A Proton Moving Freely in a Constant Magnetic Field

[myersb05]

A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.50 µs to complete one revolution. Determine the magnitude of the magnetic field.


I know that a proton has a charge of 1.6x10^-19 and a mass of 1.67x10^-27. Those are the only two relevant facts I know about a proton. The equation I was attempting to use was F=qvBsin(theta). I know that theta is 90 because the proton is perpendicular to the field. So sin(theta) will be one. Q is the charge of the proton stated above. I do not know v or B, v being velocity and B the magnitude of the force of the magnetic field. I tried numerous ways to solve for v but I can not figure out how. We do have a time but we have no radius of the circle so I do not know how to reap a distance from the given information. I know that I have not explicitly attempted a solution, but I am really really genuinely stuck on this one. Someone please help.

 

[alphysicist]

A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.50 µs to complete one revolution. Determine the magnitude of the magnetic field.


I know that a proton has a charge of 1.6x10^-19 and a mass of 1.67x10^-27. Those are the only two relevant facts I know about a proton. The equation I was attempting to use was F=qvBsin(theta). I know that theta is 90 because the proton is perpendicular to the field. So sin(theta) will be one. Q is the charge of the proton stated above. I do not know v or B, v being velocity and B the magnitude of the force of the magnetic field. I tried numerous ways to solve for v but I can not figure out how. We do have a time but we have no radius of the circle so I do not know how to reap a distance from the given information. I know that I have not explicitly attempted a solution, but I am really really genuinely stuck on this one. Someone please help.

You've got that F=qvB from the magnetic force. What does F also have to equal since that magnetic force is causing it to move in a circular path?

Also, for later in the problem, if a object is moving in a circular path of radius R with speed V, how is R and V related to the time it takes for the object to go around the circle once?

 

[myersb05]

I had tried to set F=0 because the proton has a constant speed and no acceleration. But that did not help because you can't solve 0=qvB for B which is what I need. R is related to the time it takes for a revolution by 2piR/V. However, I have no radius. If I am wrong about F being equal to zero, and I feel like I am; could you give me another nudge as to what F may be equal to. Possibly the charge of the proton since it is acting equally against that charge to hold it in circular rotation? No, I don't think so, just a guess.

 

[myersb05]

Were you referring to the fact that F would also have to equal m*(v^2/R)

 

[alphysicist]

Were you referring to the fact that F would also have to equal m*(v^2/R)

That's right; so set that equal to qvB to get your condition for circular motion in a magnetic field. Most books would solve what you get for r; what do you get?

Once you have that, find out how R, V, and T are related, and then you can get the time in your equation.

 

[myersb05]

When solving qvB=m(v^2/R) for R, I got R=mv^2/qvB and from what I said earlier, I know that 2piR/V=T. So I can solve that for R as well and get R=TV/2pi. Then I can set my two equations equal and I have mv^2/qvB=TV/2pi. So now I successfully have time in my equation. Thanks so much for the help so far. However, I still have both B and v to solve for in my equation. All other variables I know.

 

[myersb05]

Oh wait! You can solve that because the v's eventually cancel out. Thanks al, you are the man. You helped me maintain my 100 homework average for summer physics 2!

 

[alphysicist]

Sure, glad to help!