Question

An alpha particle is travelling at right angle to magnetic field with a speed of 6*10^5.the strength of the magnetic field is o.2t.calculate force and acceleration of the alpba particle

Answer 1

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

Answer 2

answer with correct explain