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Electric field and magnetic field - proton deflection

  1. Jun 20, 2008 #1
    1. The problem statement, all variables and given/known data

    when protons travelling north in a horizontal plane enter a region of uniform magnetic field of 0.8Teslas in the downward direction, they are deflected into a horizontal circle of radius 0.2 meters. what is the magnitude and direction of a uniform electric field applied over the same region of space that will allow the protons to pass through the region undelflected

    2. Relevant equations

    radius r = mv/qB where m is mass, v is velocity, q is charge, B is magnetic field

    electromagnetic force F = qE + qv X B where X indicates cross product

    electric field force F_E = qE where E is electric field

    magnetic force F_B = qv X B where B is magnetic field

    charge of electron/proton = 1.6*10^-19 coulombs

    mass proton m = 1.67*10^-27 kg

    3. The attempt at a solution

    i used :
    since the proton must not be deflected, i assumed electric field force must equal magnetic force so:

    qE = qv X B

    i then used radius eq, r = mv/qB and solved for v --> v = qBr/m and subbed it in for v to get:

    qE = q(qBr/m) X B and solved for E --> E = (qBr/m)Bsin(theta), i have q = 1.6*10^-19 coulombs, B = 0.8 Teslas, r = 0.2 meters, mass m = 1.67*10^-27 kg but what is theta?

    is my approach correct?

    cheers
     
  2. jcsd
  3. Jun 20, 2008 #2

    G01

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    Your approach for finding the magnitude of the E field is correct.

    The direction of the B-field is given in the problem, though it is worded in a somewhat confusing way. They tell you that the protons are moving north and that the B field is pointing in the "downward" direction. I believe this means "into the plane of the page."

    Considering this to be the case, what is your angle?
     
  4. Jun 20, 2008 #3
    which page? the page the problem is written on? in that case, the angle would be 90 degrees.

    correct?
     
  5. Jun 20, 2008 #4
    If you were to take a path northward over the Earth's surface, southward would be the direction opposite of northward (which is antiparallel to the motion and so the proton would be unaffected), and downward would be the direction into the Earth (which is perpendicular to the motion and is what the question is probably referring to---so you're correct, angle is 90 degrees)...heh, question was worded kind of confusingly. There is a trick to determining the angle (if I'm right about this): the particle will move in a circle if the field is completely 90 degrees perpendicular to the motion...otherwise the path is helical.
     
    Last edited: Jun 21, 2008
  6. Jun 21, 2008 #5

    G01

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    Yup. You got it.:smile:
     
  7. Jun 21, 2008 #6
    thanks, so using theta = 90 degrees and E = (qBr/m)Bsin(theta), i have q = 1.6*10^-19 coulombs, B = 0.8 Teslas, r = 0.2 meters, mass m = 1.67*10^-27 kg

    i got electric field E = 1.23*10^7 coulombs/meter
     
  8. Jun 21, 2008 #7

    G01

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    Looks good.

    Good job!:smile:
     
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