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Electromagnetism equation help

  1. Jun 4, 2005 #1
    I have three multiple choice questions on electromagnetism that i have been unable to get. The answers are there, i just can't get them.

    http://i2.photobucket.com/albums/y15/seiferseph/1.jpg

    i can't find any equation to use that doesn't have more than two unknowns. i've tried Fb = I*L*B*sinx but you don't have the force to solve.

    http://i2.photobucket.com/albums/y15/seiferseph/2.jpg

    i get the first part, but what does the resistor on the right have to do with it? how do you find anything for that?

    http://i2.photobucket.com/albums/y15/seiferseph/3.jpg

    i feel like i've tried every equation, if someone could just give me some hints as to how to use all the givens, like the turns, the 2 lengths, b field, etc.

    Thanks in advance!
     
  2. jcsd
  3. Jun 4, 2005 #2

    OlderDan

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    The first problem involves the force of gravity and the tension in the supporting string as well as the magnetic force on a current carrying wire.

    The second problem involves induction. A coil of wire in the presence of a changing magnetic field has an induced electomotive force (emf, voltage) resulting in current flow. You need to understand the rules for determining the direction of this induced emf.

    The last problem is just a current carrying wire in a constant magnetic field. The vertical parts of the loop carry currents in oppostie directions, so the forces cancel. The bottom leg of the loop provides a vertical force that reduces the tension. Each turn experiences the same force as a single wire carrying the current.
     
  4. Jun 4, 2005 #3

    ek

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    I think that second question is taken from the Physics Provincial I took in January 2004.

    I'm pretty sure I got it wrong though.

    :rofl:
     
  5. Jun 4, 2005 #4

    i don't need to know the 2nd one for now then, thats part of the next unit.
    as for the frst one, does that mean Ftcosx = mg + Fb? we still don't have the Ft, so we're left with two variables

    ok, so for the last one, do i just use Fb = I*L*B, using L to be 25 times 0.15 (or 3.75) and Fb to be 4.0 x10^-2 and B as 0.065? I did that and solved for the answer, 0.16, is that the right way?
     
    Last edited: Jun 4, 2005
  6. Jun 4, 2005 #5

    Pyrrhus

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    For the first problem,

    this vectorial equation, gives two scalar equations you need to solve, to find the magnetic field (from the definition of magnetic force).

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 [/tex]
     
  7. Jun 4, 2005 #6
    are there only the 4 forces, Ftcosx + Ftcosx + mg + Fb = 0? i'm still confused.
     
  8. Jun 4, 2005 #7

    Pyrrhus

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    There are 3 forces acting, Magnetic Force, Tension and weight of the cylinder.

    Find out what is the:

    [tex] \sum F_{x} = ? [/tex]

    [tex] \sum F_{y} = ? [/tex]
     
  9. Jun 4, 2005 #8
    i'm thinking right now i can get the tension force by using the fact that the horizontal forces are balanced, but i can't find the other force, all i see i the horitizontal component of the Ft.
     
  10. Jun 4, 2005 #9

    Pyrrhus

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    Here i'll give you a hint, the magnetic force is exactly on the x axis, if you pick your coordinate system at 0 degrees with the horizontal. The tension will have components in both x and y axis, the weight obviously on the y axis, so you can find the Tension, and then on the sum of forces on the x axis find the magnetic field.
     
  11. Jun 4, 2005 #10

    YESS!! i got it, thanks for the help. i forgot that the magnetic foce acts perpendicular to the field :rofl: i had two forces down, mg and Fb, and only the Ft up and right.
    after drawing it correctly, i saw that Ftcosx = mg, and Ftsinx = Fb. Since you know mg, you solve for Ft, and then put that into the other equation to solve for Fb. Putting that into Fb = I*L*B, you can easily solve for B.

    Thanks, unfortunately i still have some more problems. First of all, is what i did for #3 correct?
    do i just use Fb = I*L*B, using L to be 25 times 0.15 (or 3.75) and Fb to be 4.0 x10^-2 and B as 0.065? I did that and solved for the answer, 0.16, is that the right way?


    also...

    http://i2.photobucket.com/albums/y15/seiferseph/4.jpg

    maybe i'm using the wrong equation, but i'm just using B = u0 * I/2pi*R Then i just solve for I, but i get something wrong.


    http://i2.photobucket.com/albums/y15/seiferseph/5.jpg

    is this one electric induction? if so, ignore it. otherwise, i have no idea how to do it, can anyone help?


    http://i2.photobucket.com/albums/y15/seiferseph/6.jpg

    just need some help deriving the equations, how do i get anything equal to the kinetic energy, or 1/2mv^2?
     
  12. Jun 4, 2005 #11

    Pyrrhus

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    Ok for the 6th problem, you know throught magnetic force will be the centripetal force of the electron. If the kinetic energy is doubled how is the speed affected?
     
  13. Jun 4, 2005 #12
    i'm not sure how to do that type of ratio, is it sqrt(2) times v? or is it 2^2 or 4 times v? Ke = 1/2mv^2, so if 1/2 and m are constants, Ke = v^2. so doesn't 2Ke = (2v)^2

    also, i got question 5, you use B = u0*I/2pi*r to solve for the B field, then put that into Fb = I*L*B
    also got question #4, i guess i wasn't using the equation correctly.
     
    Last edited: Jun 4, 2005
  14. Jun 4, 2005 #13

    Pyrrhus

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    Ok you know the magnetic force will be the centripetal force so

    [tex] qvB = m \frac{v^2}{r} [/tex]

    [tex] qB = m \frac{v}{r} [/tex]

    and you know

    [tex] v = \sqrt{\frac{2K}{m}} [/tex]

    so

    [tex] r = \frac{mv}{qB} [/tex]

    Therefore

    [tex] r = \frac{m \sqrt{\frac{2K}{m}}}{qB} [/tex]

    Now you can see what happens to the radius when the kinetic energy is doubled.
     
  15. Jun 4, 2005 #14
    put that way its so simple :surprised thanks a lot for the help, i understand it all now :biggrin:
     
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