A beam of protons, initially at rest, is accelerated horizontally, from rest, be
ID: 1394069 • Letter: A
Question
A beam of protons, initially at rest, is accelerated horizontally, from rest, between the plates of a parallel-plate capacitor, the potential d fference (AV) between the plates is 1000V and the plate separation is 1.00 cm. The particle then enters a region where a uniform magnetic field, B 0.913 T (directed out the page). To detect the particle, a sensor is located directly below the point where the particle enters the field. What is the velocity vector for the particle just as it particle enters the magnetic field? AV Determine the magnetic force vector (in Sensor component form) exerted on the particle immediately after it enters the magnetic field In order to detect the exiting particle, at what distance below the particle's entrance point does the sensor need to be placed? Next, a beam of alpha particles is then accelerated (same AV as above) and fired into the magnetic field. To what magnitude should the magnetic field be is adjusted to so that the particles will be detected by the sensor? Note: the mass of an alpha particle is 6.64x10 27 and the charge is +2e.Explanation / Answer
a) Apply work-energy theorm to find the velocity of the partciel
workdone = gain in kinetic enrgy
q*delta V = 0.5*m*v^2
v = sqrt(2*q*delta V/m)
= sqrt(2*1.6*10^-19*1000/(1.67*10^-27))
= 4.38*10^5 i m/s (towards +x axis)
b) magnetic force = q*(v cross B)
= 1.6*10^-19*((4.38*10^5 i) cross (0.913 k))
= 6.4*10^-14 N k
c) Apply, q*v*B = m*v^2/r
r = m*v/(B*q)
= 1.67*10^-27*4.38*10^5/(0.913*1.6*10^-19)
= 0.005 m
d = 2*r
= 2*0.005
= 0.01 m
= 1 cm
so the detecter should be placed 1 cm below.
d)
for alfa particle, v' = v/sqrt(2)
m' = 4*m
q' = 2*q
r' = r
r = m'*v'/(B'*q')
B' = m'*v'/(r*q')
= 4*m*(v/sqrt(2)/(r*2*q')
= (m*v/r*q)*(2/sqrt(2))
= B*sqrt(2)
= 0.923*1.414
= 1.305 T
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.