Proton moving in a magnetic field

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SUMMARY

A proton with a mass of 1.67×10-27 kg and a charge of 1.6×10-19 C, traveling at a speed of 1×106 m/s, enters a uniform magnetic field of 1.2 T. The radius of the proton's circular path is calculated using the equation R = (mv)/(qB), resulting in a radius of 8.698×10-3 m. The time required for the proton to re-emerge into the field-free region is determined to be 2.73×10-8 seconds. Rounding errors in numerical answers can lead to incorrect submissions in online homework systems.

PREREQUISITES
  • Understanding of magnetic force on charged particles
  • Familiarity with the equation qvB = mv2/R
  • Knowledge of circular motion and radius calculations
  • Basic skills in significant figures and rounding in physics calculations
NEXT STEPS
  • Study the Lorentz force and its effects on charged particles
  • Learn about the implications of significant figures in scientific calculations
  • Explore the concept of circular motion in magnetic fields
  • Investigate common pitfalls in online physics homework systems
USEFUL FOR

Physics students, educators, and anyone involved in electromagnetism or charged particle dynamics will benefit from this discussion.

erodger
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Homework Statement



A proton (mass= 1.67×10-27 kg, charge= 1.6×10-19 C) traveling with speed 1×106 m/s enters a region of space containing a uniform magnetic field of 1.2 T.

The magnetic field is coming out of the screen.

What is the time t required for the proton to re-emerge into the field-free region?

Homework Equations



qvB = mv2/R

The Attempt at a Solution



I begin by solving for R using the above equation:

R= (mv)/(qB)

Since the proton moves in and then back out it completes a semi-circle. The distance traveled is then πR as it is half of the circumference of a circle.

The velocity does not change for the entire distance traveled so dividing distance by velocity should give me the time.

This comes out as incorrect though. I am unsure as to what I am doing wrong.
 
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in what direction is the magnetic field pointed relative to the initial velocity of the proton?
 
Oh sorry, I edited the post. It is coming out of the screen. So it is perpendicular to its motion which means it will create a net force on the proton and cause it to undergo the circular motion.
 
ah, okay. Well, what did you get for your time?
 
Well the question involved some steps and one of the steps was to find the radius. It continued to tell me I am wrong. I decided to stop using that and work on it my own using my method and I managed to come up with the correct answer, which just involved me to not round to 2 sig figs but add on more decimal digits. My answer was 2.73*10^(-8). Before I was just putting in 2.7*10^(-8).
 
Thanks for trying to help.
 
what did you get for your radius?

huh, actually that's the same time I got.
 
I got 8.698*10^(-3) for my radius. The computer kept telling me it was incorrect but I managed to get the right answer with this value so I do not know what is wrong...
 
so the time you got was correct, but the radius is not? Maybe it wants radius in terms of some unit like cm?
 
  • #10
No it was asking for metres which is what my answer is in.
 
  • #11
interesting... I don't know what to tell you :(

did you try the same thing with what you did with time? Try changing the number of decimal points, or how you round your numbers when you solve for it.Sometimes I have these same problems with online homework things. They usually accept answers within a certain range of the correct answer, but can be really stubborn sometimes :P
 

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