A beam of protons, each with rest mass about 1.6 * 10^-27 kg, are accelerated to near-light-speed velocities before entering a magnetic field of .5 Tesla. The magnetic field is directed perpendicular to the direction of the beam. Without taking account of the relativistic effect on mass, what should be the radius of curvature the protons if they are moving at 2.52 * 10^8 m/s as the beam passes through the magnetic field?
For reasons related to time dilation and length contraction, as a massive particle approaches the speed of light an observer in the 'lab frame' will conclude that under a constant applied force the particle will accelerate at a decreasing rate. The observer will therefore conclude that the mass of the particle is increasing with its velocity.
Without presenting the details, we will simply state that the 'mass increase factor' is identical to the 'time dilation factor' 1 / `sqrt( 1 - ( v / c ) ^ 2 ), so that if m is the 'rest mass' of the object (the mass as observed when stationary with respect to the observer), its mass at velocity v with respect to the observer is m' = m * [ 1 / `sqrt( 1 - ( v / c ) ^ 2 ].
The force exerted on a proton by the given magnetic field is
(note that the field and the velocity are perpendicular so sin(`theta) = 1 and we can just use F = q v B ).
If we take the relativistic mass of the proton into account we find that the force F = 2.016 * 10^-11 N acts on a mass of
The centripetal acceleration will thus be
which results in radius of curvature
As v approaches c, v^2 / c^2 approaches 1 and the relativistic mass approaches infinity. As a result the centripetal acceleration approaches zero, and the radius of the circle approaches ininity. Thus the path approaches a straight line.
At velocity v in magnetic field B, oriented perpendicular to v, `theta will be 90 degrees so a charge q will experience force
If the charged particles have rest mass m and are not moving at relativistic speeds we will thus have centripetal acceleration
Setting centripetal acceleration equal to v^2 / r we can solve for r:
so
If the velocity is relativistic then the mass is m / `sqrt( 1 - v^2 / c^2 ) so
which is 1 / `sqrt( 1 - v^2/c^2) times the radius that woud be predicted ignoring relativistic effects.
As velocity approaches c, v^2/c^2 approaches 1 so 1 - v^2/c^2 approaches zero. Thus the denominator of the mass increase factor 1 / `sqrt( 1 - v^2/c^2) approaches 0, and the factor approaches infinity. That is, there is no limit to the increase of the mass. As the mass becomes larger and larger the centripetal acceleration aCent = F / m will approach zero, and the path of the particle will approach a straight line.
"