(II) A 5.0-MeV (kinetic energy) proton enters a 0.20-T field, in a plane perpendicular to the field. What is the radius of its path?

Question

mintiruki · Answer

@abb0t

mintiruki · Answer

@abb0t nvm, someone recommended you to me but it seems you're more a chemistry guy.  Sorry about that!

souvik · Answer

|dw:1430252739131:dw|
first find the velocity of the electron...the K.E. is given
force on electron=evB.............e is the charge of the electron
and this force produces the centripetal acceleration..
so 
evB=mv^2/r 
r is the radius of the circle

isaiah.feynman · Answer

Or you can create one single equation to give you the answer in one go. ;)

mintiruki · Answer

@Isaiah.Feynman , thanks for the response! What does "k" represent? And also, why does v=sqrt2k/m?

mintiruki · Answer

@Isaiah.Feynman Nvm I get it. Thank you so much!

mintiruki · Answer

@Souvik, so I would just use the mass of a proton and plug that in to get the velocity right?

isaiah.feynman · Answer

Yes, m is the mass of the proton.

souvik · Answer

the equation is $$K.E=1/2mv^2$$
here K.E is given in MeV unit...u need to convert it to the J unit ....then find the velocity