9.3 Solutions
521
depend on the composition of emulsions, z is the charge of the ion and M its
mass in terms of proton mass. The data on R and E for protons have been
used to determine n and R. We find n = 0.6 and K = 0.2276. Using these
values of n and K , the energy for various values of R have 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
D E(MeV) 0
3.06
6.31 34.2
45.6
57.0
68.6
80.1
91.4 102.6 113.7
3 He
0 46.2
70.0
89.3 106.2 121.4 135.4 148.5 160.9 172.7 184.0
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 of B Tesla perpendicular to the orbit, the radius r in metres
is related by the formula
P = qBr → cp = qBrc J
cp = (1.6 × 10
−19 )(3 × 10
8 ) Br J
= 4.8 × 10
−11 Br J
= (4.8 × 10
−11
/1.6 × 10
−10 ) Br GeV
P = 0.3 Br GeV/c
9.68 Let the proton be at a distance d from S, the earth’s centre. Under the influence
of magnetic field it will describe an arc of a circle of radius r .
From Fig. 9.9, it is clear that
(R + r )
2
= d
2
+ r
2
Or 2 R r = d
2
− R
2
≈ d
2 (Because R << d)
Therefore r = d
2
/2R = (1, 000 R)
2
/2R
= 5 × 10
5 R
The momentum, p = 0.3 Br
P = (E
2
− m
2 )
1/2
= (10.94
2
− 0.94
2 )
1/2
= 10.9 GeV/c
B = p/0.3r = 10.9/0.3 × 5 × 10
5
× 6.4 × 10
6