Question

An accelerator produces a beam of protons with a circular cross section that is 1.6 mm in diameter and has a current of 1 mA. The current density is uniformly distributed throughout the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target.

1) What is the number density of the protons in the beam?

2) How many protons strike the target in each minute?

3) What is the magnitude of the current density J in this beam?

Answer 1

1.

Current is given by,

I = n*e*A*Vd

here, n = number density of the protons in the beam = ??

e = charge on a proton = 1.6*10^-19 C

A = Cross sectional area of a beam = pi*d^2/4

d = diameter of beam = 1.6 mm = 0.0016 m

I = current = 1 mA = 0.001 A

Vd = drift speed of proton = ?

given, kinetic energ of proton = 0.5*mp*Vd^2 = 20 MeV

where, mp = mass of proton = 1.67*10^-27 kg

So, Vd = sqrt[(20*10^6)*(1.60*10^-19)/(0.5*1.67*10^-27)]

Vd = 6.19*10^7 m/s

then, n = number density of the protons in the beam = I/(e*A*Vd) = 4*I/(e*pi*d^2*Vd)

n = 4*0.001/[(1.60*10^-19)*pi*(0.0016^2)*(6.19*10^7)]

n = 5.02*10^13 m^-3

2.

Now, current is also given by,

I = dQ/dt

here, dQ = charge pass through the beam = N*e

dt = time interval = 1 minute = 60 sec.

So, N = Number protons strike the target in each minute = I*dt/e

N = 0.001*60/(1.60*10^-19)

N = 3.7*10^17

3.

Now, current density is given by,

J = I/A

J = 4*I/(pi*d^2)

Using known values:

J = 4*0.001/(pi*0.0016^2)

J = 497 A/m^2

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