An alpha particle is travelling right angkes to the magnetic feild with a speed of 6× 10^5 m/s the strength of magnetic feild is 0.2T.calculate i) force on alpha particle ii) acceleration of alpha particle
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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
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
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