Recent content by Jeremy Higgins

Well, don't I feel dumb. I redid the above equation as: Magnetic Field = Centripetal Force / (Charge of the Particle ⋅ Velocity): .0035 = (2.977 ⋅ 10-16) / (1.6 ⋅ 10-19) ⋅ (5.255 ⋅ 105) This was marked as the correct answer! Thank you so very much!

Forgive me, I was not aware that I was supposed to look above the curve for the answer. Such a method is not obvious to me. But, it worked. The h value is 0.105m. Thank you very much for your help. I only have the magnitude of the Magnetic Field (Bz) left to find.

I tried to solve for the magnitude of the Magnetic Field (Bz) just now, by way of the following equations. Centripetal Acceleration = Velocity2 / Radius 1.783 ⋅ 1011 = (5.255 ⋅ 105)2 / 1.549 Centripetal Force = Mass ⋅ Centripetal Acceleration 2.977 ⋅ 10-16 = (1.783 ⋅ 1011) ⋅ (1.67 ⋅ 10-27)...

This is the picture that I've drawn. I realize now that the previous solution I found, 1.9 ⋅ 105, is literally Vy and not h. This explains why that was wrong. I do not know the proper way to solve the geometry of a curve like this. I tried the previous attempt again, using the initial Velocity...

So, I am now left with the challenge of finding two of the remaining parameters: 1. h, the height of the point where the particle leaves the Magnetic Field (Bz) 2. The magnitude of the Magnetic Field (Bz) itself. At this point, the magnitude of the Centripetal Force (F) is still...

Honestly, I do not know. Is it the hypotenuse of Vy/Vx? I.E. √(Vx2 + Vy2)? UPDATE: Well, I tried that as an answer and it worked! Thank you! One down, two more to go.

I suppose the Conservation of Angular Momentum applies here, and I do know that the Magnetic Field (Bz) does zero work, so that the Velocity (V) remains constant within the Magnetic Field. But I do not understand how I can draw the magnitude of the Velocity from this. There must be an equation...

My particular problem is that I don't know how to calculate the initial velocity (V), as stated in part a. I also do not know how to calculate parts b or c, but let's just start with a. I found the radius of the path from the following two equations: θ = tan-1(Vy/Vx) .3699 = tan-1(4.9⋅105 /...

Homework Statement A proton moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D in the x-direction. The proton leaves the field having a velocity vector (Vx...