# Angles and Magnetic Fields

In summary, when a particle is moving with an instantaneous velocity in a magnetic field, the magnitude of the force exerted on it by the field is proportional to sin(angle between the particle and the field). The direction of the force is perpendicular to both the velocity vector and the direction of the field.

## Homework Statement

A particle with a charge of −2.50×10-8 C is moving with an instantaneous velocity of magnitude 40.0 km/sec in the xy-plane at an angle of 50.0° counterclockwise from the +x axis.

1) What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the -x direction?

2) What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.

3) What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?

F=|q|VBsin(θ)

## The Attempt at a Solution

I thought 1) would be: 0 N because the force is in the -z direction and I have no idea conceptually what I doing wrong here.
And I thought part 3) was (-2.50x10-8)(4x105)(2)(sin(50)) but I was wrong and I'm not sure why.
Part 2) I have no idea what to do or how to get there.

I'm completely new here, but I've looked at past question and that has really helped a lot, so I was hoping more interaction would be even better.

Thanks guys, and please be as detailed as you can, I'm not sure why I'm not getting this...

Last edited:
Welcome to PF!

(have a theta: θ and try using the X2 tag just above the Reply box )

The force from a magnetic field B is q(v x B).

In particular, the magnitude in 1) isn't zero (it will never be zero unless θ is zero), and the angle in 3) isn't 50º.

Thank you for the quick link, I hope I fixed all the formatting errors.

What happens to the force if it isn't "flowing" in that direction?
I get that it's not zero but if it isn't in that direction, what's going on with it?

I know the force is F=|q|VBsin(θ), but what are you doing to the angle if the force is -z and they are asking about +z?

(just got up :zzz: …)
What happens to the force if it isn't "flowing" in that direction?
I get that it's not zero but if it isn't in that direction, what's going on with it?

Think of the particle as "cutting through" the magnetic field lines …

if it's moving parallel to them, it doesn't cut any, so the force is zero …

if it's moving perpendicular to them, the force is maximum …

generally, the force is proportional to sine of the angle.
I know the force is F=|q|VBsin(θ), but what are you doing to the angle if the force is -z and they are asking about +z?

θ = 180º, so sinθ = 0, and the force is zero, same as if they were both +z.

(but that's not the case in any of your three questions)

Errr... okay:

I know that one is: (2.5x10-8)(4x104)(2)(sin(50))
**I had messed up my km to m conversion, and feel remarkably stupid***
and
I know that three is: (2.5x10-8)(4x104)(2)(sin(90))

but I'm still unsure what to do with part 2.
I thought it would be 230° from the y-axis and was wrong and now I'm at a complete loss. So am I just getting my angles wrong?

A particle with a charge of −2.50×10-8 C is moving with an instantaneous velocity of magnitude 40.0 km/sec in the xy-plane at an angle of 50.0° counterclockwise from the +x axis.

2) What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.
… but I'm still unsure what to do with part 2.
I thought it would be 230° from the y-axis and was wrong and now I'm at a complete loss. So am I just getting my angles wrong?

Let's see …

the force will be in the direction of q(v x F),

which is the direction of -v x z,

so it's perpendicular to both v and z, either up-left or down-right.

Which?

tiny-tim said:
Let's see …

the force will be in the direction of q(v x F),

which is the direction of -v x z,

so it's perpendicular to both v and z, either up-left or down-right.

Which?
Down right is what I assumed it was, which makes sense to me, at the 230 from the +y-axis, but it wasn't right.

Down right is what I assumed it was, which makes sense to me, at the 230 from the +y-axis, but it wasn't right.

No, because it's in the direction of -v x z, which is z x v;

and z x x = y, so "z x " rotates x (and anything else in the plane) anti-clockwise …

so it'll be up-left.