00:01
Using this equation and the definition of the period, find the correct expression for the period.
00:08
So for part a, we know that the orbital radius classically is mv over qb.
00:24
So that makes the period 2 pi r not over v, which is 2 pi m, the v.
00:40
Will cancel over qb.
00:45
So in the relativistic limit, r is equal to the momentum over qb, which is the momentum is gamma mv over qb, and this is gamma r0, which gives us a period of gamma 2 pi m over qb.
01:22
Which is gamma t knot, so this is t knot.
01:29
Part b is t independent of v.
01:37
It's not.
01:39
So it depends on, it's not independent of v.
01:55
Part c, if a 10 mega electron moves in a circular path in a uniform magnetic field of magnitude 2 .2 tesla, what are the radius, according to chapter 28, the correct radius, the period, according to 28, and the real period.
02:17
Okay, so the 10 mega electron volts is the kinetic energy.
02:23
So we know that the rest mass energy of an electron is 0 .511 mega electron volts.
02:34
So the classical kinetic energy is 1 .5 mb squared, which is 1⁄2 mc squared, b squared over 6.
02:43
Squared, which is one -half -m -c -squared, beta -squared.
02:54
So if the classical kinetic energy is 10 mega -electron volts, solving for beta, and we get 6 .256, since we know the rest -mass energy is 0 .511 mega -electron volts.
03:15
But a beta factor of 6 .256 is impossible because it is higher than the speed of light.
03:24
But to find the classical radius, let's use this energy...