Deriving Larmor frequency problem

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SUMMARY

The discussion centers on the derivation of the Larmor frequency, specifically addressing the behavior of the magnetic moment vector ##\vec{\mu}## in relation to the magnetic field vector ##\vec{B}##. It is established that when ##\vec{\mu}## and ##\vec{B}## are not aligned, the time derivative ##\frac{d\mu}{dt}## is perpendicular to both vectors, resulting in circular motion of ##\vec{\mu}##. The reasoning provided illustrates that continuous addition of infinitesimal vectors perpendicular to both ##\vec{\mu}## and ##\vec{B}## leads to a circular trajectory, confirming the assumptions made in the derivation.

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fisher garry
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I don't get how they get Eq. 5. Why is the direction of ##\mu## going outwards from the direction of B? And why does the fact that ##\frac{d\mu}{dt}## is perpendicular to both ##\mu## and ##B## mean that ##\mu## goes in circle?
 
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All that they are assuming is that ##\vec{\mu}## and ##\vec{B}## are not aligned. If they are aligned, then the equation does not apply, since then the cross product is zero, so that there is no change in ##\vec{\mu}##. But they are assuming that they are not aligned, and that the equation does apply.

As for why it goes in a circle... Start with two vectors. Call them ##\vec{\mu}## and ##\vec{B}##. Now draw a third, infinitesimal, vector perpendicular to both ##\vec{\mu}## and ##\vec{B}##, call this third vector ##\vec{\delta}##. Form the sum ##\vec{\mu} + \vec{\delta} = \vec{\mu '}##. Now draw another infinitesimal vector perpendicular to ##\vec{\mu '}## and ##\vec{B}##. Add that one to ##\vec{\mu '}## to form ##\vec{\mu ''}##. Keep doing that forever. You'll find that you've drawn a circle.
 
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