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Homework Help: Can someone please explain this solution? (magnetic fields))

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle with a charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a path of radius R, find expressions for (a) its speed and (b) its mass.

    2. Relevant equations

    K=1/2mv^2 and R=mv/qB

    3. The attempt at a solution

    I couldn't think of anything to do besides solve for v and m in the second equation. I figured this couldn't be correct because that's trivial. Since this is somewhat recreational work for me, I shortly gave up and consulted a solution. The solution I found showed the method for solving the problem to be dividing K by R and then solve for v. This is then substituted into the second equation to answer the second question.

    I NEVER would have made this intuitive leap so I'm glad I didn't waste too much time on it. However, even after checking the solution, I'm confused about how or why that approach would be taken at all. Is this something that someone could explain to me?
  2. jcsd
  3. Mar 23, 2012 #2


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    Homework Helper

    I see it's your first post. Welcome to physicsforums! The intuition will come in time when you do lots of physics problems. What is it about the approach to the problem which is confusing?
  4. Mar 24, 2012 #3
    Thank you for your reply and thank you for your welcome! Although this is my first post, I've been lurking about for a while.

    What would cause a better problem-solver than myself to look at this information and think to divide the the kinetic energy by the radius?

    Usually when I'm so stuck on a problem that I'm forced to seek a solution I can usually see an approach in the solution that is enlightening for me. I can typically see some relationship or intuition or something that I can then use for another problem. I can't see that here. I don't see a relationship between the question, the kinetic energy and the radius that would cause me to ever divide R into K. Dividing one equation into the other just seems kind of random but I assume it's not...
  5. Mar 24, 2012 #4


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    Actually that is a correct and straightforward approach here, known as the substitution method. I presume you would take the resulting expression for v (or m) and substitute it into the first equation, then solve that for m (or v).

    It's probably worth your time to go ahead and try the method you initially thought of.

    After you've done stuff like this many many times, you can spot the fact that "m" appears as a factor in each equation. So dividing one equation by the other will make the m's cancel (m/m=1), leaving an equation in just one unknown, v. (EDIT: Once you get experienced and comfortable with algebra, you can see that the m's will cancel before actually writing the full expressions out, sort of like chess players who are looking a move ahead.)

    All that method does is save some steps, compared to the straightforward approach of solving one equation for m and then substituting into the other equation -- which works just fine.
  6. Mar 24, 2012 #5
    Sometimes there may be a physical relationship that would cause you to divide, but more often you just do it because that is one of the three options you have for solving a system of equations.

    1) back substitution: solve for a variable in one eq. (like you tried) and substitute it into the other
    2) add/subtract: add the LHS of the eq.s and the RHS of the eq.s to cancel a variable
    3) mult./divide: same as 2) but mult. instead

    If there is no intuition, try all of those. If those don't work, pray for intuition. In this case you might notice that since your variables are [itex]m[/itex] and [itex]v[/itex] and you have [itex]mv[/itex] in one equation and [itex]mv^2[/itex] in the other, dividing will cancel [itex]m[/itex] and one [itex]v[/itex] leaving an equation that is linear in [itex]v[/itex] and linear is nice.

    Edit: sorry to be repetitive; I'm a slow typer.
    Last edited: Mar 24, 2012
  7. Mar 24, 2012 #6
    Ah! These are wonderful replies. Thank you so much!

    I did of course notice the common variables but I just didn't make that leap. I realize now that some of my issue was that I just didn't quite understand what the problem was asking for (an issue I often encounter with physics problems). If it had said 'find an expression for speed and mass with respect to K' or something like that, I might have better made the connection. Either way...

    Thank you thank you thank you!
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