How Do You Calculate Proton's Speed and Kinetic Energy in a Magnetic Field?

AI Thread Summary
To calculate a proton's speed and kinetic energy in a magnetic field, the magnetic force formula F=qvbsin(theta) is used, yielding a speed of 400,000 m/s. The kinetic energy is calculated using K = 0.5*m*v^2, initially resulting in 7.28*10^-20 J. However, the mass of the proton was incorrectly assumed to be that of an electron, leading to an incorrect kinetic energy value of 0.45 eV. Correcting the mass to 1.67262158 × 10^-27 kg provides the accurate kinetic energy of 835 eV. This highlights the importance of using the correct mass for calculations involving different particles.
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[SOLVED] Simple Kinetic energy problem

Homework Statement



A proton traveling at 23 degrees with respect to the direction of a magnetic field of strength 2.6mT experiences a magnetic force of 6.5*10^-17 N. Calculate a) the proton's speed and b) its kinetic energy in electron-volts

Homework Equations



For part a) I simply used F=qvbsin(theta), I got the correct answer of 400,000m/s
For part b) I used K = 0.5*m*v^2

The Attempt at a Solution



For part b) I plugged in the value of 9.1*10^-31 for m and the value of v I found above(400,000m/s). I return 7.28*10^-20 (assume this in in J?). Upon conversion to eV I find a measly 0.45eV. (The answer is 835ev!)
 
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It's a proton, not an electron.
 
I had a feeling this might be it. I assumed charge of proton and electron is the same, got the right answer, so then I proceeded to assume mass was the same too.

Now if I plug in 1.67262158 × 10-27 instead, I get 835eV!

Thanks
 
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