00:01
This is chapter 27 problem number 16.
00:03
We are given a proton that is moving along a circular trajectory from a point a to b.
00:10
And we know the initial velocity.
00:14
It's 1 .41 times 10 ,000 per second.
00:18
And, well, this circular motion is due to the magnetic force acting on the proton.
00:27
So as you know, the direction has to be along the centipedal direction so that with the initial velocity, the initial speed, the particle would go along the circular trajectory.
00:40
So this is the direction of f sub b.
00:42
We are asked to calculate the magnitude and the direction of the magnetic field.
00:50
So just simply from right -hand rule, we can say that the direction of b, direction of b has to be out of the ward or the screen, right, for a positive charge.
01:08
And in order to determine the magnitude of it, as you know, your f sub b is qvb, since there is 90 degree angle between the velocity vector and the force.
01:21
I'm adding sign 90 here, which is one.
01:23
This has to be equal to from newton's second law, mv squared over r when you have a circular motion.
01:29
So from here, if we want to go to the magnitude of the magnetic field, then we divide both sides by a qv.
01:37
Qv, these are going to cancel up.
01:39
You're going to have the magnetic field to be.
01:42
These are going to cancel mv over qr...