The angular frequency of the cyclotron is 0.07 x Hz.
The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.
<h3>Calculation of angular frequency:</h3>Given,
B = 0.47 T
r = 0.68
mass of proton = 1.6x
q = 1.6 x
so, the frequency is:
f = qB/2m
f = 1.6 x x 0.47/2x3.14x1.6x
f = 0.07 x
Hence, the angular frequency of the cyclotron is 0.07 x Hz.
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Total time elapsed is =8.2y
The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:
Δt=3.9ly/0.9c
Δt=4.3y
For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,
Δτ
Δτ
Δτ=1.8ly
The total elapsed time is:
T elapsed=Δt+3.9
T elapsed=4.3+3.9
T elapsed=8.2y
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Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r = 2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q = 7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC