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Conducting loop suspended in a magnetic field

  1. Feb 4, 2014 #1

    ShayanJ

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    Consider a conducting loop with resistance R and area A suspended by a non-conducting wire in a magnetic field [itex] \vec{B}=B\hat{y} [/itex]. The wire is a torsion spring with constant k.The equilibrium state of the loop is when it resides in the yz plane and its suspension is somehow that it can rotate around z axis with moment of inertia I.The loop is rotated by a very small angle [itex] \theta_0 [/itex] and is released.
    This is how I tried to derive the equation of motion:
    [itex]
    i=-\frac{1}{R} \frac{d\phi_B}{dt}=-\frac{1}{R} \frac{d}{dt} BA\cos{\theta}=\frac{BA\dot{\theta}\sin{\theta}}{R} \Rightarrow i\approx \frac{BA\dot{\theta}\theta}{R}
    [/itex]
    [itex]
    \vec{\tau}=i \vec{A}\times\vec{B}=\frac{BA\dot{\theta}\theta}{R} A\hat{x}\times B \hat{y} \Rightarrow \vec{\tau}=\frac{B^2A^2\dot{\theta}\theta}{R}\hat{z}
    [/itex]
    There is also the restoring torque of the torsion spring and so we have:
    [itex]
    \frac{B^2A^2\dot{\theta}\theta}{R}-k\theta=I\ddot{\theta} \Rightarrow IR\ddot{\theta}+(kR-B^2A^2\dot{\theta})\theta=0
    [/itex]
    The problem is,because [itex] \dot{\theta}_0=0 [/itex], the differential equation derived above isn't describing the motion of the loop at the first moment and only when the loop gains angular speed,the above DE can be used to describe its motion and so the answer to the above DE is contradictory when one applies the initial conditions to it.
    What should I do?
    Thanks
     
  2. jcsd
  3. Feb 4, 2014 #2
    In addition to R, we must also include XL. The current I is the open circuit voltage divides by impedance Z = R + jXL.

    Claude
     
  4. Feb 4, 2014 #3

    ShayanJ

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    Oh...forgot to tell...the problem states that the inductance is negligible.
     
  5. Feb 4, 2014 #4

    mfb

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    Find the general solution, then you'll be able to apply your initial conditions to it. Don't worry about the zero initial ##\dot \theta##, the differential equation is still well-defined (and the equation simplifies to ##I R\ddot \theta + k R \theta = 0## at this point).
     
  6. Feb 4, 2014 #5

    ShayanJ

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    Its not that I just derived the DE and then thought its contradictory.I tried solving it and I saw contradictions.
    One way of solving it is assuming [itex] \frac{d\theta}{dt}= \omega [/itex] and so [itex] \frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega\frac{d\omega}{d\theta}[/itex] and the DE becomes:
    [itex]
    IR \omega \frac{d\omega}{d\theta}+(kR-B^2A^2\omega)\theta=0
    [/itex]
    We know that the initial condition is [itex] \theta(0)=\theta_0 [/itex] and [itex] \omega(0)=0 [/itex]so at time t=0 the first and last term of the DE are zero and we have [itex] kR\theta=0 [/itex] which is not right because k and R are non-zero,as well as [itex] \theta [/itex] at the starting time!
    I tried to get its solution via alpha.wolfram.com and Maple but the result was an intractable integral!(See here).
    Also the answer to the DE [itex] IR\ddot\theta+kR\theta=0 [/itex] doesn't satisfy the original equation!!!
    Maybe such contradictions mean the small angle approximation goes bad because the magnetic torque beats up the restoring torque and is always increasing the angle instead of restoring it to 0!!!

    I FOUND IT!!!​
    [itex]
    \vec{\tau}=\frac{B^2A^2\dot\theta\theta}{R}\hat{z}
    [/itex]is zero when [itex] \dot\theta=0[/itex] and so the magnetic torque is zero at t=0 and becomes non-zero at [itex]t=\varepsilon [/itex]!
    So we have:
    [itex]
    \left\{ \begin{array}{cc} IR\ddot\theta+kR\theta=0 \ \ \ \ t=0 \\ IR\ddot\theta+(kR-B^2A^2\dot{\theta})\theta=0 \ \ \ \ t>\varepsilon \end{array} \right.
    [/itex]
    And we have to attach the answers somehow...but how?
     
    Last edited: Feb 4, 2014
  7. Feb 4, 2014 #6

    mfb

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    Physicist error :p. You cannot use those hand-waving manipulations if dω/dθ is not well-defined - and that is the case exactly at your initial condition. It diverges if you go towards this point, so you get a case of "0*infinity".

    Right - of course it does not, it does not include the damping term. You can use the solution for the original DE, and it will satisfy the original DE.

    The magnetic influence just acts as damping, you have a damped harmonic oscillator (and there is a well-known general solution).

    There is no need to separate your DE in two cases, the second case works everywhere.
     
  8. Feb 4, 2014 #7

    ShayanJ

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    The equation for a damped harmonic oscillator is [itex] I\ddot\theta+\Gamma\dot\theta+k\theta=0 [/itex] but the equation for the system I described is of the form [itex] I\ddot\theta+\Gamma\dot\theta\theta+k\theta=0 [/itex],they are different!
     
  9. Feb 5, 2014 #8

    mfb

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    Ah right.
    Looking at your damping term, the evaluation of A x B should lead to a different angular dependence.
     
  10. Feb 5, 2014 #9

    ShayanJ

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    I'm really sorry...the equation I've written is wrong.Today I found my mistake and figured that the right equation is just the usual equation of a damped harmonic oscillator.
    Thanks for answers
     
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