1000 Solved Problems in Modern Physics

9.3 Solutions 523

Fig. 9.10

9.72t=

2 usinα

α

=

2 usinα

Ee

.m

Where the accelerationα=Ee/m

t=

2 × 8 × 105 sin 30◦× 9. 1 × 10 −^31

50 × 1. 6 × 10 −^19

= 91 × 10 −^9 s=91 ns

9.73 The component of the velocity,⊥to the field isvsinθ. Equating the cen-
tripetal force to the magnetic force
mv^2 ⊥=qv⊥BR
R=mv⊥/qB=vsinθ/(q/m)B

=

3 × 105 sin 30^0

108 × 0. 3

= 0. 5 × 10 −^2 m

= 0 .5cm

9.74 p= 0. 3 BR=(0.3)(10−^4 )(6. 4 × 106 )=192 GeV/c
T≈192 GeV

9.75 Cosmic ray flux=1cm−^2 s−^1 = 104 m−^2 s−^1
Earth’s surface area,A= 4 πR^2 = 4 π(64× 105 )^2
= 5. 144 × 1014 m^2
Cosmic rays incident on earth’s surface
= 5. 144 × 1014 × 104
= 5. 144 × 1018 m−^2 s−^1
Cosmic rays energy delivered to earth
= 5. 144 × 1018 ×3GeVs−^1
= 1. 5432 × 1019 GeV s−^1
= 1. 5432 × 1019 × 1. 6 × 10 −^10 Js−^1
= 2. 47 × 109 W
= 2 .47 GW.