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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? 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 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 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 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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