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2 problems conserning magnetic fields and units.

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    http://www.aijaa.com/img/b/00125/3627010.jpg [Broken]

    http://www.aijaa.com/img/b/00758/3627011.jpg [Broken]

    2. Relevant equations

    [tex]E_k=\frac{1}{2}mv^2[/tex]
    [tex]\vec{F}=m\frac{\vec{v}^2}{r}[/tex]
    [itex]\vec{F}=q\vec{v} \times \vec{B}[/itex]
    [tex]\frac{v}{r}=\frac{2\pi}{T}=\omega[/tex]

    3. The attempt at a solution

    The only difficulty in problem 2 is the unit of the mass of the proton. I can't use any info that's not given in the problem or I don't know by heart. I know the mass of the proton is about 1u, but I don't know how to convert it to kg. First I solve [tex]E_k[/tex] for [tex]v=\sqrt{\frac{2E_k}{m}[/tex] and then by setting the forces equal I get [tex]r=\frac{mv}{qB}[/tex]. Now here's where I need kg instead of u - or am I missing something.

    The thing that gets me in number 5, is the way the magnetic field is given. If we set the Earth's surface along the xy-plane so that the positive y-axis points towards north and the positive x-axis towards east. Then [tex]\vec{B_h}=18\mu T\vec{j}[/tex]. But is [tex]\vec{B_v}=-54\mu T\vec{j}[/tex]? I think in that case they would have given the magnetic field as "[tex]B=36\mu T[/tex] to south". So they only option left is [tex]\vec{B_v}=-54\mu T\vec{k}[/tex]. Which one is it and why? From here the problem is pretty straight forward: I just need to find the cross product of the vectors and calculate its length.

    I'm sorry for the non-native English, but ask if you can't understand something I've written.

    Edit: Some typos...
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 18, 2009 #2

    Delphi51

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    Homework Helper

    The mass of a proton is 1.67 x 10^-27 kg.
    You can find things like that in Wikipedia or in the back of a physics text.

    Many people do not realize that there is an "S" pole up in northern Canada and an "N" in Antarctica! You can check - a compass that points north will also point toward the S end of a magnet. I expect you will keep the full vertical component, but multiply the horizontal component of B by cos(45) to get the part that is perpendicular to the wire.
     
  4. Feb 18, 2009 #3
    But that's the whole point, if this kind of a question is in an exam, I can't use Wikipedia or any other source of information, other than the question or my memory. So I guess these kind of things has to be known by heart.

    Why should B be perpendicular to the wire? If I can just figure out [tex]\vec{B_v}[/tex], I can create [tex]\vec{B}[/tex] by adding the vertical and horizontal components together. And then calculate the cross product [tex]\vec{l} \times \vec{B}[/tex]

    Thanks for the reminder, that made things clear! Since Earth has curvature, so does the magnetic field. One of the fields components then must be towards the center of the Earth and that makes the vertical component align with [tex]-\vec{k}[/tex].
     
  5. Feb 18, 2009 #4

    Delphi51

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    Homework Helper

    Surely they put constants on the exam - e, mass of electron and proton, c, G, h, etc.
    Most people have them in their calculators - do they clear calculators for exams?
    Ideally, there would be a data sheet for the course that you use through the term and on the exam.
     
  6. Feb 18, 2009 #5
    Well, I really hope that!

    We aren't allowed to use calculators in exams. Only pens, eraser and rulers are allowed. Oh, and a slide rule, if you happen to have one. :tongue2:
     
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