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Physics 20 Online
OpenStudy (perl):

Traveling at an initial speed of 11.5 x 10^6 m/s, a proton enters region of constant magnetic field of magnitude 1.5 T. If the proton's initial velocity vector makes an angle of 30 degrees with the magnetic field, compute the protein's speed 4 seconds after entering the magnetic field.

OpenStudy (perl):

hey

OpenStudy (perl):

so far I used F = qVxB |F| = q*V*B*sin(theta)

OpenStudy (deoxna):

Ok, great. Hold on I have to get through a bit more of the problem...

OpenStudy (deoxna):

So F=(1.60*10^(-19))(11.5*10^6)(1.5)sin(30) F=1.38*10^(-12) N. Now, F=ma, so: a=(1.38*10^(-12))/(1.673*10^(-27))=8.249*10^14 m/s² Now V=at+u, so: V=(8.249*10^14)(4)+(11.5*10^6)=3.299*10^15 m/s V=3.30*10^15 (taking sigfigs into account)

OpenStudy (perl):

one moment,

OpenStudy (perl):

I got the same answer, using the same logic but look at the options http://forums.studentdoctor.net/showthread.php?t=700345

OpenStudy (perl):

maybe its a typo?

OpenStudy (deoxna):

Wait! We haven't taken vector's into account... The direction of the magnetic force is out of the paper, which means that the proton follows a kind of spiral path in its current direction, so it does decelerate...

OpenStudy (deoxna):

At least as its coming up from the loop...

OpenStudy (perl):

hmmm

OpenStudy (deoxna):

The formula for the radius of the circular path taken by a charged particle is: R=mv/qB So that must help...

OpenStudy (perl):

maybe you could post this on yahoo answers

OpenStudy (perl):

with the options

OpenStudy (perl):

dont worry about it, you did enough. spend some time constructively :)

OpenStudy (perl):

i actually havent even learned about magnetic theory, somebody just asked me. so its silly

OpenStudy (perl):

but your argument makes sense, and i used it as well. it just didnt fit the choices

OpenStudy (perl):

and the mass of a proton i also looked up 1.67 x 10^-27 kg

OpenStudy (deoxna):

Well, no it isn't silly. Its always good to learn new things for their own sake...

OpenStudy (deoxna):

I was thinking though if the force is perpendicular, and hence doesn't affect the initial speed, then what does the speed we calculated (except for the +u) even mean. Is it the "centripetal velocity"?

OpenStudy (perl):

i was confused since it was 30 degrees

OpenStudy (perl):

im going to bed, sorry , unless you want to pm me . thanks man

OpenStudy (perl):

if you have any other math questions, pm me :D

OpenStudy (deoxna):

haha sure thing! Thanks for all the help dude. Really. I'm also going to bed. its waay too late.. Anyway see ya man... :D

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