Period of proton orbit in 0.758T magnetic field

[ballahboy]

A proton moves in a circular orbit perpendicular to a uniform magnetic field of 0.758T. Find the time it takes the proton to make one pass around the orbit.

Just need someone to help me start off. All I have is that the B=0.758T, q=1.6x10^-19C and m=1.67x10^-27kg. What formulas do I need to use to find t?
Thanks

 

[dextercioby]

U can prove (i advice u to do it) that

$$r=\frac{mv}{qB}$$

Exploit the connection beteen T,v & r.

Daniel.

 

[ballahboy]

hmm so r=mv/qB is derived from mv^2/r=qvB. I got r=(1.4x10^-8)v. the thing is.. i don't kno the relationship between the time, radius and velocity. Do i use r to find the circumference and multiply that by the velocity? I am not really sure..
thanks for the response

 

[dextercioby]

Yes,that "v" is the constant tangential speed.Multiplied with the period should give exactly 2\pi r...

Daniel.

 

[ballahboy]

ok i got the correct answer now.. THanks a lot! :D