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Electron within a Magnetic Field

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

    An electron has an initial velocity of (12.0j + 15.0k) km/s and a constant acceleration of (2.00x1012 m/s2)i in a region in which uniform electric and magnetic field are present. If B=(400x10-6)i , find the electric field E.

    2. Relevant equations

    [itex]\vec {B}=qVBsin \theta[/itex]
    F=ma

    3. The attempt at a solution

    Total force on the electon is the sum of the force from the electric field and the force from the magnetic field.

    [itex]F_{tot} = F_B + F_E [/itex]
    [itex]F_{tot} = qVBsin \theta + qE[/itex]

    I don't know how to use the vectors in this notation in the above equations. For instance, if I calculate the total force in the x-direction using [itex]F=ma_x[/itex], I get a number. If I do that in the y-direction, I get zero since there's no acceleration in the y direction. Same for z. However, when breaking up the equation [itex]F=qVBsin\theta[/itex] I don't know how to break up V and B into its components.

    (I hate vectors and I hate my book.)
     
  2. jcsd
  3. Oct 18, 2009 #2
    No one else can do this either, eh?
     
  4. Oct 18, 2009 #3

    Delphi51

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    Homework Helper

    You can find F = qvB using the cross product vxB
    Or, realizing that the v vector is perpendicular to the B, find the magnitude of v and multiply it by the magnitude of v (and q) to get the magnitude of F. A bit tricky to get the direction this way, though. It is perpendicular to both v and B.
     
  5. Oct 19, 2009 #4
    I don't know how to work out the problem though with vector notation though.
     
  6. Oct 19, 2009 #5

    berkeman

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    Staff: Mentor

    There should be a way shown in your textbook to do the vector cross product directly in rectangular coordinates. Keep the whole Lorentz Force equation in vector notation in order to get the solution.
     
  7. Oct 19, 2009 #6
    There isn't. It's only "Part III - Elec & Mag", chapters 21-30 or something of the full text. Vectors were dealt with in Chapter 3 apparently, which I don't have and even the online e-book extension of this doesn't give me access to.

    This is why I'm stuck. I don't know how to do vector manipulation easily. I've looked around the internet but I can't seem to figure it out.

    I don't know how to do this.
     
  8. Oct 19, 2009 #7

    berkeman

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    Try this. Go to http://www.wikipedia.org/ and type in Vector Cross Product. That wiki page explains the cross product pretty well, including the matrix notation trick for computing it directly in rectangular coordinates.

    That should give you the tools to solve the vector equations in this problem.
     
  9. Oct 19, 2009 #8
    I don't know how to set up the problem at hand.
     
  10. Oct 19, 2009 #9

    berkeman

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    You write the F = ma vector equation in rectangular coordinates, using the total Lorentz force for F

    The vector Lorentz force (see the first equation): http://en.wikipedia.org/wiki/Lorentz_force

    And you do the vector cross product using the matrix method at the link I gave you in my previous post.

    Please try to set these equations up yourself. We cannot do your homework for you here.
     
  11. Oct 19, 2009 #10
    First of all, it's not a homework problem, but a practice problem in the book i'm doing before the homework is due next Monday, similar to what will be due. Secondly, this is an first-year physics course with no expectation that you've taken linear algebra. Thirdly, I'm 32 years old and am returning to school after TEN YEARS of absence. It's been that long since I've worked with matrices and vector equations. I'm asking for some assistance in setting up the problem, and help through the solution.

    I have the answer since it's in the back of the book but that DOESN'T MATTER TO ME. I can't use what you gave me if I don't understand HOW to use it. I'm looking for someone to walk me through it so I *do* understand how it works. I'm not looking for an answer.

    And this is a "level one" question in the chapter, which means it shouldn't be difficult or require more than a few minutes of work to solve.
     
  12. Oct 19, 2009 #11

    tiny-tim

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    To find the acceleration due to B, use i x j = k and i x k = -j (and i x i = 0) :smile:
     
  13. Oct 19, 2009 #12

    berkeman

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    Fair enough. As schoolwork, we still want you to do the bulk of the work. Also, there may be an easier way to solve the problem without using the full vector cross product version of the Lorentz force, but it's the easiest way to solve it if you can get up to speed on working with vector cross products in rectangular coordinates.

    Start using that the sum of the vector F is equal to the sum of the masses multiplied by the vector acceleration a. Then equate that sum of forces to the Lorentz force, since that is all that is creating forces on the moving electron (I think gravity is not mentioned, correct?).

    F = ma = q[ E + v X B ]

    You are given the vectors v, a, and B, and are asked to find the vector E.

    Most of that is straightforward, except you need to use the vector cross product solution technique shown in the link using the matrix method. Have you used determinants before? If so, then just set up the 3x3 determinant shown in the link. If not, here's a link to info about how to calculate a determinant. It's just 6 cross-multiplies across the 3x3 determinat matrix, keeping track of the rectangular coordinate unit vectors to group the multiplied components. Look at the example 3x3 determinant cross-multiply (red) lines here:

    http://en.wikipedia.org/wiki/Determinant

    Doing the cross product calculation this way is generally easier than doing it with the scalar qvB*sin(theta) technique, and figuring out the vector direction with the right hand rule.

    Does any of that help? I've posted a note in the Homework Helper's forum to see if anybody else can pop in to see if there's a better way to be suggesting that you do this problem.
     
  14. Oct 19, 2009 #13

    berkeman

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    Thanks TT!
     
  15. Oct 19, 2009 #14
    So to form a 3x3 matrix it would look like this?:

    [tex]
    \begin{pmatrix}
    \hat i & \hat j & \hat k\\
    0 & 12 & 15\\
    400\mu & 0 & 0 \\
    \end{pmatrix}
    [/tex]

    and solving it looks like:

    [tex]\hat i (12)(0) + \hat j (15)(400\mu) + \hat k (0)(0) - \hat k (12)(400\mu) - \hat i (15)(0) - \hat j (0)(0)[/tex]
    [tex]= 6x10^{-3}\hat j - 4.8x10^{-3}\hat k[/tex]

    Therefore:

    [tex]\vec F_b=m\vec a = q(\vec V x \vec B) + q\vec E[/tex]

    [tex]\vec E = \frac{m\vec a}{q} - (\vec V x \vec B)[/tex]

    [tex]\vec E = \frac{(9.1x10^{-31})(2x10^{12}\hat i)}{(-1.6x10^{-19})} - (6x10^{-3}\hat j - 4.8x10^{-3}\hat k)[/tex]

    [tex]\vec E = -11.4\hat i - (6x10^{-3}\hat j - 4.8x10^{-3}\hat k)[/tex]

    This *is* the right answer. I'm confused as to the creation of the matrix. Why do I use i, j, and k on the top row? More generally, why do I use the rows the way they are?

    Also, you mention that it's faster to do it this way than the other way (using sin theta). How would you do it the other way? I'm afraid that the book might be looking for that method.
     
  16. Oct 19, 2009 #15

    berkeman

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    Staff: Mentor

    AWESOME! Very awesome. Good work. Really.

    As for why the vector cross product can be expressed in that form in the determinant, I don't remember (I'm way older than you), and am too lazy to look it up. Maybe TT or another helper could post about that, or you and I could google a bit. Still, it's a nice compact form, eh?

    As for some simpler, non-vector solution to the problem, I don't think you need it. Unless you use TT's hint, which is based on the vector cross product determinant anyway, or else it can be viewed as based on the application of the right hand rule to the unit vectors in rectangular coordinates. Yeah, that might be a way to pursue a simpler solution, using the RHR to figure out how to handle the vector cross product. After all, they did put most of the vectors along the rectangular axes.

    But in the general case, the vectors will not be aligned with the axes, so that's why I don't like to waste time -- just solve the vector equations. And it looks like you are starting to understand those better. Great!
     
  17. Oct 19, 2009 #16
    The book stresses the right hand rule, images and everything. In fact, that's how I'm picturing the resulting force direction of a moving particle within the magnetic field. I just don't know how to calculate it, and get an answer that gives me all three axes.
     
  18. Oct 20, 2009 #17

    tiny-tim

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    Homework Helper

    Hi exitwound! :smile:

    Basically, it's because + and x (for 2D or 3D vectors only) obeys the distributive law

    (a + b) x c = a x c + b x c

    … so we can split any vector into Ai + Bj + Ck, and multiply it by (in this case) 400µi, and get Aixi + Bixj + Cixk,

    and you can easily check that, for example, ixk = -j comes from
    Code (Text):
    i j k
    1 0 0
    0 0 1
     
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