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I had a question from the magnetic dipole thread that was posted earlier today, but it's a bit more mundane. The torque on a magnetic dipole, using

My question is, in the general case, how is the direction of the infinitesimal rotation determined, out of the two possible choices? I apologise if I am missing something obvious.

*a right handed cross product*is ##\vec{\tau} = \vec{\mu} \times \vec{B}##. The work done during a rotation is $$W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot d\vec{\theta} \times \vec{r} = \int \vec{r} \times \vec{F} \cdot d\vec{\theta} = \int \vec{\tau} \cdot d\vec{\theta}$$Now the result of this depends on which direction we take the vector ##d\vec{\theta}## to be in, and there is only one correct result for work! We know it has to be parallel to the axis of rotation, but get the right answer ##d\vec{\theta}## also has to be in the opposite direction to ##\vec{\tau}##, i.e. if ##\vec{\tau} = \tau \hat{z}## then ##d\vec{\theta} = -d\theta \hat{z}##.My question is, in the general case, how is the direction of the infinitesimal rotation determined, out of the two possible choices? I apologise if I am missing something obvious.

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