Question

The Tevatron at Fermilab is a cyclotron capable of producing a maximum energy of 2.0 TeV (2×1012eV). Its

circumference is 6.4 km. (a) How fast would proton be moving in the Tevatron? How large of a magnetic

field is required? (b) Repeat part (a) using electrons instead

of protons.

Answer 1

a)

m = mass of proton = 1.67 x 10^-27 kg

v = speed of proton = ?

KE = kinetic energy of proton = 2 x 10¹² eV = 2 x 10¹² x 1.6 x 10^-19 J = 3.2 x 10^-7 J

Using the equation

KE = (0.5) m v²

3.2 x 10^-7 = (0.5) (1.67 x 10^-27) v²

v = 1.96 x 10¹⁰ m/s

B = magnitude of magnetic field

q = charge on proton = 1.6 x 10^-19 C

L = circumference of the circle = 6.4 km = 6400 m

r = radius = ?

Circumference of the circle is given as

L = 2r

6400 = 2 (3.14)r

r = 1019.11 m

For cyclotron

r = mv/(qB)

1019.11 = (1.67 x 10^-27) (1.96 x 10¹⁰)/((1.6 x 10^-19)B)

B = 0.2 T

b)

m = mass of electron = 9.1 x 10^-31 kg

v = speed of electron = ?

KE = kinetic energy of electron = 2 x 10¹² eV = 2 x 10¹² x 1.6 x 10^-19 J = 3.2 x 10^-7 J

Using the equation

KE = (0.5) m v²

3.2 x 10^-7 = (0.5) (9.1 x 10^-31) v²

v = 8.4 x 10¹¹ m/s

B = magnitude of magnetic field

q = charge on electron = 1.6 x 10^-19 C

L = circumference of the circle = 6.4 km = 6400 m

r = radius = ?

Circumference of the circle is given as

L = 2r

6400 = 2 (3.14)r

r = 1019.11 m

For cyclotron

r = mv/(qB)

1019.11 = (9.1 x 10^-31) (8.4 x 10¹¹)/((1.6 x 10^-19)B)

B = 0.0047 T