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Showing the force on a magnetic dipole

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    $$\vec{F} = (\vec{m} \nabla) \vec{B} $$

    3. 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 $$\vec{F} = \vec{m} \times (\nabla \times \vec{B}) +\vec{B} \times(\nabla \times \vec{m}) + (\vec{m} \cdot \nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{m}$$
     
    Last edited: Dec 12, 2016
  2. jcsd
  3. Dec 12, 2016 #2

    Charles Link

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    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} ##.
     
    Last edited: Dec 12, 2016
  4. Dec 12, 2016 #3
    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, $$\vec{F}=(\vec{m}\cdot\nabla)\vec{B}$$
     
  5. Dec 13, 2016 #4

    Charles Link

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    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.
     
    Last edited: Dec 13, 2016
  6. Dec 13, 2016 #5

    Charles Link

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    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} ##.
     
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