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Angles and Magnetic Fields

  • #1

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?

Homework Equations



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:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi inovermyhead! Welcome to PF! :smile:

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

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.
 
  • #3


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

About the problem:

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?
 
  • #4
tiny-tim
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Hi inovermyhead! :smile:

(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. :wink:
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)
 
  • #5
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?
 
  • #6
tiny-tim
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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? :smile:
 
  • #7
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? :smile:
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.
 
  • #8
tiny-tim
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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. :smile:
 
  • #9
*Bangs Head on Desk*

Okay. So it's 50...

Thank you for all of your help, I really appreciate it.
 

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