NEET Exam > NEET Questions > A cyclotron oscillator frequency is 10 MHz. I...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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'.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
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?.
Solutions 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? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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 detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup