Back to Directory
NEET PHYSICSMedium

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 .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started