Protons having kinetic energy of 5.00MeV... Protons having a kinetic energy of 5
ID: 2297546 • Letter: P
Question
Protons having kinetic energy of 5.00MeV...
Protons having a kinetic energy of 5.00 MeV are moving in the positive x-direction and enter a magnetic field of 0.050 T in the z-directions. out of the page and extending from x=0 to x=1.00m as shown in the figure below. Calculate the y Component of the proton's momentum as they leave the magnetic field. (b) Find the angle between the initial velocity vector of the proton beam and the velocity vector after the bcam emerges from the field. (Hint: l eV = l.60 times l0-19 J)Explanation / Answer
KE = 5 MeV = 5*10^6*1.6*10^-19 = 8*10^-13 J
KE = 0.5*m*v^2
==> v = sqrt(2*KE/m)
= sqrt(2*8*10^-13/1.67*10^-27)
= 3.095*10^7 m/s
time taken to cross magnetic field region,
t = x/v = 1/3.095*10^7 = 3.23*10^-8 s
a) Fy = q*v*B
= 1.6*10^-19*3.095*10^7*0.05
= 2.476*10^-13 N
Fy = m*ay
==> ay = Fy/m = 2.476*10^-13/1.67*10^-27 = 1.483*10^14 m/s^2
in y-direction
vy = uy + ay*t
vy = ay*t = 1.483*10^14*3.23*10^-8 = 4.79*10^6 m/s
Py = m*vy = 1.67*10^-27*4.79*10^6 = 8*10^-21 kg.m/s <<<<-----Answer
b) theta = tan^-1(vy/vx)
= tan^-1(4.79*10^6/3.095*10^7)
= 8.8 degrees with +x axis in clockwise direction. <<<<-Answer
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