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A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to aquire a kinetic energy of 20MeV? A) 100 B) 150 C) 200 D) 300?
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A proton is accelerating in a cyclotron where the applied magnetic fie...
Given:
- Magnetic field (B) = 2 T
- Potential gap (V) = 100 kV = 100,000 V
- Desired kinetic energy (KE) = 20 MeV

To Find:
- Number of revolutions the proton has to make between the dees (N)

Formula:
The kinetic energy of a particle in a magnetic field can be calculated using the formula:

KE = qV

Where:
- KE is the kinetic energy
- q is the charge of the particle
- V is the potential difference

The potential difference across the dees creates an electric field that accelerates the charged particle. The circular motion of the particle is due to the magnetic field, which applies a force perpendicular to the velocity of the particle.

The centripetal force required for circular motion is provided by the electric field. The centripetal force can be expressed as:

F = qvB

Where:
- F is the force
- v is the velocity of the particle
- B is the magnetic field

The centripetal force is also equal to the rate of change of momentum:

F = mv^2 / r

Where:
- m is the mass of the particle
- r is the radius of the circular path

Setting the two expressions for the force equal to each other:

qvB = mv^2 / r

Simplifying the equation:

v = qBr / m

Calculation:
Given that the potential difference (V) = 100,000 V, the kinetic energy (KE) = 20 MeV, and the charge of a proton (q) = 1.6 x 10^-19 C, we can calculate the velocity of the proton using the formula:

KE = qV

20 MeV = (1.6 x 10^-19 C)(100,000 V)

Converting MeV to joules:

1 MeV = 1.6 x 10^-13 J

20 MeV = 20(1.6 x 10^-13 J) = 3.2 x 10^-12 J

3.2 x 10^-12 J = (1.6 x 10^-19 C)(100,000 V)

Solving for V:

V = (3.2 x 10^-12 J) / (1.6 x 10^-19 C)
= 2 x 10^7 V

Substituting the known values into the equation for velocity:

v = (1.6 x 10^-19 C)(2 T)(r) / (1.67 x 10^-27 kg)

Simplifying the equation:

v = (3.2 x 10^-19 T)(r) / (1.67 x 10^-27 kg)

Now, equating the two expressions for velocity:

(1.6 x 10^-19 C)(2 T)(r) / (1.67 x 10^-27 kg) = 2 x 10^7 V

Simplifying:

r = (2 x 10^7 V)(1.67 x 10^-27 kg) / (1.6 x 10^-19 C)(2 T)
= (3.34 x 10^-20 kg m^2 s^-2) / (3.2 x
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A proton is accelerating in a cyclotron where the applied magnetic fie...
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A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to aquire a kinetic energy of 20MeV? A) 100 B) 150 C) 200 D) 300?
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A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to aquire a kinetic energy of 20MeV? A) 100 B) 150 C) 200 D) 300? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to aquire a kinetic energy of 20MeV? A) 100 B) 150 C) 200 D) 300? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to aquire a kinetic energy of 20MeV? A) 100 B) 150 C) 200 D) 300?.
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