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Physics Question on Moving charges and magnetism

A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz11\, MHz in external magnetic field of 1 T1 \,T. If the radius of its dees is 55 cm55\, cm, then its kinetic energy (in MeVMeV) is (mp=1.67×10−27 kg(m_{p}=1.67 \times 10^{-27} \, kg, e=1.6×10−19 C)e = 1.6 \times 10^{-19}\, C)

A

13.3613.36

B

12.5212.52

C

14.8914.89

D

14.4914.49

Answer

14.4914.49

Explanation

Solution

Here, υc=11 MHz=11×106 Hz\upsilon_{c}=11\, MHz = 11 \times10^{6}\,Hz B=1 T,R=55 cm=55×10−2 mB=1\,T, R=55\,cm =55\times10^{-2}\,m, e=1.6×10−19C;mp=1.67×10−27 kge=1.6\times10^{-19} C ; m_{p}=1.67 \times10^{-27}\,kg, ∴K.E.=q2B2R22m\therefore K.E. =\frac{q^{2}B^{2}R^{2}}{2m} =(1.6×10−19)2×(1)2×(55×10−2)22×1.67×10−27=\frac{\left(1.6\times10^{-19}\right)^{2}\times\left(1\right)^{2}\times\left(55\times10^{-2}\right)^{2}}{2\times1.67\times10^{-27}} =23.19×10−13 J=23.19\times10^{-13}\,J =23.19×10−131.6×10−19=\frac{23.19\times10^{-13}}{1.6\times10^{-19}} =14.49×106 eV=14.49\times10^{6}\,eV =14.49 MeV=14.49\, MeV