Answer:
[tex]r_2=21.4cm[/tex]
Explanation:
Particle 1, charge q, [tex]m_1=1.2*10^{-8}Kg:[/tex]
The particle is accelerated thanks to the difference voltage V, the electrical energy becomes kinetic energy:
[tex]q*V=\frac{1}{2}m_1v^2[/tex]
[tex]v^2=2*q*V/m_1[/tex] 1
When the particle enter into the magnetic Field, it feels a centripetal magnetic Force, that is why the particle travel on circular path:
[tex]F_{magnetic}=q*v*B=m_1*a_c=m_1*v^2/r[/tex]
[tex]r_1^2=\frac{m_1^2v^2}{q^2B^2}[/tex] 2
We replace 1 in 2:
[tex]r_1^2=\frac{2*m_1*V}{q*B^2}[/tex] 3
Particle 2, charge q, [tex]m_2=5.5*10^{-8}Kg:[/tex]
The particle is accelerated thanks to the difference voltage V, the electrical energy becomes kinetic energy:
[tex]q*V=\frac{1}{2}m_2v^2[/tex]
[tex]v^2=2*q*V/m_2[/tex] 4
When the particle enter into the magnetic Field, it feels a centripetal magnetic Force, that is why the particle travel on circular path:
[tex]F_{magnetic}=q*v*B=m_2*a_c=m_2*v^2/r[/tex]
[tex]r_2^2=\frac{m_2^2v^2}{q^2B^2}[/tex] 5
We replace 4 in 5:
[tex]r_2^2=\frac{2*m_2*V}{q*B^2}[/tex] 6
The magnetic Field B, the voltage V, the charge q, are the same for both particles. We can divide 6 with 4 and find the radius for the second particle:
[tex]\frac{r_2^2}{r_1^2}=\frac{m_2}{m_1}[/tex]
[tex]r_2=r_1*\sqrt{\frac{m_2}{m_1}}=10cm*\sqrt{\frac{5.5*10^{-8}Kg}{1.2*10^{-8}Kg}}=21.4cm[/tex]
The radius of the circular path for particle 2 is mathematically given as
r2=11.81 cm
What is the radius (in cm) of the circular path for particle 2
Question Parameter(s):
Particle 1 and particle 2 have masses of m1 = 1.2 × 10-8 kg and m2 = 5.5 × 10-8 kg,
The radius of the circular path for particle 1 is r1 = 10 cm
Generally, the equation for the Velocity is mathematically given as
[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Where
[tex]B=\frac{\sqrt{\frac{2Vm}{q}}}{r}[/tex]
Hence
[tex]\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}[/tex]
Therefore
[tex]r_2^2=\frac{4.8}{2.2}*8^2[/tex]
r2=11.81 cm
In conclusion, the radius is
r2=11.81 cm
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