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The force on a small element (of length dl) of a current carrying wire, place in a magnetic field B can be calculated using the following equation (which is simply an application of the Lorentz force equation):

d F = i dl X B

My question is, how does one calculate the force acting on a small magnetized volume d V (say magnetized in the z direction with magnetization M or remanent flux density Br) when the same is placed in a magnetic field B.

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  • $\begingroup$ I would say, this is kind of a tedious algebra which has been worked out in details in Julian Schwinger's book, Classical Electrodynamics Chapter 4-Macroscopic Electrodynamics Section 4.2. Here, first Schwinger has discussed the force on an atom due to an external electromagnetic field. Then, to find the force on a finite volume element he has integrated that force over the total volume with $n(\mathbf{r})$ as the number density of the molecules. You also need to define quantities like magnetic/electric dipole moment for the macroscopic body $\endgroup$ Jan 30, 2012 at 19:58

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Interaction energy is $U=-\int \vec{m(x)} * \vec{B(x)}d V$. You get the force by taking the negative gradient of this, $\vec{F} = -\nabla U$. The gradient is with respect to rigid displacements of the magnetized object, with the external magnetic field held fixed.

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  • $\begingroup$ Yes, the correct answer to leading approximation is the derivative of $m \cdot B$, where m is the magnetic moment. $\endgroup$
    – Ron Maimon
    May 2, 2012 at 14:59
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A bar magnet, current loop, electron etc has a magnetic moment vector associated with it. When placed in a magnetic field, the torque $\tau$ tending to align the magnetic moment $m$ with the applied magnetic field $B$ is given by $\tau = m\times B$

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  • $\begingroup$ This is a torque, not a force, -1. Please stop supplying ridiculous wrong answers to questions. The correct answer is below. How did this get upvoted and accepted? $\endgroup$
    – Ron Maimon
    May 2, 2012 at 14:58

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