Vector Potential of a Rotating Mangetic Dipole

[allison west]

Homework Statement

A rotating magnetic dipole is built by two oscillating magnetic dipole moments, one along the y-axis and one along the x-axis. Find the vector potential at a point: (0, 0, $z_0$) along the z-axis. Then find the magnetic field at $z_0$ . As the magnetic field is a function of time, determine the induced Electric field from Faraday's laws. The magnetic dipole moment is given as a function of time as ($m_0$ is a constant)
$\vec m$ = $m_0 [cos(\omega t) \hat x + sin(\omega t) \hat y]$

Homework Equations

[/B]
$\vec A_{dip} (\vec r) = \frac {\mu_0} {4 \pi} \frac 1 {r^3} \vec m \times \vec r$

$\vec B = \vec \nabla \times \vec A$

Induced electric field, Faraday's law:

$\vec E = \frac 1 {4 \pi}$ $\int$ $[({\frac {\partial \vec B} {\partial t}} {\times \mathcal {\hat r}})$ $\frac 1 {\mathcal {r^2}}]$

The Attempt at a Solution

Since I am given the magnetic dipole moment, I attempted to find the vector potential by taking the cross product of $\vec m$ with $\vec r$. I took $\vec r$ to be: ${<x \hat x, y \hat y, z \hat z>}$ which
After taking the cross product I got:

$\vec m \times \vec r = m_0 z sin(\omega t) \hat x - m_0 z cos(\omega t) \hat y + (m_0 cos(\omega t) y - m_0 sin(\omega t) x) \hat z$

Since I'm taking the potential at the point (0, 0, $z_0$) I think I put that into the cross product which would mean my vector potential would be equal to:

$\vec A_{dip} (\vec r) = (\frac { \mu_0 m_0 z_0} {4 \pi r^3})(sin(\omega t) \hat x - cos(\omega t) \hat y)$

I'm not entirely sure that this is right but when I try to take the cross product of $\vec A_{dip} (\vec r)$ to find the magnetic field, I came up with:

$\vec \nabla \times \vec A_{dip} =(\frac { \mu_0 m_0 z_0} {4 \pi r^3}) (2cos(\omega t) \hat x + 2sin(\omega t) \hat y)$

which has no dependence on the z coordinate. I'm wondering if I maybe approached this problem wrong? Help would be much appreciated thank you!

 

[kuruman]

... which has no dependence on the z coordinate.

I am not sure what you mean. Isn't there a z₀ in your expression? It looks like you have a 1/z₀² dependence because of the r³ in the denominator.

 

[allison west]

I am not sure what you mean. Isn't there a z₀ in your expression? It looks like you have a 1/z₀² dependence because of the r³ in the denominator.

Okay that seems to be what I was missing, I was not inputting $z_0$ back into my definition for $\vec r$. By putting that into the equation I got:

$\vec A_{dip} (\vec r) = (\frac { \mu_0 m_0 } {4 \pi z_0^2})(sin(\omega t) \hat x - cos(\omega t) \hat y)$

and

$\vec B = \vec \nabla \times \vec A$

$\vec B = (\frac { \mu_0 m_0 } {2 \pi z_0^3})(cos(\omega t) \hat x + sin(\omega t) \hat y)$

I believe this makes more sense.

To find the induced Electric field I have to take the time derivative of the magnetic field and cross it with $\vec R$ (the separation vector) I use a capitol R to distinguish it from the other r vectors used above.

$\vec E = \frac 1 {4 \pi}$ $\int$ $[({\frac {\partial \vec B} {\partial t}} {\times \mathcal {\vec R}})$ $\frac 1 {\mathcal {r^2}}]$

$\vec R = \vec r - \vec r'$

$\vec R = < (x-x') \hat x, (y-y') \hat y, (z-z') \hat z >$

I'm fairly sure that (x, y, z) is equal to (0, 0, $z_0$) the field point, which makes (x', y', z') the source point?

​

 

[Delta2]

Can't we simply use the equation $E=-\nabla V-\frac{\partial A}{\partial t}$ for this problem V seems to be zero or spatially constant since there are no free charge densities.

 

[kuruman]

Can't we simply use the equation ...

That should work.

 

[allison west]

That should work.

Thanks for the help!