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?

Question

contestada

Respuesta :

nuuk nuuk · Answer 1 · 2019-10-16T07:04:21+02:00

Answer:[tex]r_2=11.81 cm[/tex]

Explanation:

Given

[tex]m_1=2.2\times 10^{-8} kg[/tex]

[tex]m_2=4.8\times 10^{-8} kg[/tex]

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

[tex]qV=\frac{mv^2}{2}[/tex]

[tex]v=\sqrt{\frac{2qV}{m}}[/tex]

and we know

Force due to magnetic field will Provide centripetal Force

[tex]qvB=\frac{mv^2}{r}[/tex]

[tex]B=\frac{\sqrt{\frac{2Vm}{q}}}{r}[/tex]

and B is equal for both particles

thus [tex]\frac{m}{r^2}=constant[/tex]

[tex]\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}[/tex]

[tex]r_2^2=\frac{4.8}{2.2}\times 8^2[/tex]

[tex]r_2=11.81 cm[/tex]