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9 Particle Physics – I
9.76 p = 0.3BR
(1)
where p is in GeV/c, B in Tesla and R in metres.
Now θ ≈ L/R = 0.3 BL/ p
(2)
where L is the length of the magnet, θ is the angle of deflection, R is the radius
of curvature of the circular arc of the path in the magnetic field, and l is the
length of the straight path from the slits (Fig. 9.11).
Δθ = s/l = 0.3 BLΔ p/ p
2
Δ p/ p = 1/100
S = 0.3 × 1.2 × 1.5 × 10 × (1/100)(1/25)
= 2.16 × 10
−3 m = 2.16 mm
Fig. 9.11
9.3.11 Betatron
9.77 ΔT = eΔϕ/Δt = 1.6 × 10
−19
× 50 J
= 50 eV
9.78 (a) If N is the number of revolutions then
(2πRN )(4 f ) = c
The factor 4 arises due to the fact that the duty cycle is over a quarter of a
period.
N = c/8π Rf = 3 × 10
8
/8π × 0.9 × 50 = 2.65 × 10
5
(b) Radius, R = 0.9 m
T max = BRec = 1.2 × 0.9 × 1.6 × 10
−19
× 3 × 10
8
= 5.184 × 10
−11 J
= 5.184 × 10
−11
/1.6 × 10
−13
= 324 MeV
(c) The average energy gained per revolution
ΔT = T max /N = 324 MeV/2.65 × 10
5
= 122.2 × 10
−5 MeV
= 1.222 keV
9.3.12 Cyclotron
9.79 The resonance condition is
ω = qB /m