Electron accelerated through a potential and magnetic field

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Homework Statement


(a) What voltage will accelerate electrons to a speed of ##6.00 \times 10^{-7} m/s## ? (b) Find the radius of curvature of the path of a proton accelerated through this potential in a 0.500-T field and compare this with the radius of curvature of an electron accelerated through the same potential.

Homework Equations


$$r=\frac{mv}{qB}
\\ q\Delta V = \Delta U
\\ K = \frac{1}{2}mv^2 = -\Delta U$$

The Attempt at a Solution


For the first part I used the second equation and set it equal to the kinetic energy, K. This gave a potential difference of ##1.02 \times 10^{-24} V##. Then for the b), I used the first equation to solve for ##r = |-6.82 \times 10^{-18}| = 6.82 \times 10^{-18}##. Though I was unsure of how to solve for the radius of curvature when accelerated through the same potential, nor how to compare the two values. Also, should the r value been negative, or am I missing a negative sign in one of the values?

Thank you in advance
 

Answers and Replies

  • #2
kuruman
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The speed of the electron looks awfully small. Are you sure of the exponent? The radius should be positive. Also, although you do not mention it, you may have assumed that the proton has the same speed as the electron. That is not true. If the proton is accelerated by the same voltage as the electron, what quantity do they have in common?
 
  • #3
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From the problem, it does have a negative, though it could be a type (it would not be the first one if it is).

So I would have to solve for the velocity of the proton using the potential solved from the first part. I solve for ##v = 1.4 \times 10^{-8} m/s## Then I would use that to solve for the radius of curvature of the proton ##r_p = 9.73 \times 10^{-17} m##. Similarly, I used the values for the electron to solve for ##r_e = 2.27 \times 10^{-18} m##. Would I just subtract ##r_p - r_e## to compare them?

Also, for the radius of curvature of the electron, what value other than q be negative so that the signs cancel to get a positive radius? Would the velocity be negative?

Thank you in advance
 
  • #4
kuruman
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Comparison in this context means finding how many times a quantity is larger than another. If someone asked you to compare the mass of the Earth (6×1024 kg) to the mass of the Sun (2×1030 kg), the ratio is more informative than the difference. Differences become informative for comparison if the two quantities have the same order of magnitude. For example you can say "my brother lives 3 miles farther away from my parents than I", or you can say "the distance of my brother's house to my parents house is 1.6 times the distance of my house to their house."

All quantities in ##r=\frac{mv}{qB}## are positive. Strictly speaking if you are comparing radii for the electron and proton, the negative sign introduced by the charge of the electron means that the orbit of the electron in the magnetic field is in the opposite direction from that of the proton. Unless you are asked to specify the direction of the orbit, don't worry about the negative sign. Same thing applies to the velocity. In these equations the symbol "v" stands for "speed", the magnitude of the velocity, which is always positive.
 

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