In a cyclotron final kinetic energy of a particle revolving in a circular path of radius \(R_{exit}\) is :

Q.1 (iii) In a cyclotron final kinetic energy of a particle revolving in a circular path of radius \(R_{exit}\) is :

A. K.E. \(\propto R_{exit}\)

B. K.E. \(\propto R_{exit}^2\)

C. K.E. \(\propto \dfrac{1}{R_{exit}}\)

D. K.E. \(\propto \dfrac{1}{R_{exit}^2}\)

Doubtly Asked 28 Oct 2026 Board exam · July 2025 hsc 12th Save WhatsApp Report

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Answer: B. K.E. \(\propto R_{exit}^2\)

Answer: (b) K.E. \(\propto R_{exit}^2\)

In the cyclotron \(qvB = \dfrac{mv^2}{R}\), so \(v = \dfrac{qBR}{m}\). At exit, \(\text{K.E.} = \dfrac{1}{2}mv^2 = \dfrac{q^2B^2R_{exit}^2}{2m}\), which is proportional to \(R_{exit}^2\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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In a cyclotron final kinetic energy of a particle revolving in a circular path of radius \(R_{exit}\) is :

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