Review Conceptual Example 2 as background for this problem....

Review Conceptual Example 2 as background for this problem. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is $3.80 \times 10^{3} \mathrm{N} / \mathrm{C}$ , while the magnetic field is 0.360 $\mathrm{T}$ the electric field is turned off, the charged particle travels on a circular path whose radius is 4.30 $\mathrm{cm} .$ Find the charge-to-mass ratio of the particle.

Key Concepts

Centripetal Force

In the context of circular motion, the centripetal force is the net force required to keep an object moving in a circle at a constant speed. For a charged particle undergoing circular motion in a magnetic field, the magnetic force provides the necessary centripetal force. This equivalence allows us to relate measurable quantities such as radius, magnetic field strength, and particle velocity.

Uniform Circular Motion in Magnetic Fields

When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force that is always perpendicular to its velocity. This perpendicular force does not change the speed of the particle but continuously alters its direction, resulting in uniform circular motion. The phenomenon is central to applications such as cyclotrons and mass spectrometers.

Charge-to-Mass Ratio

The charge-to-mass ratio (q/m) quantifies the relative amount of charge on a particle per unit mass and is a crucial parameter in characterizing its motion in electromagnetic fields. By equating the magnetic force (responsible for the centripetal acceleration) to the centripetal force, and by using experimental conditions like velocity and radius of curvature, one can derive an expression to calculate q/m. This concept is foundational in fields such as mass spectrometry and accelerator physics.

Lorentz Force

The Lorentz force is the combined force experienced by a charged particle moving in electric and magnetic fields. It is given by the vector sum of the electric force (qE) and the magnetic force (qv×B), and its direction depends on both the charge sign and the directions of the velocity and magnetic field. This concept is fundamental in understanding how charged particles behave under the influence of electromagnetic fields.

Velocity Selector

A velocity selector uses perpendicular electric and magnetic fields to filter particles by speed. In such a device, particles with a particular velocity experience equal and opposite forces from the electric and magnetic fields, resulting in zero net force, which allows them to pass through unaltered. This principle is essential in experiments and instruments where specific particle velocities are required.

Transcript

00:01 Okay, a good place to start with this one is to figure out what we want and how to get there.

00:08 Basically to draw a map of where we're going.

00:11 So what we want is the charge to mass ratio of a particle.

00:17 And things we're given are we have the electric field.

00:22 We have the magnetic field.

00:25 And we have a radius of circular trajectory.

00:30 We have a radius.

00:31 So, the first step is to find an equation that relates some of these variables, or some of these numbers that we're given.

00:43 So the first equation that we have is the equation for the radius, which is r radius is equal to mass times velocity divided by charge times magnetic field.

01:01 So this is the radius in a magnetic field, which is good because what we're given in the preroyal.

01:06 The problem is when the electric field is shut off, all we're left with is the magnetic field.

01:14 Okay.

01:16 So we have r and we have b.

01:19 This is what we want, and that's in the problem.

01:22 So let's go ahead and solve for that first.

01:26 That'll help us see what direction we need to go in.

01:28 So this equation becomes, well we can first off, get the m in the q alone.

01:37 I'm going to invert the equation.

01:43 I'm going to put the q and the m over on this side.

01:46 To get it by itself, we need to multiply both sides by b and divide both sides by v.

01:52 To do that, we end up, or after doing that, we end up with r times b because we had to multiply it to get rid of it on the mq side, divided by v.

02:03 Now, we want the inverse of this.

02:07 So we just flip both sides.

02:12 So now we have q up top, m underneath, and that also flips the other side.

02:19 These are just algebraic laws.

02:23 So now we flip this too.

02:25 So now we have v up top divided by rb.

02:34 Okay, so now we have the equation for what we want.

02:38 This is what we're trying to solve for.

02:40 So we have r and we have b.

02:42 We need to find v...