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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

An alternating electric field of frequency vv is applied across the dees (radius = RR) of a cyclotron that is being used to accelerate protons (mass = mm). The operating magnetic field used in the cyclotron and the kinetic energy (KK) of the proton beam, produced by it, are given by:

A

B=mveB = \frac{m v}{e} and K=2mπ2v2R2K = 2m \pi^2 v^2 R^2

B

B=2πmveB = \frac{2\pi m v}{e} and K=m(2πvR)2K = m (2\pi v R)^2

C

B=2πmveB = \frac{2\pi m v}{e} and K=2mπ2v2R2K = 2m \pi^2 v^2 R^2

D

B=mveB = \frac{m v}{e} and K=m(2πvR)2K = m (2\pi v R)^2

Step-by-Step Solution

In a cyclotron, resonance occurs when the frequency of the alternating electric field (vv) matches the cyclotron frequency (vcv_c). According to the sources, the cyclotron frequency is vc=qB2πmv_c = \frac{qB}{2\pi m} . For a proton with charge ee, we have v=eB2πmv = \frac{eB}{2\pi m}. Rearranging this for the operating magnetic field gives B=2πmveB = \frac{2\pi m v}{e} . The maximum velocity (vmaxv_{max}) of the proton is attained at the radius of the dees (RR), where vmax=eBRm=ωRv_{max} = \frac{eBR}{m} = \omega R . Since ω=2πv\omega = 2\pi v, we get vmax=2πvRv_{max} = 2\pi v R. The kinetic energy is K=12mvmax2=12m(2πvR)2=2mπ2v2R2K = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (2\pi v R)^2 = 2m\pi^2 v^2 R^2 .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMalternatingelectricfrequencyappliedacross

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