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Compass in a solenoid, oscillating

  1. Nov 22, 2008 #1


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    1. The problem statement, all variables and given/known data

    A compass, consisting of a small bar magnet resting on a frictionless pivot through its centre, is placed in the middle of a long solenoid, which in turn is aligned with its axis pointing North-South. With a current of 1 Amp passing through the solenoid, a small displacement of the compass from its equilibrium orientation causes it to oscillate with a period of 2 seconds. If a current of 2 Amps is used, the restoring torque is zero and the period is infinite. Calculate the period of the oscillation when no current flows in the solenoid.

    2. Relevant equations

    Magnetic Field in a solenoid:

    [tex] B=\mu_0 nI [/tex]

    Where B is the magnetic field, n the number of coils, and I the current

    Magnetic Force:

    [tex] F = q(E+v\times B) [/tex]

    Torque on a magnetic dipole:

    [tex] G = m\times B [/tex]

    where G is torque


    [tex] torque = I\alpha [/tex]

    where I is the Moment of inertia


    [tex] x = A cos(\omega_0t + \phi) [/tex]
    [tex] v = A\omega sin(\omega_0t + \phi) [/tex]
    [tex] a = -A\omega^2 cos(\omega_0t + \phi) [/tex]

    [tex] \omega_0^2 = \frac{k}{m} [/tex]

    3. The attempt at a solution

    The equation doesn't give you some of the required variables for the formulas I believe you need; most importantly, it doesn't give you the number of coils, so you can work out the B field.

    So far, I have divided it into sections:

    1 Amp:

    [tex] B = \mu_0 n*1 [/tex]

    [tex] G = m\times \left(\mu_0 n*1 \right) [/tex]

    2 Amp:

    [tex] B = \mu_0 n*2 [/tex]

    [tex] G = m\times \left(\mu_0 n*2 \right) [/tex]

    0 Amp:

    [tex] B = \mu_0 n*0 = 0 [/tex]

    There is no B field with no current

    [tex] G = m\times \left(\mu_0 n*0 \right) [/tex]

    No torque from magnetic dipole

    does this look at all right?

  2. jcsd
  3. Nov 23, 2008 #2


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    Does this look correct


  4. Nov 24, 2008 #3


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    What should i be doing then?

  5. Nov 27, 2008 #4

    Your going in the right direction with this problem. I am not certain that the magnetic dipole will be zero though (don't forget the B-field of the magnet arises from all the tiny currents inside the magent, so there is a currrent and there should be a diplole). You need to apply Newtons second law using the torque due to a (magnetic) dipole and the usual expression for torque (I*alpha).

    Once you've done this you should end up with a second order O.D.E. , which you can then solve by letting theta=something.
  6. Nov 27, 2008 #5


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    okay, so:

    [tex] I \alpha = m \times B [/tex]

    [tex] B = \mu_0 n I [/tex]


    [tex] I \alpha = m \times \mu_0 n I [/tex]

    should I assume they are perpendicular to give:

    [tex] I \alpha = m \mu_0 n I [/tex]

    also, the moment of inertia for a cuboid is:

    [tex] I = \frac{m(b^2 + c^2)}{12} [/tex]

    but we aren't given the dimensions of the magnet


  7. Nov 27, 2008 #6
    Sorry you can just leave the moment of inertia as I, you'll find an expression later that takes care of the unknowns. Edit: sorry they are are not perpendicualr here! You can use the small angle approxiamtion ie: sin(theta)=theta. Also keep B as B in the expression.

    Now how are theta nd alpha realted?
  8. Nov 27, 2008 #7


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    Okay so:

    [tex] I_m \alpha = m \mu_0 n I [/tex]

    So now I need to find a way to get the period from this equation.

    Does this mean I need to put the angular version of this equation:

    [tex] x = A cos(\omega_0t + \phi) [/tex]

    into the previous one?

  9. Nov 27, 2008 #8
    Let me elaborate, you take the cross product as being equal to mBsin(theta) (ie just the magnitude of the cross product). This is a restoring torque so it's sign should be negative.
  10. Nov 27, 2008 #9


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    so it should be:

    [tex] I_m \alpha = -m \mu_0 n I [/tex]


  11. Nov 27, 2008 #10
    It should be;

    [tex] I \alpha=-mBsin(\theta)[/tex]

    [tex]=-mB\theta[/tex] for small [tex]\theta[/tex]

    can you see why?

    Note:I'm not expanding B to avoid confusion with inertia and also it's tidier ans we don't need to expand it yet.
  12. Nov 27, 2008 #11


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    I see,

    I get that, its the small angle substitution, because in radians, the sin of small angles is near enough the same as the angle.


    [tex] I \alpha = -mB\theta [/tex]

    so now I think I have to get a differential in there to make it into a second order ODE, probably:

    [tex] \alpha = ddot{\theta} [/tex]

    [tex] \alpha = \ddot{\theta} = \frac{d^2\theta}{dt^2} [/tex]

    Look right?

  13. Nov 27, 2008 #12
    Thats right. Now you can solve this equation if you let [tex]\theta=\theta*sin(\omega*t)[/tex]
  14. Nov 27, 2008 #13


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    [tex] I \alpha = -mB\theta [/tex]

    and thus:

    [tex] I \frac{d^2\theta}{dt^2} = -mB\theta [/tex]

    I'm not sure, were did you get the expression from?

  15. Nov 27, 2008 #14
    Well the compass is moving with simple harmonic motion (we assume), and so it's displacement [tex]\theta[/tex] at any point will be related to it's angular frequency and the time at which we are looking at the situation. The equation is identical to the one in your first post, except the phase angle is zero and I'm saying the amplitude is just theta (which it should be in this case).
  16. Nov 27, 2008 #15


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    I see so:

    \theta = \theta cos(\omega_0t + \phi)

    Assume no starting phase:

    \theta = \theta cos(\omega_0t)

    I see now. I was confused slightly by the fact that you had two thetas equalling each other. Maybe I should call the Amplitude theta theta_0, as to not confuse myself:

    [tex] \theta = \theta_0 cos(\omega_0t) [/tex]

    okay, so now:

    [tex] I \frac{d^2\theta}{dt^2} = -mB\theta [/tex]

    [tex] I \frac{d^2\theta}{dt^2} = -mB(\theta_0 cos(\omega_0t)) [/tex]

    now I have to rearrange the differential?

  17. Nov 27, 2008 #16
    Yeah, I should have made that clearer, sorry. Now you solve like you would any O.D.E. in which you are using a substitution: just differentiate your expression for theta to find theta double dot in terms of this substitution and then sub all new expressions into the original equation (if that makes sense?).
  18. Nov 27, 2008 #17


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    so we now have to differentiate:

    [tex] \theta = \theta_0 cos(\omega_0t) [/tex]

    [tex] \theta = \theta_0 cos(\omega_0t) [/tex]

    Edit: the above should be theta, not d theta.



    [tex] \frac{d\theta}{dt} = t \theta_0 sin(\omega_0t) [/tex]


    [tex] \frac{d\theta}{dt} = -t^2 \theta_0 cos(\omega_0t) [/tex]


    Last edited: Nov 27, 2008
  19. Nov 27, 2008 #18
    Ooops you should be differentiating wrt to t rather than omega.


  20. Nov 27, 2008 #19


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    I thought it should have been omega coming out, since that is what happens with the wave equation. so now do I insert this value of

    [tex] \frac{d^2 \theta}{dt^2} [/tex]

    into the previous equation,

    [tex] I \frac{d^2\theta}{dt^2} = -mB(\theta_0 cos(\omega_0t)) [/tex]

  21. Nov 27, 2008 #20
    Thats right, then you should be able to cancel down and get a simpler expression.
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