A proton travles with a velocity 8x10^5 m/s E=10^6 V/m and B= 1.5 T. Draw the path of the proton and what is speed of the proton when it comes out?
Here an image:
Well just replace with the values given...I'm stuck i don't know if that's correct or not... I decided do it with all variables and then replace them...
I guess i sould integrate the speed getting the position (x,y) but i'm not sure if that's correct...
I need help.. I HAVE to resolve this problem...
I need ideas you don't have to resolve all the problem.. just give me ideas!!
What is the shape of the path of a charged particle in a uniform magnetic field?
You correctly list the Lorentz force as the force that is acting on the proton in the magnetic field (there is no E field, right?). What does that Lorentz force do to the path of the proton? Draw the path, and plug in the numbers for the force. The force acts over a finite distance of the proton's trajectory -- what is the resulting trajectory when the proton leaves the B-field region?
But in the problem exist an E field... the trajectory doesn't have to be exact... i guess it's a Uniform Circular Movement.
If i'd do:
First for t=0 -> x=0 y=0.05
Finding t=? for x=0.3 y=0.05
That would be correct¿? then i replace that value of t in the formula of the speed... would i get the final speed (when it comes out) ??
The integrals are no so complicated... but in another forum someone told me that the proton never leaves and the path i guess it's not a straight line, because qE is diferent that q(vxB) [q(vxB) and qE are in opossite directions but they aren't equals to eliminate them...]