In Chapter 19 we will see that a charged particle can...

In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle's velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle's speed, $F=k v .$ Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period. (c) Show that the radius of the particle's circular path is proportional to the speed.

In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle's velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle's speed, $F=k v .$ Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period.
(c) Show that the radius of the particle's circular path is proportional to the speed.

Key Concepts

Magnetic Force Perpendicularity

In the context of a charged particle moving in a magnetic field, the force exerted by the field is always perpendicular to the particle's velocity. This perpendicular arrangement is crucial because it means the magnetic force does no work on the particle, changing only the direction of motion rather than its speed, thereby enabling uniform circular motion.

Uniform Circular Motion

Uniform circular motion refers to the motion of an object moving in a circle at constant speed. In such motion, even though the speed remains constant, the particle is continuously accelerating towards the center of the circle, because the direction of its velocity vector is changing despite the constant magnitude.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path and is directed towards the center of the circle. By applying Newton's second law, this force equates to the mass of the object times its centripetal acceleration, which is v^2/r, and it is provided in this scenario by the magnetic force acting on the charged particle.

Period Independence of Speed

When the magnetic force acting on a charged particle is directly proportional to its speed, using F = k*v and setting it equal to the centripetal force (m*v^2/r) leads to the result that the period of the circular motion (the time required to complete one full circle) becomes independent of speed. This is because the increase in speed is offset by a proportional increase in the radius, keeping the ratio of the circumference to speed constant.

Radius Proportionality to Speed

By equating the magnetic force (F = k*v) to the required centripetal force (m*v^2/r), one derives that the radius of the circular motion is proportional to the particle's speed (r = m*v/k). This means that as the speed of the particle increases, its circular path's radius increases proportionally, a result directly stemming from the formulation of the magnetic force in this context.

Transcript

00:01 So in this problem, we're going to be investigating the motion of a charged particle in a magnetic field.

00:07 And we don't really need to know anything about magnetic fields or charged particles yet, but we just need to know that there's some mass and it's undergoing uniform circular motion.

00:19 Now, if that's true, we know that there needs to be some sort of force pointing towards the center of our circular motion.

00:30 And in this case, that will be the magnetic force, since we have no other forces.

00:36 That we know of.

00:39 So we can say that since the magnetic force points towards the center of the circle all the time, and the velocity is always tangential to the curve of the circle, that v is always perpendicular to magnetic force, or that the angle between them is 90 degrees or pi over 2.

01:11 Now, let's investigate something else.

01:15 In particular we're going to look at the period of one of these charged particles and we're going to show that it's actually independent of the velocity of the particle.

01:29 So if we apply newton second here, some of all forces in the radial direction, well the only force we have is the magnetic force and that ought to be equal to m times the radial acceleration since we're we have uniformed circular motion here or mv squared over r now the magnetic force is just proportional to the velocity so the magnetic force is some constant times velocity which is equal to mv squared over r and actually since we're going to be talking about the period here we'll use the expression for the radial acceleration omega squared r and of course omega can be written as 2 pi over the period.

02:45 So this will make it easy for us to get the period involved in our equation here.

02:49 So if we substitute that in, we'll have that kv equals m times 2 pi over t squared times r.

03:08 Now if we solve for the period of kv let's try to make a little more, be a little more efficient here, kv equals m4 x squared over t squared r multiply both sides by t squared t squared kv equals m45 squared r and then divide both sides by kv excellent now we have t squared isolated and if we substitute the relationship between v and omega in here, that is v equals r omega, we can simplify things even more.

04:12 T squared equals 4 pi squared m r over k r omega.

04:25 So we cancel out.

04:27 An r there, again, i probably should have waited to substitute for omega here because i ended up with another one...

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