Question

Particle 1 and particle 2 have masses of m1 = 1.5×10-8 kg and m2 = 6.2×10-8 kg, but they carry the same charge q. The two particles accelerate from rest through the same electric potential difference V and enter the same magnetic field, which has a magnitude B. The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is r1 = 12 cm. What is the radius (in cm) of the circular path for particle 2?

Answer 1

m₁v₁²/2 =qU => v₁ =sqrt(2qU/m₁)
m₂v₂²/2 =qU => v₂=sqrt(2qU/m₂)
v₁/v₂=sqrt(2qU/m₁)/sqrt(2qU/m₂)=sqrt(m₂/m₁)
in magnetic field:
ma=qvBsinα
a=v²/R, α=90°, sin90°=1
m v²/R=qvB
R=mv/qB
R₁/R₂=(m₁v₁/qB)/ (m₂v₂/qB)=
=m₁v₁/m₂v₂=(m₁/m₂)•sqrt(m₂/m₁)=
= sqrt(m₁/m₂)
R₂=R₁•sqrt(m₂/m₁)=…