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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?
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A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 c...
Calculating the Magnetic Field for Accelerating Protons in a Cyclotron
Given parameters:
- Cyclotron oscillator frequency = 10 MHz
- Radius of dees = 60 cm
- Charge of a proton = e = 1.6 x 10^-19 C
- Mass of a proton = mp = 1.67 x 10^-27 kg
Formula to use:
- Cyclotron frequency = (qB)/(2πm)
where q is the charge of the particle, B is the magnetic field, and m is the mass of the particle.
Step-by-Step Solution:
1. Convert the cyclotron oscillator frequency to angular frequency:
ω = 2πf = 2π x 10 MHz = 2π x 10^7 Hz
2. Substitute the given parameters into the formula:
2π x 10^7 Hz = (eB)/(2πmp)
3. Solve for the magnetic field, B:
B = (2πmpω)/e
B = (2π x 1.67 x 10^-27 kg x 2π x 10^7 Hz)/(1.6 x 10^-19 C)
B = 0.66 T
Therefore, the magnetic field required to accelerate protons with a cyclotron oscillator frequency of 10 MHz and a radius of 60 cm is 0.66 T. The correct answer is option D.
Given parameters:
- Cyclotron oscillator frequency = 10 MHz
- Radius of dees = 60 cm
- Charge of a proton = e = 1.6 x 10^-19 C
- Mass of a proton = mp = 1.67 x 10^-27 kg
Formula to use:
- Cyclotron frequency = (qB)/(2πm)
where q is the charge of the particle, B is the magnetic field, and m is the mass of the particle.
Step-by-Step Solution:
1. Convert the cyclotron oscillator frequency to angular frequency:
ω = 2πf = 2π x 10 MHz = 2π x 10^7 Hz
2. Substitute the given parameters into the formula:
2π x 10^7 Hz = (eB)/(2πmp)
3. Solve for the magnetic field, B:
B = (2πmpω)/e
B = (2π x 1.67 x 10^-27 kg x 2π x 10^7 Hz)/(1.6 x 10^-19 C)
B = 0.66 T
Therefore, the magnetic field required to accelerate protons with a cyclotron oscillator frequency of 10 MHz and a radius of 60 cm is 0.66 T. The correct answer is 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 x10-27 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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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 x10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 x10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 x10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer?.
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 x10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 x10-27 kg)a)0.11 Tb)0.22 Tc)0.44 Td)0.66 TCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 x10-27 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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