In the Superconducting Super Collider that was planned to have been built in Tex

Question

In the Superconducting Super Collider that was planned to have been built in Texas this past decade, protons would be bent by a magnetic field and would move in a circle of radius 12.88 km. Magnetic fields of magnitude 4.44 T would have been used to bend the protons. What is the magnitude of the momentum, p, of a proton that moves at a radius of 12.88 km in this field? You may assume that the magnetic field involved is perpendicular to the motion of the charges (that's the most efficient configuration).

Explanation / Answer

here,

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

Charge on proton, q = 1.6 * 10^-19 C

radius of path, r = 12.88 km = 12880 m

Magnatic field, b = 4.44 T

From newton Second Law, SUM(F) = 0

Magnatic force - Cenrtripital force = 0
qvB = mv^2/r

solving for Velocity, v

v = qB*r/m
v = ( 1.6 * 10^-19 * 4.44 * 12880)/ ( 1.67 * 10^-27 )
v = 5.479*10^12 m/s

So, momentum = mass * Velocity
P = m*v
p = 1.67 * 10^-27 * 5.479*10^12
P = 9.15*10^-15 Kg.m/s

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