Question

An antiproton is identical to a proton except it has the opposite charge, −e. To study antiprotons, they must be confined in an ultrahigh vacuum because they will annihilate−producing gamma rays−if they come into contact with the protons of ordinary matter. One way of confining antiprotons is to keep them in a magnetic field. Suppose that antiprotons are created with a speed of 4.0 × 104 m/s and then trapped in a 4.0 mT magnetic field.

Answer 1

Answer:

d = 1.13*10^{-4}m = 0.113mm

Explanation:

To find the minimum diameter, that allow to antiproton circulate in the chamber without touching the walls, you use the following formula for the radius of the trajectory of a charged particle in a constant magnetic field.

$r=\frac{mv}{qB}$   (1)

r: radius of the trajectory

m: mass of the antiproton = 9.1*10-31 kg

v: velocity of the antiproton = 4.0*10^4 m/s

B: magnitude of the magnetic field = 4.0mT = 4.0*10^-3 T

q: charge of the antiproton = +1.6*10^{-19}C

You replace the values of the parameters in (1):

$r=\frac{(9.1*10^{-31}kg)(4.0*10^4m/s)}{(1.6*10^{-19}C)(4.0*10^{-3}T)}\\\\r=5.68*10^{-5}m$

Then, the diameter of the chamber must be, at least:

d=2r = 2(5.68*10^-5) = 1.13*10^{-4}m = 0.113mm

Answer 2

The minimum diameter that should vacuum chamber have to permit is 0.113mm.

Calculation of minimum diameter:

Here the below formula should be used.

r = mv/qB

Here

r means the trajectory radius

m means the antriproton mass i.e. $9.1\times 10^{-31} kg$

v means antiproton velocity i.e. $4.0\times 10^{4} m/s$

B means magnetic field magnitude i.e. 4.0mT = $4.0\times 10^-3$

Tq means antriprotion charge = $+1.6\times 10^{-19}C$

Now the diameter is

$= (9.1\times 10^{-31}) \times (4.10\times 10^{4}) \div (1.6\times 10^{-10}C) (4.0\times 10^{3}T)\\\\= 5.68\times 10^{-5} m$

Since diameter = 2r

So, radius is = $2(5.68\times 10^{-5}) = 1.13\times 10^{-4}m$

= 0.113mm

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