Question

Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at 30 to the field direction. Name the trajectories followed by them.

Answer 1

We know that when a charged particle enters a uniform magnetic field the force exerted on it,

F = q(v X B) = qvb sin∅

Case I : when the particle enters perpendicular,∅ = 90

In this case since sin∅ = 1 is the maximum value and the direction of the force is perpendicular to the direction of motion of particle, hence the particle starts moving in a CIRCULAR path with radius,

r= mv/bq

Case II : the particle enters at an angle of 30⁰:

In this case the force acting on it has two components, one parallel to the motion which will cause a linear motion, the other component will act perpendicular to motion which will cause circular motion.

Hence due to the resultant of these two component the particle will move along a HELICAL path.