A particle with mass m and charge q is projected into a region having a perpendicular magnetic field B . Determine the angle of deviation of the particle when it exits the magnetic field, if the width d of the region is very slightly smaller than 2mv qB .

Step-by-step solution

The radius of the circular path of a charged particle in a perpendicular magnetic field is given by R = mv qB . When the particle enters the magnetic field perpendicularly to its boundary, the maximum horizontal distance it can penetrate into the region is equal to its radius R . The given width of the magnetic field region is d 2mv qB = 2R . Since the width of the region d is greater than the maximum penetration depth R ( d > R ), the particle will not be able to cross the magnetic field region. Instead, it will c