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I want to find an expression for the force acting upon a magnetic dipole with dipole moment $\mathbf{m}$ if that dipole is positioned in a stationary, external magnetic field $\mathbf{B}$. The expression given for the force is the following (assuming that $\nabla \times\mathbf{B}=0$):

$$\mathbf{F}=(\mathbf{m}\cdot\nabla)\mathbf{B}\quad(1)$$

My question is mostly whether the expression above is equivalent to:

$$\begin{bmatrix} \frac{\partial \mathbf{B}}{\partial x} & \frac{\partial \mathbf{B}}{\partial y}& \frac{\partial \mathbf{B}}{\partial z} \end{bmatrix}\mathbf{m} \quad (2)$$

or equivalent to:

$$\begin{bmatrix} \frac{\partial \mathbf{B}}{\partial x} & \frac{\partial \mathbf{B}}{\partial y}& \frac{\partial \mathbf{B}}{\partial z} \end{bmatrix}^T\mathbf{m} \quad (3)$$

I basically found these two expressions ($(2)$ $(3)$) for the force from two different sources, so one of them must be wrong. I derived the first expression in the following way:

$$(\mathbf{m}\cdot\nabla)\mathbf{B}=(m_1\frac{\partial }{\partial x}+m_2\frac{\partial }{\partial y}+m_3\frac{\partial }{\partial z})\begin{bmatrix} B_1\\ B_2\\ B_3 \end{bmatrix}=\begin{bmatrix} m_1\frac{\partial B_1}{\partial x} + m_2\frac{\partial B_1}{\partial y} + m_3\frac{\partial B_1}{\partial z} \\ m_1\frac{\partial B_2}{\partial x} + m_2\frac{\partial B_2}{\partial y} + m_3\frac{\partial B_2}{\partial z} \\ m_1\frac{\partial B_3}{\partial x} + m_2\frac{\partial B_3}{\partial y} + m_3\frac{\partial B_3}{\partial z} \end{bmatrix}$$

The last expressions can be interpreted as the Matrix product $(2)$. Is that correct or am I missing something obvious?

Thanks!

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When in doubt, use coordinates and index notation; the expression

$$\mathbf{F}=(\mathbf{m}\cdot\nabla)\mathbf{B}~~~(*) $$

can be written in cartesian coordinates in this way:

$$ F_i = m_k \partial_k B_i $$

If you're wondering how do we know that the right-hand side of (*) expands this way, it is actually the definition of the shorthand $\mathbf m \cdot \nabla$.

If force coordinates are put into a row $\mathbf F^T$, then this row can be obtained as left multiplication of the matrix $\mathbf G$ with coordinates $G_{ki} = \partial_k B_i$ by the magnetic moment row $\mathbf m^T$:

$$ \mathbf F^T = \mathbf m^T \cdot \mathbf G. $$

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  • $\begingroup$ Thanks for your response but I think your result might have an incorrect dimension since $\mathbf{m}^T\mathbf{G} \in \mathbb{R}^{1,3}$ and $\mathbf{F}\in\mathbb{R}^{3,1}$. $\endgroup$ – Mantabit Feb 13 at 7:39
  • $\begingroup$ @Mantabit you are right, the force on the left-hand side also has to be written as a row, because $i$ is a column index. It is fixed now. $\endgroup$ – Ján Lalinský Feb 13 at 20:48
  • $\begingroup$ Okay thanks, I checked your solution and it agrees with mine. $\endgroup$ – Mantabit Feb 13 at 21:38

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