A particle with charge 9.45 ×10 ^ - 8 C is moving in a region where there is a uniform magne

A particle with charge 9.45 ×10 ^ - 8 C is moving in a...

A particle with charge $9.45 \times 10 ^ { - 8 } \mathrm { C }$ is moving in a region where there is a uniform magnetic field of 0.650 T in the $+ x$ -direction. At a particular instant of time the velocity of the particle has components $v _ { x } = - 1.68 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } , v _ { y } = - 3.11 \times$ $10 ^ { 4 } \mathrm { m } / \mathrm { s } ,$ and $v _ { z } = 5.85 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } .$ What are the components of the force on the particle at this time?

A particle with charge $9.45 \times 10 ^ { - 8 } \mathrm { C }$ is moving in a region where there is a uniform magnetic field of 0.650 T in the
$+ x$ -direction. At a particular instant of time the velocity of the
particle has components $v _ { x } = - 1.68 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } , v _ { y } = - 3.11 \times$ $10 ^ { 4 } \mathrm { m } / \mathrm { s } ,$ and $v _ { z } = 5.85 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } .$ What are the components of
the force on the particle at this time?

Key Concepts

Lorentz Force Law

The Lorentz force law describes the force experienced by a charged particle moving in electric and magnetic fields. In the case of a purely magnetic field, the force is given by F = q (v × B), indicating that the force is proportional to the charge, the velocity of the particle, and the magnetic field. This force acts perpendicular to both the velocity of the particle and the magnetic field, leading to circular or helical motion depending on the initial velocity direction.

Vector Cross Product

The cross product is a mathematical operation on two vectors that results in a vector perpendicular to both of the original vectors. In the context of a charged particle moving in a magnetic field, this operation is used to determine the direction and magnitude of the magnetic force acting on the particle. The properties of the cross product, such as its anticommutativity and the use of the right?hand rule for determining direction, are pivotal in solving problems involving magnetic forces.

Transcript

00:01 So in this question, we have a particle with charge 9 .45 times 10 to the minus 8 coulums.

00:07 It's moving in a region where there is a magnetic field of 0 .65 teslas in the positive x direction.

00:17 At a particular instance of time, the velocity of the particle has components vx, vy, and vz given in the question.

00:26 And we're asked to calculate the components of the force on the.

00:31 The particle at this particular time.

00:34 So again, we're trying to find the magnetic force, which is, again, kind of a weird force because it is based on the cross product.

00:45 So it's based on the cross product of the magnetic field, which we know is 0 .65 tesla's in the positive x direction.

00:57 So i'm going to write i had here because that's the unit vector in the, the in the positive x direction.

01:04 So the cross product between b and v as well.

01:08 So i'm also going to write v here as a vector with the individual unit vectors, i hat, j hat, and k hat, just because i find it a little bit easier to work with that way.

01:28 So we've got that x component i hat plus the y component j hat plus the z component k hat.

01:46 So this is the full velocity written out using those unit vectors i j and k.

01:52 Of course, you could also write it out as a tuple of three numbers as well.

02:00 But i just prefer to do it that way.

02:04 And then we can calculate the magnetic force by taking the cross product of v and b and multiplying by q.

02:16 Now there's a couple different ways that you can do the cross product of two different vectors.

02:21 I prefer it to do it with the unit vectors as i've written here.

02:26 But you also have the method where you write out the different vectors and then you cross, do the cross multiplication and things like that.

02:39 You can do it like that.

02:39 But i prefer to do it with the unit vectors, especially if one of the vectors only has one component like b does here.

02:50 So i'm just going to leave q as q for now, so i don't have to keep writing it.

02:54 And then i'm going to fill in my v and my b.

03:00 So v is a bit cumbersome.

03:03 I'm just going to leave out the meters per second at this point, because we all know there's meters per second there.

03:12 Okay, and then that is crossed with the b field.

03:22 So 0 .65 tesla's ihat.

03:27 So what happens here when you do the cross product, you can just kind of distribute the cross product in here, right? so you're going to get q, and then the first thing will be negative 1 .68 times 10 to the 4 times the magnitude of the b field and then that's going to be i -hat cross i -hat right because we're taking that b field and we're just distributing it inside here so that's going to be the consequence of that and then we'll do this one and this one so the next one will be negative 3 .11 times 10 to the 4 0 .65 teslas, j -hat, cross -i -hat.

04:22 And then we have one more...

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