# Showing the force on a magnetic dipole

• rebc
In summary: This demonstrates that the magnetization ## \vec{M} ## and the magnetic field ## \vec{B} ## vary through space and the force on the magnet is directly proportional to the gradient of the magnetic field. So when the magnetic field is uniform, the gradient of the magnetic field is zero and there is no net force on the magnet (but there might be a torque). For a permanent magnet in a non-uniform magnetic field, there is a force on the magnet that is proportional to the gradient of the magnetic field.Let me edit the last equation to show the ## \vec{x} ##-dependence: ## d\vec{F}=(\vec{M}(\vec{x}) \cdot
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 $$\vec{F} = \vec{m} \times (\nabla \times \vec{B}) +\vec{B} \times(\nabla \times \vec{m}) + (\vec{m} \cdot \nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{m}$$

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rebc said:

## 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} ##.

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

rebc said:
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.

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rebc
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} ##.

rebc

## What is a magnetic dipole?

A magnetic dipole is a fundamental concept in magnetism that represents the smallest unit of magnetic moment. It is a small magnet with a north and south pole, similar to a bar magnet.

## How is the force on a magnetic dipole shown?

The force on a magnetic dipole can be shown through the use of a magnetic field. When a magnetic dipole is placed in a magnetic field, it experiences a force that aligns its north and south poles with the direction of the field. This can be demonstrated through experiments or visual representations.

## What factors affect the force on a magnetic dipole?

The force on a magnetic dipole is affected by the strength of the magnetic field, the orientation of the dipole with respect to the field, and the distance between the dipole and the field source. Additionally, the magnetic properties of the material of the dipole can also impact the force.

## Can the force on a magnetic dipole be calculated?

Yes, the force on a magnetic dipole can be calculated using the equation F=μ∇(B⃗), where F is the force, μ is the magnetic moment of the dipole, and ∇(B⃗) is the gradient of the magnetic field. This equation takes into account the factors that affect the force on a dipole.

## What are some real-life applications of showing the force on a magnetic dipole?

The understanding of the force on a magnetic dipole is crucial in many real-life applications, such as in the design of electric motors and generators, magnetic levitation systems, and magnetic resonance imaging (MRI) machines. It is also used in compasses and navigation systems to detect and measure magnetic fields.

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