Physics Asked by king_geedorah on August 3, 2021
I was studying recently and noticed that the torque on a magnetic moment due to a field is defined as $tau = mu times B$. Classically, as far as I’m aware, the magnetic moment is given by $gamma L$ where $gamma$ is the gyromagnetic ratio (which is classically equal to $frac{q}{2m}$, so that we have:
$$tau = frac{q}{2m} L times B = frac{q}{2m} (r times mv) times B = frac{q}{2} (r times v) times B$$
By the Lorentz force, we also have a force $F = q(v times B)$. Therefore, we have a torque:
$$tau_{Lorentz} = q r times (v times B)$$
However, since the cross product is not associative, these two torques are not equal (even disregarding the constant). Is there any relationship between these two torques? And if I were to properly model the equations of motion of a moving charge in a magnetic field, would I need to consider both of these?
If we have an charge $q$ travelling in a loop of radius $r$ and velocity $v$, then it generates current:
$$I= frac{ev}{2 pi r}$$
We also know that the torque on this loop is given by :
$$tau= I (A times B)$$
Where $A$ is a vector with size proportional to the area of the loop, facing positive perpendicular w.r.t the normal of the loop and the current convention (right hand rule). This is a trivial result the explanation for which can be found many places.
We also define the magnetic moment $mu$ such that $tau = mu times B$
Therefore we can say: $$mu = IA= frac{ev}{2 pi r} pi r^2 = frac{evr}{2} = frac{e}{2m}mvr = gamma L$$
Notice how we are $textbf{not}$ talking about the charge itself but about the loop the charge goes around, if you consider simply that the direction of the electron is perpendicular to the normal vector of the loop it goes around you should see the discrepancy. Think about how magnetic fields cannot do work.
Answered by xXx_69_SWAG_69_xXx on August 3, 2021
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