Particle 1 and particle 2 have masses of m1 = 1.7 × 10-8 kg and m2 = 4.7 × 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 = 11 cm. What is the radius (in cm) of the circular path for particle 2?
Answers
Answer:
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Explanation:
Particle 1, charge q, m_1=1.2*10^{-8}Kg:m
1
=1.2∗10
−8
Kg:
The particle is accelerated thanks to the difference voltage V, the electrical energy becomes kinetic energy:
q*V=\frac{1}{2}m_1v^2q∗V=
2
1
m
1
v
2
v^2=2*q*V/m_1v
2
=2∗q∗V/m
1
1
When the particle enter into the magnetic Field, it feels a centripetal magnetic Force, that is why the particle travel on circular path:
F_{magnetic}=q*v*B=m_1*a_c=m_1*v^2/rF
magnetic
=q∗v∗B=m
1
∗a
c
=m
1
∗v
2
/r
r_1^2=\frac{m_1^2v^2}{q^2B^2}r
1
2
=
q
2
B
2
m
1
2
v
2
2
We replace 1 in 2:
r_1^2=\frac{2*m_1*V}{q*B^2}r
1
2
=
q∗B
2
2∗m
1
∗V
3
Particle 2, charge q, m_2=5.5*10^{-8}Kg:m
2
=5.5∗10
−8
Kg:
The particle is accelerated thanks to the difference voltage V, the electrical energy becomes kinetic energy:
q*V=\frac{1}{2}m_2v^2q∗V=
2
1
m
2
v
2
v^2=2*q*V/m_2v
2
=2∗q∗V/m
2
4
When the particle enter into the magnetic Field, it feels a centripetal magnetic Force, that is why the particle travel on circular path:
F_{magnetic}=q*v*B=m_2*a_c=m_2*v^2/rF
magnetic
=q∗v∗B=m
2
∗a
c
=m
2
∗v
2
/r
r_2^2=\frac{m_2^2v^2}{q^2B^2}r
2
2
=
q
2
B
2
m
2
2
v
2
5
We replace 4 in 5:
r_2^2=\frac{2*m_2*V}{q*B^2}r
2
2
=
q∗B
2
2∗m
2
∗V
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:
\frac{r_2^2}{r_1^2}=\frac{m_2}{m_1}
r
1
2
2
2= m
1m
2r_2=r_1*\sqrt{\frac{m_2}{m_1}}=10cm*\sqrt{\frac{5.5*10^{-8}Kg}{1.2*10^{-8}Kg}}=21.4cmr
2=r
1m
1m
2=10cm∗
1.2∗10
−8Kg
5.5∗10
−8Kg
=21.4cm