This question describes a charged particle (with charge $q$ and mass $m$) moving with an initial velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$, where the velocity is perpendicular to the magnetic field. The magnetic force acting on the particle causes it to move in a circular path, and the radius of this path is denoted as $R$.

The formula for the radius $R$ of the circular path of a charged particle moving perpendicular to a magnetic field is:

$R = \frac{mv}{|q|B}$

This equation is derived from the Lorentz force, which is responsible for the circular motion of the particle, and is given by the magnetic force:

$F = qvB$

Since the magnetic force acts as a centripetal force, we can equate it to the centripetal force formula $F = \frac{mv^2}{R}$, and solve for $R$:

$qvB = \frac{mv^2}{R} \quad \Rightarrow \quad R = \frac{mv}{|q|B}$

Thus, the correct answer is C:

$R = \frac{mv}{|q|B}$

Would you like further details or have any questions about this concept?

Here are some related questions for you:

1. What is the role of the Lorentz force in the motion of a charged particle?
2. How does the direction of the velocity affect the radius of the path?
3. What happens to the radius if the charge is doubled?
4. How does the magnetic field strength influence the radius of the circular path?
5. What is the equation for the period of the particle's motion?

Tip: To better understand circular motion in magnetic fields, it's helpful to practice deriving the centripetal force and equating it to the magnetic force. This will reinforce your understanding of how magnetic fields influence charged particle trajectories.