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 1.22 × 106 V and then enters a uniform magnetic field whose magnitude is 1.80 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

Answer:

Explanation:

charge on alpha particle q = 2 x 1.6 x 10⁻¹⁹ C

mass m = 6.64 x 10⁻²⁷

potential diff V = 1.22 x 10⁶ v

magnetic field B = 1.8 T

speed of particle = v

kinetic energy will be acquired by charged particle in electric field

1/2 m v² = q V

1/2 x 6.64 x 10⁻²⁷ x v² = 1.22 x 10⁶ x  2 x 1.6 x 10⁻¹⁹

v² = 1.176 x 10¹⁴

v = 1.08 x 10⁷ m /s

speed of particle v = 1.08 x 10⁷ m /s

b ) magnetic force in magnetic field

= Bqv

= 1.8 x 2 x 1.6 x 10⁻¹⁹ x  1.08 x 10⁷

= 6.22 x 10⁻¹² N

c )

radius of circular path r

m v² / r = Bqv

r =   mv / Bq

= 6.64 x 10⁻²⁷ x 1.08 x 10⁷ / (1.8 x 2 x 1.6 x 10⁻¹⁹)

= 7.17 x 10⁻²⁰ / 5.76 x 10⁻¹⁹

= .124 m

12.4 cm