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(B) The kinetic energy $K$ of a particle of charge $q$ and mass $m$ in a cyclotron with magnetic field $B$ and radius $r$ is given by the formula:$K =

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$18$

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(B) The kinetic energy $K$ of a particle of charge $q$ and mass $m$ in a cyclotron with magnetic field $B$ and radius $r$ is given by the formula:$K = \frac{q^2 B^2 r^2}{2m}$Given:$q = 1.6 	imes 10^{-19}\,C$$B = 1.0\,T$$r = 60\,cm = 0.6\,m$$m = 1.6 	imes 10^{-27}\,kg$Substituting the values:$K = \frac{(1.6 	imes 10^{-19})^2 	imes (1.0)^2 	imes (0.6)^2}{2 	imes 1.6 	imes 10^{-27}}$$K = \frac{2.56 	imes 10^{-38} 	imes 0.36}{3.2 	imes 10^{-27}}$$K = 0.8 	imes 10^{-11} 	imes 0.36 = 0.288 	imes 10^{-11}\,J$To convert Joules to $MeV$,divide by $1.6 	imes 10^{-13}$:$K = \frac{0.288 	imes 10^{-11}}{1.6 	imes 10^{-13}} = 0.18 	imes 10^2 = 18\,MeV$

$A$ cyclotron is used to accelerate protons. If the operating magnetic field is $1.0\,T$ and the radius of the cyclotron 'dees' is $60\,cm$,the kinetic energy of the accelerated protons in $MeV$ will be.
[use $m_{p} = 1.6 \times 10^{-27}\,kg, e = 1.6 \times 10^{-19}\,C$]

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$A$ cyclotron is used to accelerate protons. If the operating magnetic field is $1.0\,T$ and the radius of the cyclotron 'dees' is $60\,cm$,the kinetic energy of the accelerated protons in $MeV$ will be.[use $m_{p} = 1.6 \times 10^{-27}\,kg, e = 1.6 \times 10^{-19}\,C$]