An alpha particle (a Helium nucleus with 2 neuturons and 2 protons) has a charge

Question

An alpha particle (a Helium nucleus with 2 neuturons and 2 protons) has a charge of +2e and a mass four times larger than a proton, or 6.64×1027 kg. It is moving at a velocity of 2.8×106 m/s as shown in the figure when it enters a region (shaded in grey) of constant magnetic field. It moves in a semicircle of radius, R = 20 cm. Begin by drawing a free body diagram of the alpha particle showing the direction of the force just as it enters the grey area of constant field. (Ignore gravity)

A. What is the magnitude of the acceleration of the alpha particle just as it enters the grey area?

B. What is the direction of the acceleration of the alpha particle just as it enters the grey area?

C. What is the magnitude of the force on the alpha particle just as it enters the grey area?

D. What is the direction of the force on the alpha particle just as it enters the grey area?

E. What is the magnitude of the magnetic field inside the grey area?

F. What is the direction of the magnetic field inside the grey area?

G. If a single proton (charge = +e, mproton=1.67×1027 kg ~ malpha/4 ) enters the grey area with exactly the same velocity, what is the radius of its motion?

Explanation / Answer

a)

For centripetal motion,

a = v²/R

a = (2.8 x 10^6 m/s)² / (0.02 m)

a = 3.92 x 10^14 m/s²

b) It's accelerating down in the first instant, towards the center of the circle.

c) F = ma = (6.64 x 10^-27 kg) (3.92 x 10^14 m/s²) = 2.602 x 10^-12 N

d) Down, in the direction of the acceleration.

e)

F = q v B, so

B = F / qv = (2.602 x 10^-12 N) / (3.2 x 10^-19 x 2.8 x 10^6) = 2.9040 T

f) Using the right hand rule: force is down (at the first instant), velocity is to the right (at the first instant), so the field is pointing out of the page/screen.

g) Work backwards through the equations we used. B and v are the same. If q is halved then F must be halved. If F is halved and m is quartered then a must be doubled. v is the same. If a is doubled then R must be halved. The radius is 1/2 R.

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