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?

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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards