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A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?
  • a)
    0.72 T
  • b)
    0.65 T
  • c)
    0.39 T
  • d)
    0.12 T
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A cyclotron is operated at an oscillator frequency of 12 MHz and has a...
To find the magnitude of the magnetic field needed for a proton to be accelerated in a cyclotron, we can use the equation:

B = (q * F) / (2 * π * m * R * f)

Where:
B = Magnetic field
q = Charge of the proton
F = Centripetal force
m = Mass of the proton
R = Dee radius of the cyclotron
f = Oscillator frequency

Centripetal force is given by the equation:

F = (m * v^2) / R

Where:
v = Velocity of the proton

We know that the velocity of the proton can be calculated using the equation:

v = 2 * π * R * f

Now let's substitute the values into the equations and calculate the magnetic field:

1. Calculate the velocity of the proton:
v = 2 * π * R * f
= 2 * π * 0.5 m * 12 * 10^6 Hz
= 12 * 2 * π * 0.5 * 10^6 m/s
≈ 37.7 * 10^6 m/s

2. Calculate the centripetal force:
F = (m * v^2) / R
= (1.67 * 10^-27 kg * (37.7 * 10^6 m/s)^2) / 0.5 m
= (1.67 * 10^-27 kg * 1.42 * 10^15 m^2/s^2) / 0.5 m
≈ 4.69 * 10^-11 N

3. Calculate the magnetic field:
B = (q * F) / (2 * π * m * R * f)
= (1.6 * 10^-19 C * 4.69 * 10^-11 N) / (2 * π * 1.67 * 10^-27 kg * 0.5 m * 12 * 10^6 Hz)
= (7.04 * 10^-30 C * N) / (4 * π * 1.67 * 0.5 * 12 * 10^-21 kg * Hz)

Simplifying, we get:
B ≈ 0.72 Tesla

Therefore, the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron is approximately 0.72 T.
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A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?a)0.72 Tb)0.65 Tc)0.39 Td)0.12 TCorrect answer is option 'A'. Can you explain this answer?
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