Understanding the Force on a Magnetic Dipole in Different Orientations

In summary, the problem is solved by using the Spherical to Cartesian conversion and obtaining the force between the dipoles as expected.
  • #1
Mr_Allod
42
16
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##?
 
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  • #2
You have the correct expression for ##\vec B## and you have ##\vec m##.
Mr_Allod said:
After much rearranging and trig
Why make things difficult (when, with a little effort, you can make them nigh on impossible :wink: ) ? 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.

Mr_Allod said:
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 !
 
  • #3
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.
 
  • #4
BvU said:
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 :wink: ) ? 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?
 
  • #5
$$\hat x = \sin \theta \cos \phi\; \hat r + \cos \theta \cos \phi\; \hat \theta - \sin \theta \; \hat \phi$$ where does that come from ? Surely you want to pick the orientation such that ##\hat x = \hat r## ?

Anyway,
Mr_Allod said:
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.
 
  • #6
BvU said:
$$\hat x = \sin \theta \cos \phi\; \hat r + \cos \theta \cos \phi\; \hat \theta - \sin \theta \; \hat \phi$$ 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.
 
  • #7
Mr_Allod said:
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
Mr_Allod said:
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
BvU said:
Certainly not ! the r dependence has a different exponent !
because it's all ##r^{-3}## -- lesson learned !

##\ ##
 
  • #8
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
BvU said:
" 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.
 

FAQ: Understanding the Force on a Magnetic Dipole in Different Orientations

1. What is a magnetic dipole?

A magnetic dipole is a small magnet that has two poles, a north pole and a south pole. These poles create a magnetic field around the magnet.

2. What is the force on a magnetic dipole?

The force on a magnetic dipole is the force exerted on the magnet by an external magnetic field. This force is dependent on the strength of the external magnetic field and the orientation of the dipole relative to the field.

3. How is the force on a magnetic dipole calculated?

The force on a magnetic dipole can be calculated using the formula F = m x B, where F is the force, m is the magnetic dipole moment, and B is the external magnetic field. The direction of the force is perpendicular to both the dipole moment and the magnetic field.

4. What is the torque on a magnetic dipole?

The torque on a magnetic dipole is the rotational force exerted on the magnet by an external magnetic field. It is calculated using the formula τ = m x B, where τ is the torque, m is the magnetic dipole moment, and B is the external magnetic field. The direction of the torque is perpendicular to both the dipole moment and the magnetic field.

5. How does the force on a magnetic dipole change with distance?

The force on a magnetic dipole decreases with distance from the external magnetic field. This is due to the inverse-square law, which states that the force is inversely proportional to the square of the distance between the dipole and the field. As the distance increases, the force decreases exponentially.

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