A proton (q = 1.6 X 10 -19 C, m = 1.67 X 10 -27 kg) moving with constant velocit
ID: 2138986 • Letter: A
Question
A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.78 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (2.3 X 105 m/s, 1.7 X 105 m/s).
1)
What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field?
2)
What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?
3)
What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field?
4)
What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.
A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.78 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (2.3 X 105 m/s, 1.7 X 105 m/s). What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field? What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field? What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.Explanation / Answer
a) same as vx=2.3 X 105 m/s.because there is no force in x direction therefore intial and final velocity in x direction must be same.
b)we know that the proton in constant magnetic field experiences a force of F=qv*b and this force provides a centripetal acceleration, therefore by the formula qvb=mv^2/r which gives r=mv/qb. the valu of b (magnetic field) is unknown here.we can fin b by the eqn .....v=u+at here t=D/vx=.339*10^-5sec,u=0hence a=v/t=6.87*10^10m/s^2and F(magnetic)= ma=11.47*10^-17N=qub hence r=.77m
c)y coordinate can be given by the eqn...s=ut+.5at^2,here u=0therefore s=.3989mts(t=339*10^-5sec,a=6.87*10^10m/s^2)
d)b=f(as calculated above)/vq=11.47*10^-17N/(2.3*10^5*(1.6*10^-27))=3.11*10^5T
plzzz rate my answer as best it took me half an hour to solve and type the answer plzzz....
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