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Poobel
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Proton in Magnetic field -- am I correct?
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=1.3x10^5 eV=1.3x10^5
Fellow physisists is my solution correct?
Thank you
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=1.3x10^5 eV=1.3x10^5
Fellow physisists is my solution correct?
Thank you