Recent content by woaname

yes BruceW, it just took a little doodling and experimenting, but i figured out the question. thanks for your input :D, and CWatters, thanks to you too

i found θ=24 using arctan(vy/vx), and the trace of the radius with the center of the circle in the top left of the square, as the picture shows.

i realized that i didn't post the questions that needed answering, so here they are: 1)What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? 2)What is h, the y co-ordinate of the proton as it leaves the region conating the...

but getting back to the question, how do i get around solving the questions? so the velocity is towards positive x direction, force is towards positive y, and magnetic field is pointing into screen (negative z direction). if i use a consolidated equation v = (rQB)/m , i have two unknowns.

OH! FLASHBACK! i just remembered an unconsciously learned moment from class. our prof actually DID say that negative and positive charges will change the RHR directions. thanks sandy.bridge for reminding me. won't forget it for my test tomorrow :D

using this picture, the magnetic field is represented by the middle finger. but before, in other questions, the direction of the thumb indicated the magnetic field. why does the representation change?

it was found to be negative.

is there an alternate version to the right hand rule? i pointed my index finger towards velocity vector, and the force vector inwards using my middle finger, which gave me the magnetic field pointing outward. i understand that the positioning would be different each time, but is there any...

lols sorry, I am using excel to do my math. got the numbers sorted out. but i had a general question: shouldn't the negatives be a part of the answer by default, or would i add the negative direction using intuition? I'm asking because the same sign issue carried over to the final question...

so, B will point out of the screen, and F in positive y direction (initially). i've tried using pytagoras' theorem to get a value for Vnet, but that would be for the exit velocity, not the velocity on entrance...?

1.72E+03 m/s was the answer marked correct for a preceding question. using F= mvsquared/r, i got a force magnitude 1.33E-17. the x component was 1.15E-17, and y component=6.67E-18. i manually added the negative directions.

i can't even tell what direction the magnetic field is in

i gather both components will have negative direction then? sorry, our prof isn't very helpful and he didn't even give us a hint on thinking like this, with triangles. so i got Fnet= (m*v^2/r), and multiplied by cos(30) and sin(30) for the x and y components, respectively, but my answers are...

Homework Statement A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) 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 = 0.7 m in the x-direction. The...

in fact it is this first step where I'm unclear. i can't apply a square coordinate system, since the movement is circular. but winging it, if i set (90 degrees/ x)=(565/188.3), i get x = 30 degrees. is that close?