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An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64\cdot10^{-27} kg) traveling horizontally at 35.6 km/

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asked 2021-02-27
An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(\displaystyle{6.64}\cdot{10}^{{-{27}}}\) kg) traveling horizontally at 35.6 km/s enters a uniform, vertical, 1.10 T magnetic field.
A) What is the diameter of the path followed by this alpha particle?
B) What effect does the magnetic field have on the speed of the particle?
C) What are the magnitude of the acceleration of the alpha particle while it is in the magnetic field?
D) What are the direction of the acceleration of the alpha particle while it is in the magnetic field?

Answers (1)

2021-03-01

(a) We know that in the magnetic field
magnetic force = Centrifugal force
\(\displaystyle{q}{V}{B}={\frac{{{m}{V}^{{2}}}}{{{R}}}}\)
\(\displaystyle{R}={\frac{{{m}{V}}}{{{q}{B}}}}\)
Where R is radius
m is mass of helium \(\displaystyle={6.64}\cdot{10}^{{-{27}}}\)
V is speed \(\displaystyle={35.6}\cdot{10}^{{3}}\frac{{m}}{{s}}\)
q is charge on particle =\(\displaystyle{2}\cdot{1.6}\cdot{10}^{{-{19}}}\) C
B is magnetic field=1.1 T
\(\displaystyle{R}={\frac{{{6.64}\cdot{10}^{{-{27}}}\cdot{35.6}\cdot{10}^{{3}}}}{{{2}\cdot{1.6}\cdot{10}^{{-{19}}}\cdot{1.1}}}}={6.72}\cdot{10}^{{-{4}}}\)
Now the diameter (D)=2R=\(13.44 \cdot 10^{-4}\) m
(b) Magnetic field has no effect on speed but it bound the particle to move in circular motion.
(c) We know that radial acceleration \(\displaystyle{\left({a}\right)}={\frac{{{V}^{{2}}}}{{{R}}}}\)
\(\displaystyle={\frac{{{\left({35.6}\cdot{10}^{{3}}\right)}^{{2}}}}{{{6.72}\cdot{10}^{{-{4}}}}}}={1.89}\cdot{10}^{{{12}}}\frac{{m}}{{s}^{{2}}}\)
(d) The direction of the acceleration will be toward center.

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