A proton, moving in a uniform magnetic field, moves in a circle perpendicular to the field. If the proton's speed tripled, its time to go around its circular path would be
a)double.
b)one third.
c)triple.
d)one half.
e)the same

And would you give me a brief explanation, please.

Question

A proton, moving in a uniform magnetic field, moves in a circle perpendicular to the field. If the proton's speed tripled, its time to go around its circular path would be
a)double.
b)one third.	
c)triple.
d)one half.
e)the same

And would you give me a brief explanation, please.

schrodingers_cat · Answer

Remember the force acting on a particle in a uniform magnetic filed is f=q(v x B) 
(v is crossed with b) Since the field is acting perpendicularly to v the angle between them is 90.

So, this reduces to f=q3VBsin(90) (V and B being the magnitude of the v and b)

As such, this force equates to mv^2/r for circular motion. So, v= q3rB/m.

So, if the velocity is three times greater you will get there 3 times quicker so I would say the answer is c) triple. 

Hope this helps :)

anonymous · Answer

I appreciate the explanation. The only problem is I've tried triple already, and also 1/3 since the question asked for the time to go around 3x times =1/3 the time. I only have one shot left and I am starting to think his key is wrong, or we're missing something.

schrodingers_cat · Answer

Oops my bad I didn't even look at my own equation. The time would remain the same due to the fact the radius r will also be tripled along with the velocity. If you look at the ratio of V/r it will equal a constant set of values BQ/m. So, for those values to remain constant if you triple the velocity you must triple the radius. Also, my equation should be 3V/r = qrb/m not v= q3rB/m. That's what I get for not looking over my answer more throughly :P. 

So, the correct answer is will be e) not c).

:)