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Q.
In the question number $34$, the kinetic energy (in $MeV$) of the proton beam produced by the accelerator is (radius of dees $= 60 \,cm)$
Moving Charges and Magnetism
Solution:
Final velocity of proton is
$v=r \times2\pi\upsilon=0.6\times2\times3.14\times12\times10^{6}$
$=4.5\times10^{7}\,m \, s^{-1}$
$\therefore K=\frac{1}{2} mv^{2}=\frac{1}{2}\times\frac{1.67\times10^{-27}\times\left(4.5\times10^{7}\right)^{2}}{1.6\times10^{-13}}$
$=10.6\, MeV$