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A cyclotron is accelerating proton, where the applied magnetic field is 2 T, the potential gap is 100 KV, then how much turn the proton has to move between the dees to acquire a kinetic energy 20 Me V
- a)200
- b)300
- c)150
- d)100
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A cyclotron is accelerating proton, where the applied magnetic field i...
Given, Magnetic field B=2T
Potential gap ΔV=100kV
We know that, r= mv/qB
⇒V=100�10^3 =10^5 volts
K1=2eV
So, KE gained in each revolution, Kn =e(V)+e(V)=2e(V)=2e x 10^5
Thus, the number of revolutions, N= Kn/K1
N = (20 x 10^6) / (2 x 10^5)
=100revolution
Most Upvoted Answer
A cyclotron is accelerating proton, where the applied magnetic field i...
To calculate the number of turns a proton has to move between the dees of a cyclotron to acquire a kinetic energy of 20 MeV, we can use the following formula:
ΔV = qV
where ΔV is the potential difference between the dees, q is the charge of the proton, and V is the potential gap.
- Given:
- Magnetic field (B) = 2 T
- Potential gap (V) = 100 kV
- Kinetic energy (KE) = 20 MeV
- Constants:
- Charge of a proton (q) = 1.6 x 10^-19 C
- Mass of a proton (m) = 1.67 x 10^-27 kg
- Speed of light (c) = 3 x 10^8 m/s
1. Calculating the potential difference (ΔV):
ΔV = 100 kV = 100,000 V
2. Calculating the velocity of the proton (v):
KE = (1/2)mv^2
20 MeV = (1/2)(1.67 x 10^-27 kg)(v^2)
v^2 = (2 x 20 x 10^6 eV) / (1.67 x 10^-27 kg)
v^2 = 2.4 x 10^13 m^2/s^2
v ≈ 1.55 x 10^6 m/s
3. Calculating the radius of the path (r):
qvB = mv^2 / r
r = mv / (qB)
r = (1.67 x 10^-27 kg)(1.55 x 10^6 m/s) / (1.6 x 10^-19 C)(2 T)
r ≈ 5.13 x 10^-2 m
4. Calculating the circumference of the path (C):
C = 2πr
C = 2π(5.13 x 10^-2 m)
C ≈ 0.322 m
5. Calculating the distance traveled between the dees in one revolution:
Distance traveled = C
Distance traveled ≈ 0.322 m
6. Calculating the number of turns (n) required to acquire the desired kinetic energy:
Distance traveled in n turns = ΔV
n * Distance traveled = ΔV
n = ΔV / Distance traveled
n = 100,000 V / 0.322 m
n ≈ 310,559
Therefore, the proton has to move approximately 310,559 turns between the dees to acquire a kinetic energy of 20 MeV.
ΔV = qV
where ΔV is the potential difference between the dees, q is the charge of the proton, and V is the potential gap.
- Given:
- Magnetic field (B) = 2 T
- Potential gap (V) = 100 kV
- Kinetic energy (KE) = 20 MeV
- Constants:
- Charge of a proton (q) = 1.6 x 10^-19 C
- Mass of a proton (m) = 1.67 x 10^-27 kg
- Speed of light (c) = 3 x 10^8 m/s
1. Calculating the potential difference (ΔV):
ΔV = 100 kV = 100,000 V
2. Calculating the velocity of the proton (v):
KE = (1/2)mv^2
20 MeV = (1/2)(1.67 x 10^-27 kg)(v^2)
v^2 = (2 x 20 x 10^6 eV) / (1.67 x 10^-27 kg)
v^2 = 2.4 x 10^13 m^2/s^2
v ≈ 1.55 x 10^6 m/s
3. Calculating the radius of the path (r):
qvB = mv^2 / r
r = mv / (qB)
r = (1.67 x 10^-27 kg)(1.55 x 10^6 m/s) / (1.6 x 10^-19 C)(2 T)
r ≈ 5.13 x 10^-2 m
4. Calculating the circumference of the path (C):
C = 2πr
C = 2π(5.13 x 10^-2 m)
C ≈ 0.322 m
5. Calculating the distance traveled between the dees in one revolution:
Distance traveled = C
Distance traveled ≈ 0.322 m
6. Calculating the number of turns (n) required to acquire the desired kinetic energy:
Distance traveled in n turns = ΔV
n * Distance traveled = ΔV
n = ΔV / Distance traveled
n = 100,000 V / 0.322 m
n ≈ 310,559
Therefore, the proton has to move approximately 310,559 turns between the dees to acquire a kinetic energy of 20 MeV.
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A cyclotron is accelerating proton, where the applied magnetic field is 2 T, the potential gap is 100 KV, then how much turn the proton has to move between the dees to acquire a kinetic energy 20 Me Va)200b)300c)150d)100Correct answer is option 'D'. Can you explain this answer?
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