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I have seen two different ways of expressing the magnetic force on a dipole moment in a non-uniform magnetic field:
$\vec{F_a}=(\vec{m}\cdot\nabla)\vec{B}$
or
$\vec{F_b}=\nabla(\vec{m}\cdot \vec{B})$

Are these two formulas equivallent? When I expand them, I get different results. For example, if we assume $\vec{m}=\begin{bmatrix}m_x\\m_y\end{bmatrix}$, $\vec{B}=\begin{bmatrix}B_x(x,y)\\B_y(x,y)\end{bmatrix}$, $\nabla=\begin{bmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\end{bmatrix}$, the forces are:
$\vec{F_a}= (\vec{m}\cdot\nabla)\vec{B}= (m_x \frac{\partial}{\partial x} + m_y \frac{\partial}{\partial y}) \begin{bmatrix}B_x(x,y)\\B_y(x,y)\end{bmatrix}= \begin{bmatrix}(m_x \frac{\partial }{\partial x} B_x(x,y)+ m_y \frac{\partial }{\partial y}B_x(x,y))\\ (m_x \frac{\partial }{\partial x} B_y(x,y)+ m_y \frac{\partial }{\partial y}B_y(x,y))\end{bmatrix}$

$\vec{F_b}= \nabla(\vec{m}\cdot \vec{B})= \begin{bmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\end{bmatrix}(m_x B_x(x,y) + m_y B_y(x,y)) = \begin{bmatrix}(m_x \frac{\partial }{\partial x} B_x(x,y)+ m_y \frac{\partial }{\partial x}B_y(x,y))\\ (m_x \frac{\partial }{\partial y} B_x(x,y)+ m_y \frac{\partial }{\partial y}B_y(x,y))\end{bmatrix}$

These are not equivallent, for example, in the $x$ direction, $\vec{F_a}$ has derivatives by $x$ and by $y$ of $B_x$, but $\vec{F_b}$ has derivatives by $x$ only of both $B_x, B_y$.

Is my understanding of the original formulas incorrect? How is it that these are both used?

Note that here $B_x$ is the $x$ component of $\vec{B}$, not its partial derivative w.r.t. $x$.

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1 Answer 1

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We have, $\nabla(\vec{m}\cdot \vec{B}) = (\vec{m}\cdot\nabla)\vec{B} +(\vec{B}\cdot \nabla)\vec{m} + \vec{m}\times(\nabla\times \vec{B})+ \vec{B}\times (\nabla \times \vec{m})$

When $\vec{m}$ is constant, this reduces to $(\vec{m}\cdot\nabla)\vec{B} + \vec{m}\times(\nabla\times \vec{B})$.

Applying Ampere-Maxwells' law, this is $ (\vec{m}\cdot\nabla)\vec{B} + \vec{m}\times(\mu_0\vec{j}+\mu_0 \epsilon_0\frac{\partial \vec{E}}{\partial t})$.

So in free space, in steady state, the second term is zero, and then both are equal.

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  • $\begingroup$ I still get different results if I calculate using the example: The magnetic field at a point $(x,y,0)$ generated by single loop of wire with radius $R$ placed on the yz plane where the coil center is placed at $(0,0,0)$. From equation, we see that the contribution of this coil loop to the magnetic field is: \begin{equation} B_x=\frac{\mu_0 I R}{4\pi}\int_0^{2\pi}\frac{R-y \sin{\phi '}}{(R^2+y^2+x^2-2yR\sin{\phi '})^{1.5}}d\phi ' \end{equation} \begin{equation} B_y=\frac{\mu_0 I R x}{4\pi}\int_0^{2\pi}\frac{\sin{\phi '}}{(R^2+y^2+x^2-2yR\sin{\phi '})^{1.5}}d\phi ' \end{equation} $\endgroup$
    – Ariella
    May 6, 2019 at 17:43
  • $\begingroup$ Unless I did not understand the way in which you mean that I should calculate $(\vec{m}\cdot\nabla)\vec{B}$ or $\nabla(\vec{m}\cdot\vec{B})$. P.S. Source for magnetic field example in my previous comment is from appendix of: <ocw.mit.edu/courses/physics/…> $\endgroup$
    – Ariella
    May 6, 2019 at 17:50
  • $\begingroup$ Can you please elaborate on what you are trying to say? It is not very clear to me $\endgroup$ May 6, 2019 at 17:52
  • $\begingroup$ If you take $(\vec{m}\cdot\nabla)\vec{B}$ or $\nabla(\vec{m}\cdot\vec{B})$ using the $\vec{B}$ specified in the first comment (which comes from a situation where we are in free space in steady state), the resultant force is not equal from the 2 ways of calculating. $\endgroup$
    – Ariella
    May 6, 2019 at 17:55
  • $\begingroup$ I have not done the calculation, but they must be equal at any point other than on the current loop, because in free space, for static fields,$ \nabla \times \vec {B} = 0$, from Maxwell's equations, and therefore $\frac{\partial B_x}{\partial y} = \frac{\partial B_y}{\partial x}$ $\endgroup$ May 6, 2019 at 18:08

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