A magnet produces a 0.30-T field between its poles, directed...

A magnet produces a 0.30-T field between its poles, directed to the east. A dust particle with charge $q=$ $-8.0 \times 10^{-18} \mathrm{C}$ is moving straight down at $0.30 \mathrm{~cm} / \mathrm{s}$ in this field. What is the magnitude and direction of the magnetic force on the dust particle?

Key Concepts

Lorentz Force Law

This fundamental principle states that a charged particle moving in a magnetic field experiences a force given by F = q(v × B), where q is the charge, v is the velocity vector, and B is the magnetic field vector. The magnitude of the force depends on the charge's speed, the magnetic field strength, and the sine of the angle between the velocity and the field. This concept is central to understanding the behavior of particles in magnetic fields across physics and engineering.

Vector Cross Product

The magnetic force is determined by the cross product of the velocity and magnetic field vectors. The cross product not only provides the magnitude—via the product qvB sin?—but also the direction of the force, which is perpendicular to both the velocity and the magnetic field. This operation is a key mathematical tool in vector calculus and physics when dealing with rotational and directional phenomena.

Right-Hand Rule and Charge Sign

The direction indicated by the cross product is determined using the right-hand rule: when the fingers point in the direction of the velocity and curl toward the magnetic field, the thumb points in the direction of the force for a positive charge. For negative charges, the force direction is opposite. Understanding this concept is crucial for predicting how the sign of a charge affects the movement of particles in a magnetic environment.

Unit Conversion and Magnitude Computation

Accurate calculation of the magnetic force requires converting units where necessary (e.g., converting centimeters per second to meters per second) and correctly computing the multiplication of the charge magnitude, speed, and magnetic field strength. This ensures that the computed force has the proper magnitude and units, an essential practice in all physics problem-solving.

Transcript

00:01 In this question, we know that the magnetic field has a magnitude of 0 .30 tesla's.

00:10 And they did tell us that the magnetic field is directed east.

00:16 So i'm just going to draw several magnetic field lines directed east so that i can create a diagram for this problem later.

00:26 And they also tell us that there is going to be a charge of magnitude 8 .0 times 10 to the negative 18th coulum.

00:38 And they did specify that this is a negative charge.

00:42 And they also said that this charge is moving with a velocity of 0 .30 centimeters per second, which is converted to meters per second as 0 .00.

00:56 And they said that this velocity is straight downward.

01:03 So if i were to draw this charge q, then it has a velocity vector that is straight down.

01:11 And they're asking for the magnitude and the direction for the magnetic force.

01:17 So because there is a 90 degree angle between the velocity vector and the magnetic field vector, it simplifies the formula for magnetic force to be q times v times b.

01:35 And we're only looking for the magnitude so far with this equation.

01:40 So i'm not going to worry about the negative sign as i plug things in.

01:44 So q is just going to be 8 .0 times 10 to the negative 18th.

01:49 And v will be 0 .003 meters per second.

01:53 And b will be 0 .3.

01:57 And so to get all of that into the calculator, i find that the magnitude of the magnetic force is 7 .2 times 10 to the negative 21st newton's.

02:09 This is only magnitude.

02:11 How do i figure out the direction? well, that's where our diagram here will help us...