- #1
myersb05
- 14
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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.
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.