How Is Force Calculated on a Proton in a Magnetic Field?

  • Thread starter Thread starter jimmyting
  • Start date Start date
  • Tags Tags
    Force Vectors
AI Thread Summary
The discussion focuses on calculating the force exerted on a proton in a magnetic field of 1.47 T, with the proton's velocity given in the x-y plane. The initial approach involved using the sine of the velocity vector, leading to confusion about the correct angle to use for the cross product. Participants clarified that the angle should be between the magnetic field direction (z-axis) and the x-y plane, not simply derived from the velocity components. A suggestion was made to draw a three-dimensional diagram to better visualize the vectors involved. The conversation highlights the importance of accurately representing vector angles in physics calculations.
jimmyting
Messages
13
Reaction score
0
[SOLVED] Force dealing with vectors

Homework Statement


A uniform magnetic field of magnitude 1.47 T acts in the positive z direction.
Find the magnitude of the force exerted by the field on a proton if the proton's velocity is <3Mm/s i + 3Mm/s j>. Answer in pN

Homework Equations


F=QV \times B

The Attempt at a Solution



Being a cross product, I thought that it would just take the sin of the velocity vector. Since it is going at the same speed in x and y, the angle I calculated was 45, so it is 135 from the direction of the field. So after doing all that calculation, I get .7056. Where am I doing this wrong?
 
Last edited:
Physics news on Phys.org
The magnetic field is in the z direction; the velocity is in the x-y plane. So the angle you need is the angle between the z direction and the xy plane. What do you get?
 
jimmyting said:
Since … the angle I calculated was 45, so it is 135 from the direction of the field.

Hi jimmyting! :smile:

Usually I'd recommend "draw a diagram"

but I guess that it's your diagram that's misled you.

You've got three arrows for, say v y and z, and you've correctly labelled the angles vy and yz as 45º and 90º. :smile:

And then you've just added them, because your diagram doesn't look three-dimensional.

hmm … take art classes … or see http://en.wikipedia.org/wiki/Perspective_(graphical) :smile:
 
Haha, sadly Art isn't my forte, neither is music for that matter, but thank you guys for your help
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top