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PhysicsEMI/ACMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A cyclotron's oscillator frequency is $10 M H z$ and the radius of its dees is $60 c m$ , then the kinetic energy of the proton beam produced by the accelerator is [ $m_{proton} = 1.67 \times 10^{- 27} kg$ ]

Options:

Answer:

Solution:

$KE = \frac{q^{2} B^{2} R^{2}}{2 m}$

Given, $f = 10 M H z = 10^{7} H z$

$R = 60 c m = 0 .6 m$

$f = \frac{q B}{2 π m}$

$∴ K E = 2 π^{2} m f^{2} R^{2}$

$= 2 \times {(3 .14)}^{2} \times 1.67 \times 10^{- 27} \times 10^{14} \times 0. 36 J$

$\approx 7 M e V$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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