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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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Verified Answer
A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 c...
v = 10 MHz = 10 x 106 Hz = 107 Hz
R = rm = 60 cm = 0.6 m
e = 1.6 x 10-19 C
m = 1.67 x 10-27 kg
resonance v = Be/2πm
or B = 2πmv/e
= 2 x 3.14 x 1.67 x 10 −27 x 107 /1.6 x 10 −19
= 0.66 tesla
R = rm = 60 cm = 0.6 m
e = 1.6 x 10-19 C
m = 1.67 x 10-27 kg
resonance v = Be/2πm
or B = 2πmv/e
= 2 x 3.14 x 1.67 x 10 −27 x 107 /1.6 x 10 −19
= 0.66 tesla
Most Upvoted Answer
A cyclotron oscillator frequency is 10 MHz. If radius of its dees 60 c...
Introduction
To find the magnetic field needed for a cyclotron oscillator frequency, we can use the relationship between frequency, charge, mass, radius, and magnetic field. The formula for the cyclotron frequency (\(f\)) is given by:
Formula for Cyclotron Frequency
\[ f = \frac{qB}{2\pi m} \]
Where:
- \(f\) = frequency (10 MHz = \(10 \times 10^6\) Hz)
- \(q\) = charge of proton (\(1.6 \times 10^{-19}\) C)
- \(B\) = magnetic field (T)
- \(m\) = mass of proton (\(1.67 \times 10^{-27}\) kg)
Rearranging the Formula
To find \(B\), rearranging the formula gives:
\[ B = \frac{2\pi mf}{q} \]
Substituting Values
Now, substitute the known values:
- \(m = 1.67 \times 10^{-27}\) kg
- \(f = 10 \times 10^6\) Hz
- \(q = 1.6 \times 10^{-19}\) C
Calculating \(B\):
\[ B = \frac{2\pi (1.67 \times 10^{-27} \, \text{kg})(10 \times 10^6 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]
Performing the calculation:
1. Calculate the numerator:
- \(2\pi \times 1.67 \times 10^{-27} \times 10 \times 10^6 \approx 1.05 \times 10^{-19}\)
2. Divide by the charge:
- \(B \approx \frac{1.05 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.65625 \, T\)
Final Result
Thus, the magnetic field \(B \approx 0.66 \, T\), which corresponds to option 'D'.
To find the magnetic field needed for a cyclotron oscillator frequency, we can use the relationship between frequency, charge, mass, radius, and magnetic field. The formula for the cyclotron frequency (\(f\)) is given by:
Formula for Cyclotron Frequency
\[ f = \frac{qB}{2\pi m} \]
Where:
- \(f\) = frequency (10 MHz = \(10 \times 10^6\) Hz)
- \(q\) = charge of proton (\(1.6 \times 10^{-19}\) C)
- \(B\) = magnetic field (T)
- \(m\) = mass of proton (\(1.67 \times 10^{-27}\) kg)
Rearranging the Formula
To find \(B\), rearranging the formula gives:
\[ B = \frac{2\pi mf}{q} \]
Substituting Values
Now, substitute the known values:
- \(m = 1.67 \times 10^{-27}\) kg
- \(f = 10 \times 10^6\) Hz
- \(q = 1.6 \times 10^{-19}\) C
Calculating \(B\):
\[ B = \frac{2\pi (1.67 \times 10^{-27} \, \text{kg})(10 \times 10^6 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]
Performing the calculation:
1. Calculate the numerator:
- \(2\pi \times 1.67 \times 10^{-27} \times 10 \times 10^6 \approx 1.05 \times 10^{-19}\)
2. Divide by the charge:
- \(B \approx \frac{1.05 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.65625 \, T\)
Final Result
Thus, the magnetic field \(B \approx 0.66 \, T\), 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?
Question Description
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? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET 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 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? covers all topics & solutions for NEET 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 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?.
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? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET 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 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? covers all topics & solutions for NEET 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 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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