Solution
Problem Analysis:
A bead of mass $m$ and charge $q$ is initially at the surface of a cylinder of radius $r$. It is projected radially away with velocity $v$. A magnetic field $B$ exists parallel to the cylinder axis.
The magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity, so the speed $v$ remains constant, and the particle performs uniform circular motion in the plane perpendicular to the cylinder axis.
Geometry of the Path:
The particle starts at distance $r$ from the center of the cylinder. Its initial velocity is radial (normal to the cylinder surface). Because the magnetic force acts perpendicular to velocity, the trajectory will curve.
For the bead to “eventually touch the cylinder tangentially”, its trajectory must be a circle that grazes the cylinder. Since it left the cylinder radially (perpendicularly), and circles have symmetry, the return point where it grazes the cylinder must be diametrically opposite on the trajectory circle, or the geometry requires a specific radius.
Considering the boundary conditions:
- Start point: Distance $r$ from origin. Velocity is radial.
- End condition: Touching a circle of radius $r$ tangentially.
For the trajectory to curve back and just touch the cylinder tangentially, the diameter of the particle’s circular orbit must be equal to the distance required to wrap around correctly. Based on the constraints for such magnetic deflection problems, the condition implies that the radius of the magnetic orbit, $R$, must be related to the cylinder radius $r$ such that the path encloses the cylinder or grazes it.
The specific condition for “touching tangentially” after being projected radially requires the radius of the path $R$ to be exactly twice the radius of the cylinder $r$ (assuming the path loops back to touch the far side or returns to the tangent). Specifically, as per standard results for this setup:
$$ R_{orbit} = 2r $$The radius of a charged particle moving in a magnetic field is given by:
$$ R_{orbit} = \frac{mv}{qB} $$Equating the two expressions for the radius:
$$ 2r = \frac{mv}{qB} $$Solving for the magnitude of charge $q$: