Showing the force on a magnetic dipole

[rebc]

Homework Statement

We're given a situation where the circuit can move under the influence of a magnetic force. Now I need to show that the force on the magnetic dipole is

Homework Equations

$$\vec{F} = (\vec{m} \nabla) \vec{B} $$

The Attempt at a Solution

Could I start from a given Force for a loop with a dipole moment in the B-field, where
$$\vec{F} = \nabla (\vec{m} \cdot \vec{B})$$
and apply the identity $$\vec{F} = \vec{m} \times (\nabla \times \vec{B}) +\vec{B} \times(\nabla \times \vec{m}) + (\vec{m} \cdot \nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{m}$$

 

[Charles Link]

Homework Statement

We're given a situation where the circuit can move under the influence of a magnetic force. Now I need to show that the force on the magnetic dipole is

Homework Equations

$$\vec{F} = (\vec{m} \nabla) \vec{B} $$

The Attempt at a Solution

Could I start from a given Force for a loop with a dipole moment in the B-field, where
$$\vec{F} = \nabla (\vec{m} \cdot \vec{B})$$
and apply the identity $$\vec{F} = \vec{m} \times (\nabla \times \vec{B}) +\vec{B} \times(\nabla \times \vec{m}) + (\vec{m} \cdot \nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{m}$$

I think your approach is more accurate than the problem statement which needs a dot product in it. And yes, you got the vector identity correct for the gradient of a dot product. $\vec{m}$ is a constant so vector derivative operations on it are zero. Also $\nabla \times \vec{B} =0$ for the steady state where no currents are present. (Comes from Maxwell's $\nabla \times \vec{B}=\mu_o \vec{J}+\mu_o \epsilon_o \dot{\vec{E}}$ . Any currents from $\vec{m}$ don't count as part of $\vec{J}$ because $\vec{B}$ is the external field applied to the $\vec{m}$.) Thereby, you have successfully showed the necessary result which is $\vec{F}=(\vec{m} \cdot \nabla ) \vec{B}$.

 

[rebc]

I think your approach is more accurate than the problem statement which needs a dot product in it. And yes, you got the vector identity correct for the gradient of a dot product. $\vec{m}$ is a constant so vector derivative operations on it are zero. Also $\nabla \times \vec{B} =0$ for the steady state where no currents are present. (Comes from Maxwell's $\nabla \times \vec{B}=\mu_o \vec{J}+\mu_o \epsilon_o \dot{\vec{E}}$ . Any currents from $\vec{m}$ don't count as part of $\vec{J}$ because $\vec{B}$ is the external field applied to the $\vec{m}$.) Thereby, you have successfully showed the necessary result which is $\vec{F}=(\vec{m} \cdot \nabla ) \vec{B}$.

The problem statement does not necessarily say though that the circuit is a loop. The first equation explicitly states that it applies to a loop in a magnetic dipole. Can I still apply such equation for some circuit where a circuit element of it is under some magnetic field?
And thank you, and yes, $$\vec{F}=(\vec{m}\cdot\nabla)\vec{B}$$

 

[Charles Link]

The problem statement does not necessarily say though that the circuit is a loop. The first equation explicitly states that it applies to a loop in a magnetic dipole. Can I still apply such equation for some circuit where a circuit element of it is under some magnetic field?
And thank you, and yes, $$\vec{F}=(\vec{m}\cdot\nabla)\vec{B}$$

A magnetic dipole is a loop of current. $\vec{m}=I \vec{A}$ where $I$ is the current and $\vec{A}$ is the area (points normal to the loop), sometimes with an extra constant factor depending upon the system of units. The equation will also work for a loop of current in a circuit, but works better for smaller loops. If the field changes too quickly across the loop/electrical circuit, it's more accurate to then compute the force by integrating $d\vec{F} =I \, d\vec{l} \times \vec{B}$ around the loop. $\\$ The equation also applies to the force on a permanent magnet which is a large (sometimes nearly uniform) distribution of magnetic dipoles throughout the material. The magnetization $\vec{M}$ is the density of magnetic dipoles per unit volume. As long as the magnetization $\vec{M}$ is unaffected (approximately) by the applied field, you can use it to compute the force on a permanent magnet in a magnetic field. In this case you need to sum/integrate the forces $d \vec{F}=(\vec{M} \cdot \nabla) \vec{B} \, d^3x$ over the volume of the magnet.

 

[Charles Link]

A magnetic dipole is a loop of current. $\vec{m}=I \vec{A}$ where $I$ is the current and $\vec{A}$ is the area (points normal to the loop), sometimes with an extra constant factor depending upon the system of units. The equation will also work for a loop of current in a circuit, but works better for smaller loops. If the field changes too quickly across the loop/electrical circuit, it's more accurate to then compute the force by integrating $d\vec{F} =I \, d\vec{l} \times \vec{B}$ around the loop. $\\$ The equation also applies to the force on a permanent magnet which is a large (sometimes nearly uniform) distribution of magnetic dipoles throughout the material. The magnetization $\vec{M}$ is the density of magnetic dipoles per unit volume. As long as the magnetization $\vec{M}$ is unaffected (approximately) by the applied field, you can use it to compute the force on a permanent magnet in a magnetic field. In this case you need to sum/integrate the forces $d \vec{F}=(\vec{M} \cdot \nabla) \vec{B} \, d^3x$ over the volume of the magnet.

Let me edit the last equation to show the $\vec{x}$-dependence: $d\vec{F}=(\vec{M}(\vec{x}) \cdot \nabla )\vec{B}(\vec{x}) \, d^3 \vec{x}$.