1000 Solved Problems in Modern Physics

522 9 Particle Physics – I

= 1. 13 × 10 −^11 T

= 1. 13 × 10 −^7 G

Fig. 9.9

9.69 Electric force=qE
Magnetic force=qvB
If these two forces cross, that is
qE=qvB
then the condition for null deflection is
v=E/B
Nowv = (2T/m)^1 /^2 = c(2T/mc^2 )^1 /^2 = c(2× 20 / 940 × 1 ,000)^1 /^2 =
0. 00652 c
E= 500 V/cm= 5 × 104 V/m
ThereforeB=E/v= 5 × 104 / 0. 00652 × 3 × 108 = 0 .0256 T=256 G

9.70 p^2 =^2 mT=^2 mqV=^2 mqEd (1)

Alsop=qBr (2)

Combining (1) and (2)

m=qB^2 r^2 / 2 Ed (3)

Furtherd=at^2 / 2 =qE t^2 / 2 m (4)

Using (3) in (4) and solving fort,

t=Br/E

9.71 Drop a perpendicular DA at D and extend the path DP to meet the extension
of the initial path HC in E. From the geometry of the figure (Fig. 9.10) angle
CAD=θ. Drop a perpendicular DG on CA. In the triangle AGD, sinθ=
GD/AD=d/R, whereRis the radius of curvature.

R=

√

2 mK

qB

=

√

2 × 1. 67 × 10 −^27 × 5 × 105 × 1. 6 × 10 −^19

1. 6 × 10 −^19 × 0. 51

≈ 0 .2m.

sinθ=d/R= 0. 1 / 0. 2 = 0. 5

θ= 30 ◦