JEE MainPhysicsMagnetic Effects of Current
A particle of mass m and charge q is accelerated from rest by a potential difference V . It then enters a region of uniform magnetic field B that has a finite width d . The initial velocity of the particle is perpendicular to both the magnetic field and the boundaries of the region. The maximum value of V such that the particle does not emerge from the opposite side of the magnetic field region is :
Options
- Aq B^2 d^2 m
- Bq B^2 d^2 2m
- C2m B^2 d^2 q
- Dq B d m
Correct answer
B. q B^2 d^2 2m
Step-by-step solution
The kinetic energy gained by the particle is K = qV . The radius R of its circular trajectory in the magnetic field is: R = 2mqV qB = 1 B 2mV q For the particle to turn back and not emerge from the opposite side of the region, its radius of curvature must be less than or equal to the width of the region. The maximum accelerating potential corresponds to the limiting case where the radius equals the width of the region ( R = d ). Setting R = d : d = 1 B 2mV q Squaring both sides: d^2 = 2mV qB^2 Rearranging to solve