how large must B be for the particle to avoid colliding with the far wall of the box?

Question

   A particle with charge q and mass m is shot with kinetic energy, K, through a hole in one side of a box as shown in the figure. If the magnetic field, B, in the box is uniform and points into the page, how large must B be for the particle to avoid colliding with the far wall of the box?

Answer 1

The charge q, if positive, and perpendicular to the magnetic field B experience a force always perpendicular to both the velocity and magnetic field vectors. This being the case, the trajectory of the charge will be circular.

The force on such a charge is given by F = q x v x B and must be equal to the centripetal force of the charge.

Therefore q x v x B = m x v² / d where d is the largest allowable radius of the charge's motion.

Solving for B = m x v /d x q. As v isn't given, it can be obtained from K = 1/2 x m x v² or

v = √ 2 x K / m.

Substituting in the equation for B gives B = (m / (d x q)) x √ (2 x K / m). This can be rearranged to other expressions, but they can be obtained from what is given. The expression represents the lowest value of B to avoid colliding with the wall.