Force on a Magnetic Dipole

[Mr_Allod]

Homework Statement:

Use $\vec F = \nabla (\vec m \cdot \vec B)$ to find the force of attraction between two magnetic dipoles $m_1$ at (0,0,0) and $m_2$ at (r,0,0), both of which are pointing in the +x direction

Relevant Equations:

$\vec F_{21} = \nabla (\vec m_2 \cdot \vec B_1)$
$\vec m_2 = m_2\hat x$
$\vec B_1 = \mu_0 \frac {m_1}{4\pi r^2} (2\cos \theta \hat r + \sin \theta \hat \theta)$

Hi there, I approached this problem by making use of the fact that a dipole can be modeled as a small current loop with the magnetic field $\vec B_1 = \mu_0 \frac {m_1}{4\pi r^2} (2\cos \theta \hat r + \sin \theta \hat \theta)$ which is the far-field approximation for a regular circular current loop. After much rearranging and trig. identities I found the force $\vec F_{21}$ to be:
$$\vec F_{21} = \mu_0 \frac {m_1m_2} {4\pi r^4} (\sin 2\theta (\cos \phi + \frac 1 2 \sin \phi) \hat r + 2\cos 2\theta (\cos \phi + \frac 1 2 \sin \phi) \hat \phi + 2\cos \theta (\frac 1 2 \cos \phi - \sin \phi) \hat \phi) $$
Now I know that if both dipoles were on the z-axis as opposed to x the force should simplify to:
$$\vec F_{21} = -3\mu_0 \frac {m_1m_2} {2\pi r^4} \hat z$$
With this in mind would it be correct to assume that since the only difference is the axis on which the dipoles lie that the magnitude of the force would be the same? As in, in this problem it would point in $-\hat x$ instead of $-\hat z$?

 

[BvU]

You have the correct expression for $\vec B$ and you have $\vec m$.

After much rearranging and trig

Why make things difficult (when, with a little effort, you can make them nigh on impossible ) ? You bring in a $\phi$ you have no use for in the given configuration.

$\vec B$ and $\vec m$ align, so $\vec m\cdot \vec B = mB(x)$ and the result is straightforward! See also force between dipoles , where the $\nabla$ has been executed already.

With this in mind would it be correct to assume that since the only difference is the axis on which the dipoles lie that the magnitude of the force would be the same?

Certainly not ! the r dependence has a different exponent !

 

[Mr_Allod]

The $\phi$ dependence came in when I converted to $\vec m_2 = m_2 \hat x$ to spherical polar coordinates:
$\hat x = \sin \theta \cos \phi \hat r + \cos \theta \cos \phi \hat \theta - \sin \theta \hat \phi$

It seemed easier at the time to convert the magnetic moment to Spherical Polar than converting the magnetic field to Cartesian. I see your point though, since it is just the $\hat x$ that matters in this case I would have saved myself a lot of trouble by finding $B(x)$ instead.

 

[Mr_Allod]

You have the correct expression for $\vec B$ and you have $\vec m$.
Why make things difficult (when, with a little effort, you can make them nigh on impossible ) ? You bring in a $\phi$ you have no use for in the given configuration.

$\vec B$ and $\vec m$ align, so $\vec m\cdot \vec B = mB(x)$ and the result is straightforward! See also force between dipoles , where the $\nabla$ has been executed already.

Certainly not ! the r dependence has a different exponent !

I've tried the method you've suggested however I run into a problem during some of the simplifying. This is what I get:

Spherical to Cartesian conversion:
$\hat r = \sin \theta \cos \theta \hat x +\sin \theta \sin \phi \hat y + \cos \theta \hat z$
$\hat \theta = \cos \theta \cos \phi \hat x + \cos \theta \sin \phi \hat y - \sin \theta \hat z$

Then: $B(x) = \mu_0 \frac {m_1}{4\pi r^3} (2\cos \theta (\sin \theta \cos \phi) + \sin \theta (\cos \theta \cos \phi)) = 3\mu_0 \frac {m_1}{4\pi r^3} \sin \theta \cos \theta \cos \phi$

In Spherical Polar $x = r \sin \theta \cos \phi$

So $B(x) = 3x \mu_0 \frac {m_1}{4\pi r^4} \cos \theta$

But since on the positive x-axis $\theta = \frac {\pi} {2}$ that would make $B(x) = 0$! On the other hand using the equation from wikipedia I obtain:
$$
\vec F_{21} = -3\mu_0 \frac {m_1m_2} {2\pi r^4} \hat x
$$
as expected. Can you see something wrong with my procedure that would cause the discrepancy?

 

[BvU]

$$\hat x = \sin \theta \cos \phi\; \hat r + \cos \theta \cos \phi\; \hat \theta - \sin \theta \; \hat \phi$$ where does that come from ? Surely you want to pick the orientation such that $\hat x = \hat r$ ?

Anyway,

I've tried the method you've suggested

I certainly did not suggest using spherical coordinates. In Cartesian coordinates it is as simple as can be.

 

[Mr_Allod]

$$\hat x = \sin \theta \cos \phi\; \hat r + \cos \theta \cos \phi\; \hat \theta - \sin \theta \; \hat \phi$$ where does that come from ? Surely you want to pick the orientation such that $\hat x = \hat r$ ?

Anyway,
I certainly did not suggest using spherical coordinates. In Cartesian coordinates it is as simple as can be.

It's the relationship between Cartesian and Spherical unit vectors you can see the others in matrix notation here: Vector fields in cylindrical and spherical coordinates

Yes I understand cartesian is the way to go, however I was under the impression that I had to first convert $B_1$ into it's cartesian equivalent using these unit vector relations since by default it is given in spherical polar.

 

[BvU]

It's the relationship between Cartesian and Spherical unit vectors you can see the others in matrix notation here: Vector fields in cylindrical and spherical coordinates

Ah, but there $\theta$ is the angle between the positive z-axis and the vector in question, which is a bad choice for this exercise !

Yes I understand cartesian is the way to go, however I was under the impression that I had to first convert $B_1$ into it's cartesian equivalent using these unit vector relations since by default it is given in spherical polar.

Actually, $$\vec B(\vec r) = {\mu_0\over \pi} \Biggl[ {3\vec r (\vec m \cdot\vec r )\over r^5} - {\vec m\over r^3}\Biggr ] $$ is independent of the choice of coordinate system !

To boot: the link continues with
" Equivalently, if ${\displaystyle {\hat {r}} }$ is the unit vector in the direction of ${\displaystyle \vec {r} ,}$ ... " -- i.e. here the x-axis --
$$\vec B(\vec r) = {\mu_0\over \pi} \Biggl[ {3\hat r (\vec m \cdot\hat r )- \vec m\over r^3}\Biggr ] $$
And then contiues

" In spherical coordinates with the magnetic moment aligned with the z-axis, if we use ${\displaystyle {\hat {r}} \cos \theta - {\hat {z}} =\sin \theta {\; {\hat {\theta }}}}$, then this relation can be expressed as $$

{\displaystyle \vec {B} ({\vec {r} })={\frac {\mu _{0}|\vec {m} |}{4\pi r^{3}}}\left(2\cos \theta \,\vec {\hat {r}} +\sin \theta \,{ {\hat {\theta }}}\right).} \ "$$

but in the excercise the magnetic moment is aligned with the x-axis. You would have been all right with your

much rearranging and trig. identities

if you had simply moved everything on the z-axis and aligned with that. I.e. read "z" in the problem statement instead of "x" (and, in the final answer, write "x" instead of "z"),
then $\theta = 0$ leading to $\hat r = \hat z$ and a simple expression for $\vec B$ and hence $\vec F$.

Doing only one of the two wrongfooted you.

----------------------

Incidentally, I should correct my own

Certainly not ! the r dependence has a different exponent !

because it's all $r^{-3}$ -- lesson learned !

$\$

 

[Mr_Allod]

Ah I see so my original expression for $\vec B_1$ is only applicable for $\vec m_1 = m_1 \hat z$. Thank you for clarifying that. I would have thought that since

" In spherical coordinates with the magnetic moment aligned with the z-axis, if we use ${\displaystyle {\hat {r}} \cos \theta - {\hat {z}} =\sin \theta {\; {\hat {\theta }}}}$, then this relation can be expressed as
$$
{\displaystyle \vec {B} ({\vec {r} })={\frac {\mu _{0}|\vec {m} |}{4\pi r^{3}}}\left(2\cos \theta \,\vec {\hat {r}} +\sin \theta \,{ {\hat {\theta }}}\right).} \ "
$$

is a far field approximation of the magnetic field of a current loop it would be applicable regardless of the orientation of the magnetic moment but I guess not, good to know.