Question

An α-particle has a charge of +2e and a mass of 6.64 × 10^-27 kg. It is accelerated from rest through a potential difference that has a value of 1.45 × 10⁶ V and then enters a uniform magnetic field whose magnitude is 1.75 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Answer 1

Electrostatic potential energy is transformed into the kinetic energy of the -particle
qV = 1/2 mv²
Where q is the charge, V is the potential difference, m is the mass and v is the final velocity.
v² = 2qV/m
v = SQRT[2qV/m]
= SQRT[2 x (2 x 1.6 x 10^-19) x (1.45 x 10⁶) / (6.64 x 10^-27)]
= 1.18 x 10⁷ m/s

b)
Magnetic force, F = Bqv
Where B is the magnetic field strength
F = 1.75 x (2 x 1.6 x 10^-19) x (1.18 x 10⁷)
= 6.62 x 10^-12 N

c)
Radius, r = mv/Bq               [By equating magnetic force with centripetal force]
= [(6.64 x 10^-27) x (1.18 x 10⁷) / 1.75 x (2 x 1.6 x 10^-19)]
= 0.14 m