A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu
1 answer:
The angular frequency of the cyclotron is 0.07 x Hz.
- Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
- It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.
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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The magnetic field strength is 17 Weber.
<h3>How can we find the value of magnetic field?</h3>We know, The magnetic field equation is given by:
Where, we know,
=
= It is the magnetic constant or the permeability of free space.
n = The turn density which is equivalent to number of turns per unit length = 3300 turns per meter = 3300
I= the amount of current applied to it = 4. 1 kA = 4 × A.
We have to find the magnetic field strength = B
Substituting the known values into the equation gives the strength of the magnetic field, which is:
[We know,
]
So, from the above calculation we can see that, The magnetic field strength is 17 Weber.
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