A proton moving at speed v=1.00 ·10^6 m / s enters a region in space where a magnetic field giv

step 1 You have a proton moving at a speed 10 to 6 meter per second. And it enters a region where the m

A proton moving at speed v=1.00 ·10^6 m / s enters a region...

A proton moving at speed $v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}$ enters a region in space where a magnetic field given by $\vec{B}=$ $(-0.500 \mathrm{~T}) \hat{z}$ exists. The velocity vector of the proton is at an angle $\theta=60.0^{\circ}$ with respect to the positive $z$ -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, $r$, of the trajectory projected onto a plane perpendicular to the magnetic field (in the $x y$ -plane). c) Calculate the period, $T,$ and frequency, $f$, of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

A proton moving at speed $v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}$ enters a region in space where a magnetic field given by $\vec{B}=$ $(-0.500 \mathrm{~T}) \hat{z}$ exists. The velocity vector of the proton is at an angle $\theta=60.0^{\circ}$ with respect to the positive $z$ -axis.
a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only).
b) Calculate the radius, $r$, of the trajectory projected onto a plane perpendicular to the magnetic field (in the $x y$ -plane).
c) Calculate the period, $T,$ and frequency, $f$, of the motion in that plane.
d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

Key Concepts

Pitch

This is the linear distance a charged particle travels along the direction of the magnetic field during one complete cycle of its circular motion. It is determined by the component of the particle's velocity parallel to the magnetic field and the period of the circular motion.

Cyclotron Motion

This refers to the circular motion of a charged particle in a plane perpendicular to a uniform magnetic field. The radius and period of this motion are determined by the balance between the centripetal force required for circular motion and the magnetic force, leading to definitions of cyclotron frequency and radius that characterize the motion.

Lorentz Force

This is the fundamental force experienced by a charged particle moving in electric and magnetic fields, described by F = q(v × B). It explains how the magnetic field causes the charged particle to accelerate perpendicular to both its velocity and the magnetic field direction, resulting in curved or circular motion without doing work on the particle.

Helical Motion

When a charged particle moves at an angle to a uniform magnetic field, its velocity can be decomposed into two components: one parallel to the magnetic field and one perpendicular to it. The perpendicular component undergoes circular motion, while the parallel component moves linearly. The combination of these two motions results in a helical trajectory.

Transcript

00:01 You have a proton moving at a speed 10 to 6 meter per second.

00:08 And it enters a region where the magnetic field is pointing in the minus z hat direction.

00:17 And we want to analyze the motion of the proton.

00:22 So let's just write down some constants real quick.

00:25 So we know that v equals 1 times 10 to the 6 meters per second.

00:35 We know that b equals minus 0 .5 tesla in the z hat direction.

00:52 And then that theta equals 60 degrees as well.

00:58 Okay, so i've drawn a picture here.

01:00 And we have our proton coming up.

01:07 I've drawn it off of the x -axis.

01:11 Doesn't really say in the problem description but the motion will be similar so part a we're supposed to just analyze the motion of the proton and describe this trajectory so let's think about the force on the proton so the force is equal to charge times the velocity crossed with the magnetic field okay so let's say that the velocity is as we have it drawn here.

01:52 So if we do this cross product with a right -hand rule of qv, cross b, which means that the force would be in plus y direction, or into the page in this case.

02:08 And then as it moves in that direction, the force will be constantly changing such that the proton will spiral.

02:17 It will form a circle in the xy plane, but it will also continue moving up in the z direction.

02:26 So the motion will be a helix.

03:11 And the reason that the velocity component in the z direction isn't affected, the force acting on the proton only acts in the x, y plane.

03:27 It doesn't affect the z motion.

03:30 If you do the cross product with the just thinking about the component in the z direction you would see that there's actually no force because of that component of velocity okay so we have it's circular in the x -flip plane as it moves in the plus z -hat direction okay and calculate the radius of the trajectory projected onto the x -way plane okay so to find the radius we want to look at the forces on the proton we'll ignore gravity here so some of the forces equal to mass and acceleration.

04:21 In this case the acceleration is centripetal because it's moving in a circle.

04:26 So the force is kv cross b and that's equal to the mass times the velocity squared over the radius.

04:44 Okay, and this velocity here is the velocity in the xy plane.

04:49 So this is not the velocity in the z direction.

04:54 So this is some component of velocity that will take care of in a second.

04:59 Okay.

05:04 So let's write down our velocity is what we wrote down already.

05:17 Multiply by, so it's one point, or one times 10 to the 6 meter per second.

05:23 And then in the x direction, it would be times cosine theta.

05:31 We'll say initially it has no component in the y direction and then in the z direction it would be sine theta so let's do this cross product okay so we have qv cross with b okay so in the x -hat direction we have qv cosine theta in the y hat direction zero and in the z -hat direction we have qv sine theta, and we have our magnetic field there.

06:45 Okay.

06:47 So if we do this cross product, we get bqv cosine theta...

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