Question

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 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 = 8 cm. What is the radius (in cm) of the circular path for particle 2?

Answer 1

Answer:
`r_2=11.81 cm`

Step-by-step explanation:

Given

`m_1=2.2* 10^(-8) kg`

`m_2=4.8* 10^(-8) kg`

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

`qV=(mv^2)/(2)`

`v=\sqrt{(2qV)/(m)}`

and we know

Force due to magnetic field will Provide centripetal Force

`qvB=(mv^2)/(r)`

`B=\frac{\sqrt{(2Vm)/(q)}}{r}`

and B is equal for both particles

thus
`(m)/(r^2)=constant`

`(m_1)/(r_1^2)=(m_2)/(r_2^2)`

`r_2^2=(4.8)/(2.2)* 8^2`

`r_2=11.81 cm`