7
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. :cry:

gabbagabbahey said:
Careful, you can definitely say terms with [itex]\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}[/itex] is negligible...As an example, if [itex]s=10^{-6}[/itex] and [itex]r=10^{-4}[/itex], then [itex]|z'|\leq s\ll r[/itex] as demanded by the problem statement, but [itex]\frac{z'}{r^2}\leq\frac{s}{r^2}=100[/itex] won't be negligible.

You will need to keep the first order terms in your expansions, and use your second approximation ([itex]s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them.

You're right. I didn't get any [itex]\frac{z'}{r^2}[/itex] 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!
 

1. What is a vector potential due to an oscillating dipole?

The vector potential due to an oscillating dipole refers to the mathematical representation of the magnetic field produced by a dipole that is undergoing oscillatory motion. It is a vector quantity that describes the direction and magnitude of the magnetic field at any given point in space.

2. How is the vector potential related to the dipole moment?

The vector potential is directly proportional to the dipole moment of the oscillating dipole. This means that an increase in the dipole moment will result in a corresponding increase in the vector potential and therefore, the magnetic field strength.

3. What is the equation for calculating the vector potential due to an oscillating dipole?

The equation for calculating the vector potential due to an oscillating dipole is given by A = (μ₀/4πr) * (cosθ/r - i sinθ/r), where A is the vector potential, μ₀ is the permeability of free space, r is the distance from the dipole, and θ is the angle between the dipole moment and the direction of observation.

4. How does the vector potential change with distance from the dipole?

The vector potential decreases with distance from the dipole, following an inverse square law. This means that as the distance from the dipole increases, the vector potential and magnetic field strength decrease accordingly.

5. Can the vector potential due to an oscillating dipole be measured?

Yes, the vector potential due to an oscillating dipole can be measured experimentally using magnetic field sensors or instruments such as a magnetometer. These measurements can provide valuable insights into the behavior and properties of the dipole and its surrounding magnetic field.

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