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Help! Mechanics Qs!

  1. Dec 4, 2005 #1
    Hey everyone,

    I'm currently studying maths at Uni and this year I have to do a compulsary mechanics module. For the record, I've not done physics for a long time and I'm having a lot of trouble with some questions! Here they are as printed on the sheet:

    "The force acting on a particle of mass m moving under the influence of a constant magnetic field B is equal to v x B where v is the velocity of the particle relative to the origin 0 of an inertial frame. Suppose that at time t = 0, r=0 and v = V.

    Show that mv = r x B + mV"

    it gets better

    "Let i, j and k be unit vectors in the direction of the axes of the inertial frame. In this case B = Bk and V = v1i + v2k, show that the path of the particle is given by

    r = (mv1)/B sin (Bt/m)i + (mv1)/B(cos(Bt/m) - 1)j + v2tk

    Describe carefully the path of the particle.

    [if x(.) = cy + d and y(.)= -cx (c and d constants) then x(..) = -c^2x]"

    ( x(.) denotes first derivative.)

    My lecturer is hopeless and I'm currently failing this course. I would be really grateful for any replies!


  2. jcsd
  3. Dec 4, 2005 #2


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    Actually, the magnetic force (or Lorentz-force) on a particle with charge q is [itex]q(\vec v \times \vec B)[/itex]. I don't know why that wasn't specified in the question, but let's consider q=1 for this particular case.

    Since you are given the mass and the magnetic field everywhere (B is constant) you know the force acting on the particle everywhere. With the specified initial conditions the motion of the particle is completely determined by Newton's 2nd law:

    [tex]\vec F = \frac{d}{dt}(m\vec v)=\vec v\times \vec B[/tex]
    Now you have the time derivative of mv on the left side. Since you are given the answer, all you have to do is check if the answer satisfies this equation of motion and the initial conditions.
  4. Dec 4, 2005 #3
    Right, I think I can follow the above in that the overall force is just determined by Newton's second law.

    I'm still puzzled as to how to come to the conclusion that mv = r x B + mV using the conditions t = 0, r = 0 and v = V. I mean where does mV come from?

    This is the first time I've ever done this type of physics. You may have to dumb it down in order for me to understand.
  5. Dec 4, 2005 #4


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    The answer gives mv=r x B+mV.

    The equation of motion is d/dt(mv)=v x B.

    So you have to check whether the given answer
    mv=r x B + mV obeys the equation of motion AND see if it satisfies the initial conditions: If r=0, you have v=V.

    The initial condition is easily verified (just insert r=0). For the equation of motion, take the time derivative of mv = r x B + mV.

    This is just one way to do this problem which I find to be the path of least resistance. Usually, if you are given the answer it's easier to check that it is right then to derive it in the first place.
  6. Dec 4, 2005 #5
    Ok, thanks for that. Seems to work out fine.

    Now for the rest of the question:

    Not entirely sure where to start with this one. It may just be a case of rearranging the previous equations. Not sure about all the rules for using unit vectors. Although I'm fairly certain you have to introduce polar coordinates at some stage.
  7. Dec 5, 2005 #6
    I've started this one off.

    Looking at mv = r x B + mV.

    So if you expand r x B, you get |r||B|sin (a)n

    Substituting this into the above equation I get

    mv = |r||B|sin (a)n + mV

    and rearranged gives me

    |r| = mv/(|B|sin(a)n + mV)

    Dunno what to do next.
  8. Dec 5, 2005 #7


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    You can get the desired result by direct integration:

    [tex]\int_0^t \frac {d\vec v}{dt'} = \int_0^t \frac {d\vec r}{dt'} dt' \times \vec B[/tex]

    [tex]\vec v(t) - \vec v(0) = (\vec r(t) - \vec r(0)) \times \vec B[/tex]

    and apply the given initial conditions.
  9. Dec 6, 2005 #8
    Ok, so if I substitute in the original conditions I get
    v(t) - V = (r(t) - 0) x B

    so v(t) - V = |r(t)| |B| sin a

    I'm still confused as to how to change this to

    r = (mv1)/B sin (Bt/m)i + (mv1)/B(cos(Bt/m) - 1)j + v2tk

  10. Dec 6, 2005 #9


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    Take the magnetic field to be in the z direction so that if

    [tex]\vec v = V + \vec r \times \vec B[/tex]

    then [itex] dx/dt = V_x + yB[/itex] and [itex]dy/dt = V_y - xB[/itex] and [itex]dz\dt = V_z[/itex] which you can solve easily.
  11. Dec 7, 2005 #10
    Ok, I'm with you there. Looking at each vector along the x,y and z axis. Sorry, but I still don't understand how to get to the final answer from there though. Where do the sin and cosine come from? And also how come t's appear in the final answer?

    I feel I'd really benefit my understanding of this if I could see how this all came together in the end.
  12. Dec 7, 2005 #11


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    [tex]\vec v = V + \vec r \times \vec B[/tex]

    and taking B to be constant and along the z direction then

    [tex]\frac {dx}{dt} = V_x + Bx[/tex]


    [tex]\frac {dy}{dt} = V_y - By[/tex]

    The z equation is trivial so I'll ignore it here. To solve these equations, differentiate the first one (e.g.) to find

    [tex]\frac {d^2x}{dt^2} = B \frac {dy}{dt}[/tex]

    from which

    [tex]\frac {d^2x}{dt^2} = B(V_y - Bx)[/tex]


    [tex]\frac {d^2x}{dt^2} = -B^2(x - \frac {V_y}{B})[/tex]

    Notice that this is equivalent to

    [tex]\frac {d^2}{dt^2}\left(x-\frac {V_y}{B}\right) + B^2\left(x-\frac {V_y}{B}\right)=0[/tex]

    which I am sure you can solve. Use the result to find y(t).
  13. Dec 8, 2005 #12
    I think I follow you to a certain extent. Although I do have some queries still about applying vector calculus. I'm not sure about the rules for taking a time derivative. Throughout schooling, I was taught calculus along the lines of for example.

    if y = x^2 + x

    then dy/dx = 2x + 1

    I'm not sure how to go about calculating dx/dt for certain things especially for the problem I'm working on above. I still don't see what's going on and why.

    Please bear with me on this one....
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