Proton Acceleration in Uniform Magnetic Field: Homework Solution

[mlbqd3]

Homework Statement

A proton travels horizontally with speed 4.0E6 m/s in a vertical uniform magnetic field of 1.5 tesla. Determine the magnitude of proton's acceleration. Describe it' s path.

Homework Equations

F=qvB
F=ma
Mass of proton=1.67E-27kg
Charge of proton=1.60E-19C

The Attempt at a Solution

This is what I did and want to know if I am thinking correctly.
F=qvB
F=(1.6E-19C)(4.0E6m/s)(1.5T)
F=9.6E-13

then I used Newtons 2nd Law to solve for acceleration
F=ma
a=F/m
a=9.6E13N/1.67E-27kg
a=5.752E14

Is my thought process correct? Or am I completely confused? Thanks for the help!

 

[Shooting Star]

You are going in the right direction (I haven't checked the numbers). But how does the path look like?

 

[mlbqd3]

I think that the path of the proton (+ charge) would be headed into the computer screen (or paper) RIGHT? the field is vertical and the initial path of the proton is horizontal, so if I am using the right hand rule correctly, it should be headed into the paper. Right?

 

[Shooting Star]

I think that the path of the proton (+ charge) would be headed into the computer screen (or paper) RIGHT? the field is vertical and the initial path of the proton is horizontal, so if I am using the right hand rule correctly, it should be headed into the paper. Right?

Sort of right. You have not specified the particular horizontal direction of the proton's initial velocity. Let's assume that the initial velocity is directed towards the positive x axis.

Note that the magnetic field does no work on the proton. So, even though the direction of the velocity changes, the speed remains constant. When you say that the path of the proton would be headed into the paper (or out of it), that would be for only one instant. What we are interested in is the general shape of the path. Remember, the velocity of the proton is always perpendicular to the direction of the magnetic field. Give it a try. Read up a bit if necessary. You are doing great.