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Asked by Akash Last Modified
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 the electron = eV
Where,
v = velocity of the electron
(a) Magnetic force 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 is given by the relation,
B ev
Centripetal force 
From equations (1) and (2), we get
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
(b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,
From equation (2), we can write the expression for new radius as:
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
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