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One of my Physics ISU questions is as follows:

- A proton moves in a circular path perpendicular to a 1.10 T magnetic field. The radius of the path is 4.5 cm. calculate energy of the proton in eV

Well i had hard time doing it but here we go: my solution

mV^2/r=QrB V=QrB/m

V= (1.6x10^-16) x o.o45 m x 1.10 T) / 1.67x10^-27 kg

V= 4.7x10^6 m/s

Ek= 1/2mV^2 = 1/2 x 1.67x10^-27 kg x (4.7x10^6 m/s)^2

= 1.8x10^-14 J

Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

eV=1.3x10^5eV=1.3x10^5

Fellow physisists is my solution correct?

Thank you

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# Homework Help: Proton in Magnetic field - am I correct?

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