# Force on Magnetic Dipole - Equivalent Formulas?

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. $$x$$.

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.

• 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: $$B_x=\frac{\mu_0 I R}{4\pi}\int_0^{2\pi}\frac{R-y \sin{\phi '}}{(R^2+y^2+x^2-2yR\sin{\phi '})^{1.5}}d\phi '$$ $$B_y=\frac{\mu_0 I R x}{4\pi}\int_0^{2\pi}\frac{\sin{\phi '}}{(R^2+y^2+x^2-2yR\sin{\phi '})^{1.5}}d\phi '$$ May 6, 2019 at 17:43
• 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/…> May 6, 2019 at 17:50
• Can you please elaborate on what you are trying to say? It is not very clear to me May 6, 2019 at 17:52
• 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. May 6, 2019 at 17:55
• 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 $\frac{\partial B_x}{\partial y} = \frac{\partial B_y}{\partial x}$ May 6, 2019 at 18:08