Question

An alpha particle is traveling at right angles to magnetic field

Answer 1

Explanation:

maybe you would not completed the question so I try my best

I) Force on the alpha particle :

F = qvB sinθ

Given,

v= 6×10^5 m/s

B= 0.2 T

θ= 90°

Now,the charge of an alpha particle

= 2×charge of a proton

=2×(1.6×10^-19) C

=3.2×10^-19 C

Therefore, from the formula,

F= (3.2×10^-19) × (6×10^5) × (0.2) × sin90°

=3.84×10^-14 N

ii) Acceleration = F/m

=(3.84×10^-14)/(6.44×10^-27)

=5.96×10^12