Solution
Physical Situation:
A bead has mass $M$, charge $q$, and spin angular momentum $\vec{L}$. It has an associated magnetic dipole moment $\vec{\mu}$. The bead enters a magnetic field $\vec{B}$ with velocity $\vec{v}$.
We are given two crucial conditions:
- The velocity vector $\vec{v}$ is in the direction of the angular momentum vector $\vec{L}$ (i.e., $\vec{v} \parallel \vec{L}$).
- During the motion, $\vec{v}$ and $\vec{L}$ remain parallel to each other.
Dynamics of Motion:
The particle moves in a magnetic field. The Lorentz force acts on the charge, causing the velocity vector $\vec{v}$ to rotate (circular motion). The torque due to the interaction of the magnetic dipole with the field acts on $\vec{L}$, causing it to precess.
1. Rotation of Velocity ($\vec{v}$):
The rate at which the velocity vector rotates in the magnetic field is the Cyclotron frequency (angular velocity of the particle in its orbit):
$$ \vec{\omega}_{cyclotron} = \frac{qB}{M} $$2. Precession of Angular Momentum ($\vec{L}$):
The torque on the dipole is given by $\vec{\tau} = \vec{\mu} \times \vec{B}$.
From rotational dynamics, torque is the rate of change of angular momentum:
$$ \vec{\tau} = \frac{d\vec{L}}{dt} = \vec{\omega}_{precession} \times \vec{L} $$Assuming the magnetic moment is parallel to the angular momentum, we can write $\vec{\mu} = \gamma \vec{L}$, where $\gamma$ is the gyromagnetic ratio. Thus:
$$ \vec{\tau} = \gamma \vec{L} \times \vec{B} $$Comparing magnitudes for the precession frequency:
$$ \omega_{precession} = \frac{\mu B}{L} $$Conclusion:
For $\vec{v}$ and $\vec{L}$ to remain parallel at all times, they must rotate at exactly the same rate. Therefore, the cyclotron frequency must equal the precession frequency:
$$ \omega_{cyclotron} = \omega_{precession} $$ $$ \frac{qB}{M} = \frac{\mu B}{L} $$Solving for the magnetic dipole moment $\mu$:
$$ \frac{q}{M} = \frac{\mu}{L} $$