A cyclotron's oscillator frequency is 10 text | vedclass

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(C) The cyclotron frequency is given by $f = \frac{qB}{2\pi m}$.The maximum velocity $v$ of the proton at the radius $r$ is $v = \frac{qBr}{m} = 2\pi

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$7$

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(C) The cyclotron frequency is given by $f = \frac{qB}{2\pi m}$.The maximum velocity $v$ of the proton at the radius $r$ is $v = \frac{qBr}{m} = 2\pi fr$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(2\pi fr)^2 = 2\pi^2 mf^2r^2$.Given values: $f = 10 	imes 10^6 	ext{ Hz}$, $r = 0.6 	ext{ m}$, $m = 1.67 	imes 10^{-27} 	ext{ kg}$.Substituting these values:$K = 2 	imes (3.14)^2 	imes (1.67 	imes 10^{-27}) 	imes (10^7)^2 	imes (0.6)^2$$K = 2 	imes 9.8596 	imes 1.67 	imes 10^{-27} 	imes 10^{14} 	imes 0.36$$K \approx 1.185 	imes 10^{-12} 	ext{ J}$.To convert to $	ext{MeV}$, divide by $1.6 	imes 10^{-13} 	ext{ J/MeV}$:$K = \frac{1.185 	imes 10^{-12}}{1.6 	imes 10^{-13}} \approx 7.4 	ext{ MeV}$.Rounding to the nearest given option, the kinetic energy is $7 	ext{ MeV}$.

$A$ cyclotron's oscillator frequency is $10 \text{ MHz}$ and the operating magnetic field is $0.66 \text{ T}$. If the radius of its dees is $60 \text{ cm}$, then the kinetic energy of the proton beam produced by the accelerator is: (in $\text{ MeV}$)

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