Respuesta :
Answer:
Alpha particle
Explanation:
Initially, the particle moves in a straight line. This means that the electric force and the magnetic force are equal:
[tex]qE=qvB[/tex]
where q is the charge, E is the electric field, v is the speed and B is the magnetic field.
In this problem,
E = 1.5 kV/m = 1500 V/m
B = 0.034 T
Solving the equation for v we find the speed of the particle
[tex]v=\frac{E}{B}=\frac{1500 V/m}{0.034 T}=4.41\cdot 10^4 m/s[/tex]
Now the electric field is turned off, so the particle starts moving in a circular motion, where the magnetic force acts as centripetal force:
[tex]qvB=m\frac{v^2}{r}[/tex]
where
r = 2.7 cm = 0.027 m is the radius of the circular path
Solving the problem for q/m, we find charge-to-mass ratio of the particle
[tex]\frac{q}{m}=\frac{v}{Br}=\frac{4.41\cdot 10^4 m/s}{(0.034 T)(0.027 m)}=4.8\cdot 10^7 C/kg[/tex]
And this corresponds to the q/m ratio of an alpha particle, which has:
[tex]q=2e=3.2\cdot 10^{-19}C\\m=4a.m.u.=6.64\cdot 10^{-27} kg[/tex]
The mass of the particle depends on the type of particle, and the radius of
the path of the particle depend on its mass.
The particle might be an alpha particle.
Reasons:
The given parameters are;
Electric potential of the field, E = 1.5 kV/m
Magnetic field strength, B = 0.034 T
Radius of the circular path, r = 2.7 cm
Required:
Finding the type of particle in the question
Solution:
When moving in a straight line, we have;
B·q·v = F = q·E
Therefore;
E = B·v
When the particle moves in a circular path, we have;
[tex]q\cdot E = B \cdot q \cdot v = \mathbf{\dfrac{m \cdot v^2}{r}}[/tex]
Therefore;
[tex]v = \dfrac{B \cdot q \cdot r }{m}[/tex]
Which gives;
[tex]E = \dfrac{B \cdot q \cdot r }{m} \times B = \dfrac{B^2 \cdot q \cdot r }{m}[/tex]
[tex]\mathrm{The \ mass \ of \ the \ particle, \, m} = \mathbf{\dfrac{B^2 \cdot q \cdot r }{E}}[/tex]
q = n × e
Therefore;
[tex]m = \mathbf{\dfrac{B^2 \cdot n \cdot e \cdot r }{E}}[/tex]
Where;
n = The number of electron charge
Which gives the mass in kilograms as follows;
[tex]m = \dfrac{0.034^2 \times n\times 1.6 \times 10^{-19} \times 0.027 }{1500} = \mathbf{3.32928 \times 10^{-27} \times n}[/tex]
The particle is larger than a subatomic particle
Therefore;
[tex]\dfrac{m}{n} = 3.32928 \times 10^{-27}[/tex]
For an alpha particle, we have;
m ≈ 6.645 × 10⁻²⁷ kg
n = 2
n = 2
Therefore;
[tex]\dfrac{6.645 \times 10^{-27}}{2} \approx \mathbf{3.3225 \times 10^{-27}} \approx 3.32928 \times 10^{-27}[/tex]
Therefore, the particle might be an alpha particle.
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https://brainly.com/question/15398619
