Question -
Answer -
Magnetic field strength, B =0.15 T
Charge on the electron, e =1.6 × 10−19 C
Mass of the electron, m =9.1 × 10−31 kg
Potential difference, V =2.0 kV = 2 × 103 V
Thus, kinetic energy of theelectron = eV
Where,
v =velocity of the electron
(a) Magneticforce on the electron provides the required centripetal force of the electron.Hence, the electron traces a circular path of radius r.
Magnetic force on the electron isgiven by the relation,
B ev
Centripetalforce 
From equations (1) and (2), weget
Hence, the electron has acircular trajectory of radius 1.0 mm normal to the magnetic field.
(b) Whenthe field makes an angle θ of 30° with initial velocity, theinitial velocity will be,
From equation (2), we can writethe expression for new radius as:
Hence, the electron has a helicaltrajectory of radius 0.5 mm along the magnetic field direction.