(13%) Problem 7: Consider a 25-MeV proton moving perpendicularly to a 1.45 T fie

Question

(13%) Problem 7: Consider a 25-MeV proton moving perpendicularly to a 1.45 T field in a cyclotron. Find the radius of curvature, in meters, of the path of the proton while oving through the cyclotron. Grade Summarv Deductions Potential 490 9690 sin( | ( tan( ) acos) cos Submissions Attempts remaining: 15 (5% per attempt) detailed view cotanasin(O atan acotan sih() cosh)tanhcotanh(O END Degrees Radians Submit I give up! Hints: 2 for a 4% deduction. Hints remaining: 0 Feedback: 2% deduction per feedback. The energy of the proton tells you how fast it is going The speed can be used to find the radius of curvature, since the velocity is what determines the force the protons feels

Explanation / Answer

mass of proton, m = 1.67 x 10^-27 kg

Kinetic energy of the proton E = 25 x 10^6 eV = 25 x 10^6 x 1.602 x 10^-19 J

B = 1.45 T

now -

E = (1/2)mv^2

=> v = sqrt(2E/m)

again rotation in a circular path -

F = mv^2 / r

and in a magnetic field -

F = q*v*B

equalize the two -

mv^2 / r = q*v*B

=> r = mv / (q*B) = [m*sqrt(2E/m)] / (q*B) = [sqrt(2Em)] / (q*B)

put the values -

r = [sqrt(2x25 x 10^6 x 1.602 x 10^-19x1.67 x 10^-27)] / (1.602 x 10^-19 x 1.45)

= sqrt(133.767 x 10^-40) / (2.3229 x 10^-19) = 4.98 x 10^-1 m = 0.498 m

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