1
$\begingroup$

Suppose I have a magnetic dipole oriented vertically in the $\hat{z}$ direction so

$$ \vec{m} = m\hat{z}$$

And the position vector to it is $\vec{r} = z\hat{z}$. It's entirely on the z-axis.

We're using cylindrical coordinates and I have a ring-like current loop with radius $a$ at the origin and the current ($I$) is going in the $\hat{\phi}$ and obviously the magnetic dipole is right above the center of it.

I calculated the magnetic field of the current loop using Bio-Savart's Law and got

$$\vec{B}_{loop}(\vec{r} = z\hat{z}) = \frac{\mu_o Ia^2}{2\pi(a^2 + z^2)^{3/2}}$$

Thus the force that the dipole feels should be $$\vec{F}_{dipole} = \nabla(\vec{m} \cdot \vec{B}_{loop})$$

But then I get that $\vec{F}_{dipole} < 0$ so it's pointing downwards towards the loop. Since the dipole and magnetic field are parallel to each other and aligned, shouldn't the force be upwards? Hence, a positive value?

Improve this question
$\endgroup$
1
  • $\begingroup$ Possibly helpful. It is helpful to think of a dipole aligned with the local field as a “strong-field seeker,” and a dipole anti-aligned with the local field as a “weak-field seeker.” $\endgroup$
    – rob ♦
    Commented Nov 25, 2021 at 15:46

1 Answer 1

Reset to default
2
$\begingroup$

You can view magnetic dipole just as yet another current loop which is parallel to the existing one, with the currents being in the same direction. Thus you basically have two parallel conductors and you know (hopefully) that there is always attractive force between them when currents are parallel.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.