 #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 farfield 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 zaxis 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##?
$$\vec F_{21} = \mu_0 \frac {m_1m_2} {4\pi r^4} (\sin 2\theta (\cos \phi + \frac 1 2 \sin \phi) \hat r + 2\cos 2\theta (\cos \phi + \frac 1 2 \sin \phi) \hat \phi + 2\cos \theta (\frac 1 2 \cos \phi  \sin \phi) \hat \phi) $$
Now I know that if both dipoles were on the zaxis 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##?