Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.0k views
in Physics by (75.3k points)

The cyclotron's oscillator frequency is equal to v = 10 MHz. Find the effective accelerating voltage applied across the dees of that cyclotron if the distance between the neighbouring trajectories of protons is not less than Δr = 1.0 cm, with the trajectory radius being equal to r = 0.5 m.

1 Answer

+1 vote
by (69.1k points)
selected by
 
Best answer

We know that for a charged particle (proton) in a magnetic field,

On the other hand ΔE = 2 eV, where V is the effective acceleration voltage, across the Dees, there being two crossings per revolution. So,

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...