9.3 Solutions 521
depend on the composition of emulsions,zis the charge of the ion andMits
mass in terms of proton mass. The data onRandEfor protons have been
used to determinenandR. We findn= 0 .6 andK= 0 .2276. Using these
values ofnandK, the energy for various values ofRhave been determined
for deuterons and^3 He as tabulated below and the corresponding graphs are
drawn, Fig. 9.8.Fig. 9.8
R(μm) 0 50 100 150 200 250 300 350 400 450 500
DE(MeV) 0 3.06 6.31 34. 245. 657. 068. 680. 191 .4 102.6 113. 7
(^3) He 0 46. 270. 089 .3 106.2 121.4 135.4 148.5 160.9 172.7 184. 0
9.3.10 Motion of Charged Particles in Magnetic Field .......
9.67 For a circular orbit of a charged particle of charge q and momentum p moving
in a magnetic field ofBTesla perpendicular to the orbit, the radiusrin metres
is related by the formula
P=qBr→cp=qBrcJ
cp=(1. 6 × 10 −^19 )(3× 108 )BrJ
= 4. 8 × 10 −^11 BrJ
=(4. 8 × 10 −^11 / 1. 6 × 10 −^10 )BrGeV
P= 0. 3 BrGeV/c
9.68 Let the proton be at a distancedfromS, the earth’s centre. Under the influence
of magnetic field it will describe an arc of a circle of radiusr.
From Fig. 9.9, it is clear that
(R+r)^2 =d^2 +r^2
Or 2Rr=d^2 −R^2 ≈d^2 (BecauseR<<d)
Thereforer=d^2 / 2 R=(1, 000 R)^2 / 2 R
= 5 × 105 R
The momentum,p= 0. 3 Br
P=(E^2 −m^2 )^1 /^2 =(10. 942 − 0. 942 )^1 /^2
= 10 .9GeV/c
B=p/ 0. 3 r= 10. 9 / 0. 3 × 5 × 105 × 6. 4 × 106