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Magnetic/Elec Fields (RHR)

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data

    Please refer to attachment for solution.

    Can you help with the following (please refer to attached). If I understand the problem, both magnetic fields B1 & B2 need to change direction, due to the change in current (I) by replacing the positive with a negative ion.

    Your thoughts,

    2. Relevant equations

    Right hand rule for Current, Magnetic Field, Force.

    3. The attempt at a solution
    Please refer to attached for solution.
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2015 #2

    BvU

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    Hello Ps,

    I understand your reasoning for ##\vec B_2##.

    Then it becomes confusing: Either you change the sign of I or you change the direction; not both.
    And just changing "the sign of the current" (i.e. the charge of the ions) does not change the direction of ##\vec B_1## !
    And I also don't see how changing only the "the sign of the current" (i.e. the charge of the ions) does not change the direction of ##\vec F_E##.

    You also want to keep in mind what the function of ##\vec E## and ##\vec B_1## is: together they act as a velocity selector: particles with a certain ##|v|## experience no force whatsoever and pass straight through S3.


    Perhaps you'll do well to base your thinking on the expression for the Lorentz force $$
    \vec F = q\;(\vec E + \vec v \times \vec B)
    $$which is very, very fundamental, instead of on all these derived equations that require memorizing and can become confusing (as happens here). Two threads: here and here , might help (in all modesty... ahem :rolleyes:).
     
  4. Mar 9, 2015 #3
    So what your saying, If the cross product of v & B is equivalent to a force. Then the electric field must be reversed, when a negative ion is introduced, to have zero net forces acting on the ion. please find attached.
     

    Attached Files:

  5. Mar 9, 2015 #4

    BvU

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    No! :nb)

    v is the actual velocity of the particles (the time derivative of the position). No sign confusion because of charge possible (that is sitting in the q).


    If the force has to be zero for a certain v so that these particles go straight through S3, and the force is ##
    \vec F = q\;(\vec E + \vec v \times \vec B)
    ## for a charge q, then what is the force for a charge -q ?

    Your last sentence is weird: both E and B2 ? We are discussing E and B1 !

    We already concluded B2 must be reversed.
     
    Last edited: Mar 9, 2015
  6. Mar 9, 2015 #5
    B2 needs to be reversed I agree. And we are discussing E and B1, but I was referring to all the fields that needed to be changed, mentioned in the question.
     
  7. Mar 10, 2015 #6

    BvU

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    The maker of the exercise took care to not include an answer where all three are reversed (something many ignorant folks would pick without much thought) ; that way the exercise has a bit more selective power :)

    Is it now really clear to you that reversing bot E and B1 isn't necessary, even though it will work allright ?
     
  8. Mar 10, 2015 #7
    view attachment :mad:
     

    Attached Files:

  9. Mar 10, 2015 #8

    BvU

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    Pictures look splendid. Accompanying text: ? ? (if it's adjusted for a positive particle with speed v, then it will also let through a negative particle with the same speed v).
     
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