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A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 cm, then magnetic field for accelerating the protons is (Given e = 1.6 x 10-19C, mp = 1.67 x 10-27 kg)
  • a)
    0.11 T
  • b)
    0.22 T
  • c)
    0.44 T
  • d)
    0.66 T
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 c...
Frequency (f) =q*B/2pie *m so =10*1000=1.6*10-19*B/2*3.14*1.67*10-27 so that B =0.66
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Community Answer
A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 c...

Given data:
- Cyclotron oscillator frequency = 10 MHz
- Radius of dees = 60 cm = 0.6 m
- Charge of a proton (e) = 1.6 x 10^-19 C
- Mass of a proton (mp) = 1.67 x 10^-27 kg

Formula:
The frequency of a cyclotron is given by:
f = qB / (2πm)

Where:
- f is the frequency of the cyclotron oscillator
- q is the charge of the particle (proton in this case)
- B is the magnetic field strength
- m is the mass of the particle

Calculations:
Given f = 10 MHz = 10 x 10^6 Hz
q = 1.6 x 10^-19 C
m = 1.67 x 10^-27 kg
r = 0.6 m

Substitute the values into the formula:
10 x 10^6 = (1.6 x 10^-19) * B / (2π * 1.67 x 10^-27)

Solving for B:
B = (10 x 10^6) * 2π * 1.67 x 10^-27 / (1.6 x 10^-19)
B ≈ 0.66 T

Therefore, the magnetic field required for accelerating the protons in the cyclotron is approximately 0.66 T, which corresponds to option D.
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A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 cm, then magnetic field for accelerating the protons is (Given e = 1.6 x 10-19C, mp = 1.67 x 10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer?
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